Answer key · Unit 8

Momentum.

Each lesson’s practice questions, their correct answers and the mistakes they spot. Answers read off the Scene use the same setting the learner sees.

Misconception catalogue

Answer key 4

Lesson 1 of 4 · About 20 minutes

Mass times velocity

What momentum is: mass times velocity, its unit, a heavy slow cart against a light fast one, and why momentum has a direction.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
What momentum is
Practice · Choose
Question
Momentum is mass times velocity, $p = mv$ ⁠. What is its SI unit?

Correct answer
- Correct kg·m/s
Setup, choices and feedback
The Scene at this step
Mass of cart A
1 kg
Speed of cart A
0.6 m/s
Carts / Momentum and energy
Momentum and energy
Choices
Shown in a random order.
- Correct kg·m/s
- Wrong kg·m/s²
  Misconception: Treats momentum as a force: gives it the newton, kg·m/s².
  Feedback from the catalogue
  Momentum is not a force. Force, $F = ma$ ⁠, is measured in kg·m/s², the newton. Momentum is $p = mv$ ⁠: mass times velocity, so its unit is kg·m/s, without the square. A cart rolling at a steady speed has momentum even while no force acts on it. A force is what changes the momentum.
- Wrong kg·m²/s²
  Misconception: Mixes up momentum with kinetic energy: works out ½mv² instead of mv.
  Feedback from the catalogue
  $\tfrac12 mv^2$ ⁠ is the kinetic energy, measured in joules. Momentum is $p = mv$ ⁠, with no half and no square, measured in kg·m/s. Both grow with the mass and the speed, but they are different: momentum has a direction, and kinetic energy does not.
- Wrong m/s
  Misconception: Takes momentum to be speed alone: the faster object has more momentum, whatever its mass.
  Feedback from the catalogue
  Momentum is mass times velocity, $p = mv$ ⁠. A high speed is not enough: a 10 g bullet at 400 m/s carries 4 kg·m/s, and a one-tonne car creeping at 1 m/s carries 1000 kg·m/s. Stopping the car takes an impulse 250 times as large.
Feedback for a right answer
Kilograms times metres per second: kg·m/s. Cart A, 1 kg at 0.6 m/s, carries 0.6 kg·m/s, the number the board shows on cart A’s row before the hit.

Hint
Write down the unit of mass and the unit of velocity, and multiply them.

Worked solution
1. $p = mv$ ⁠: mass times velocity.
2. Mass is measured in kg, and velocity in m/s.
3. The product: kg × m/s = kg·m/s.
If you chose kg·m/s², you multiplied mass by acceleration: that is the newton, the unit of force. The kg·m²/s², the joule, is the unit of kinetic energy, $\tfrac12 mv^2$ ⁠.
Step 2 of 5
Open this step
Working out momentum
Practice · Calculate
Question
Cart A has a mass of 2 kg and rolls at 0.8 m/s. What is its momentum before the collision?

Correct answer
1.6 kg·m/s ± 0.01 kg·m/s
Setup, choices and feedback
The Scene at this step
Mass of cart A
2 kgfixed
Speed of cart A
0.8 m/sfixed
Carts / Momentum and energy
Cartsfixed
Accepted within
± 0.01 kg·m/s
Computed from the Scene at this step’s setting
Total momentum before the collision: cart A's, massA × speed/nodes/carts/outputs/momentumBefore
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 0.64 kg·m/s ± 0.01 kg·m/s
  Misconception: Mixes up momentum with kinetic energy: works out ½mv² instead of mv.
  Feedback from the catalogue
  $\tfrac12 mv^2$ ⁠ is the kinetic energy, measured in joules. Momentum is $p = mv$ ⁠, with no half and no square, measured in kg·m/s. Both grow with the mass and the speed, but they are different: momentum has a direction, and kinetic energy does not.
- 0.8 kg·m/s ± 0.01 kg·m/s
  Misconception: Takes momentum to be speed alone: the faster object has more momentum, whatever its mass.
  Feedback from the catalogue
  Momentum is mass times velocity, $p = mv$ ⁠. A high speed is not enough: a 10 g bullet at 400 m/s carries 4 kg·m/s, and a one-tonne car creeping at 1 m/s carries 1000 kg·m/s. Stopping the car takes an impulse 250 times as large.
Feedback for a right answer
$p = 2 \times 0.8 = 1.6$ ⁠, that is 1.6 kg·m/s, to the right. A cart of twice the mass at the same speed carries twice the momentum, and so does a cart of the same mass at twice the speed.

Hint
$p = mv$ ⁠.

Worked solution
1. The givens: $m = 2\ \mathrm{kg}$ ⁠, $v = 0.8\ \mathrm{m/s}$ ⁠.
2. The formula: $p = mv$ ⁠.
3. Substitute: $p = 2 \times 0.8 = 1.6$ ⁠.
4. The momentum: 1.6 kg·m/s, to the right.
If you got $0.64$ ⁠, you worked out $\tfrac12 mv^2$ ⁠: that is the kinetic energy, in joules, not the momentum. If you answered $0.8$ ⁠, you gave the speed alone.
Step 3 of 5
Open this step
Heavy and slow, light and fast
Practice · Reach the target
Question
A 4 kg cart at 0.3 m/s carries 1.2 kg·m/s of momentum. Mass of cart A is now fixed at 1 kg. Change Speed of cart A until it carries 1.2 kg·m/s too, then press Check.

Goal
- 1.2 kg·m/s ± 0.01 kg·m/s
  At the start of the step: 0.3 kg·m/s
Setup, choices and feedback
The Scene at this step
Mass of cart A
1 kgfixed
Speed of cart A
0.3 m/s
Carts / Momentum and energy
Momentum and energy
Read from the Scene at
Total momentum before the collision: cart A's, massA × speed/nodes/carts/outputs/momentumBefore
One setting that reaches it
Speed of cart A
1.2 m/s
Feedback for a right answer
At 1.2 m/s: $1 \times 1.2 = 1.2$ ⁠. A light, fast cart and a heavy, slow one can carry exactly the same momentum, so stopping either takes exactly the same push.

Feedback for a wrong answer
Not yet. Divide the momentum you want by the mass: $v = \dfrac{p}{m}$ ⁠.

Hint
A cart four times lighter needs four times the speed.

Worked solution
1. You want $p = 1.2\ \mathrm{kg \cdot m/s}$ ⁠ with $m = 1\ \mathrm{kg}$ ⁠.
2. From $p = mv$ ⁠: $v = \dfrac{p}{m}$ ⁠.
3. Substitute: $v = \dfrac{1.2}{1} = 1.2$ ⁠.
4. Set Speed of cart A to 1.2 m/s.
If you left it at 0.3 m/s, the heavy cart’s speed, the light cart carries a quarter of the momentum: 0.3 kg·m/s.
Step 4 of 5
Open this step
Car or bullet
Practice · Choose
Question
Which has more momentum: a 1000 kg car creeping through a car park at 1 m/s, or a 10 g bullet flying at 400 m/s?

Correct answer
- Correct The car
Setup, choices and feedback
The Scene at this step
Mass of cart A
4 kg
Speed of cart A
0.3 m/s
Carts / Momentum and energy
Momentum and energy
Choices
Shown in a random order.
- Correct The car
- Wrong The bullet
  Misconception: Takes momentum to be speed alone: the faster object has more momentum, whatever its mass.
  Feedback from the catalogue
  Momentum is mass times velocity, $p = mv$ ⁠. A high speed is not enough: a 10 g bullet at 400 m/s carries 4 kg·m/s, and a one-tonne car creeping at 1 m/s carries 1000 kg·m/s. Stopping the car takes an impulse 250 times as large.
- Wrong They have the same momentum
  Feedback
  Work each one out: $1000 \times 1 = 1000$ ⁠ and $0.01 \times 400 = 4$ ⁠. Far from equal.
Feedback for a right answer
The car carries 1000 kg·m/s, and the bullet only 4 kg·m/s. The bullet does have more kinetic energy, 800 J against 500 J, which is why it pierces things. But stopping the car takes an impulse 250 times as large.

Hint
First convert the bullet’s mass to kilograms.

Worked solution
1. The bullet: $10\ \mathrm{g} = 0.01\ \mathrm{kg}$ ⁠, so $p = 0.01 \times 400 = 4\ \mathrm{kg \cdot m/s}$ ⁠.
2. The car: $p = 1000 \times 1 = 1000\ \mathrm{kg \cdot m/s}$ ⁠.
3. $1000 > 4$ ⁠: the car has $250$ ⁠ times as much momentum.
If you chose the bullet, you looked only at the speed. In momentum the mass counts exactly as much as the speed.
Step 5 of 5
Open this step
Momentum has a direction
Practice · Calculate
Question
Now cart B is twice as heavy, and the collision is perfectly bouncy. After the hit cart A, 1 kg, rolls back, to the left, at 0.2 m/s. Right is positive. What is cart A’s momentum after the hit?

Correct answer
-0.2 kg·m/s ± 0.01 kg·m/s
Setup, choices and feedback
The Scene at this step
Mass of cart A
1 kgfixed
Speed of cart A
0.6 m/sfixed
Carts / Momentum and energy
Cartsfixed
Other settings
Mass ratio/nodes/carts/props/massRatio = 2
Other settings
Bounciness/nodes/carts/props/bounciness = 1
Accepted within
± 0.01 kg·m/s
Computed from the Scene at this step’s setting
Cart A's momentum just after the collision/nodes/carts/outputs/momentumAfterA
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 0.2 kg·m/s ± 0.01 kg·m/s
  Misconception: Drops the sign of a velocity or momentum that points left, as if every motion were to the right.
  Feedback from the catalogue
  Momentum and velocity have a direction. On the track right is taken as positive, so a cart rolling left, for instance after bouncing back, gets a minus sign: 0.2 m/s to the left is −0.2 m/s. Without the sign the books do not balance: the momentum of a cart coming back takes away from the total.
Feedback for a right answer
$p = 1 \times (-0.2) = -0.2$ ⁠, that is −0.2 kg·m/s. The sign is part of the answer: it says the momentum points left.

Hint
A velocity to the left gets a minus sign.

Worked solution
1. Right is positive, so A’s velocity after the hit is $v_A' = -0.2\ \mathrm{m/s}$ ⁠.
2. $p = mv$ ⁠.
3. Substitute: $p = 1 \times (-0.2) = -0.2$ ⁠.
4. The momentum: −0.2 kg·m/s, to the left.
If you answered $0.2$ ⁠, you dropped the direction. A momentum of +0.2 kg·m/s belongs to a cart carrying on to the right, not to one coming back.

Lesson 2 of 4 · About 30 minutes

Impulse

How a force changes momentum: impulse, force times time, the equal and opposite pushes of two colliding carts, why a bounce pushes harder than a stick, and why an airbag protects.

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The 7 steps and their answers

Step 1 of 7
Open this step
Force times time
Practice · Calculate
Question
A steady force of 3 N pushes a cart for 0.2 s. By how much does its momentum change?

Correct answer
0.6 N·s ± 0.01 N·s
Setup, choices and feedback
The Scene at this step
Mass ratio B : A
1 ×
Bounciness
0.5
Show momentum and energy
On
Accepted within
± 0.01 N·s
A fixed value in the question
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 15 N·s ± 0.01 N·s
  Misconception: Turns the time upside down: divides the force by the time to get the impulse, or multiplies the impulse by the time to get the force.
  Feedback from the catalogue
  Impulse is force times time, $J = F\,\Delta t$ ⁠, so its unit is the N·s. From it, the average force is impulse divided by time, $F = \dfrac{\Delta p}{\Delta t}$ ⁠. Check with the units: $\dfrac{\mathrm{N \cdot s}}{\mathrm{s}} = \mathrm{N}$ ⁠. And by common sense: a force acting for longer changes the momentum more, and the same change of momentum in a shorter contact takes a bigger force.
Feedback for a right answer
$\Delta p = F\,\Delta t = 3 \times 0.2 = 0.6$ ⁠, that is 0.6 N·s. That is exactly cart A’s momentum in the Scene: 1 kg at 0.6 m/s.

Hint
The change of momentum equals the impulse: the force times the time.

Worked solution
1. The givens: $F = 3\ \mathrm{N}$ ⁠, $\Delta t = 0.2\ \mathrm{s}$ ⁠.
2. The impulse: $J = F\,\Delta t$ ⁠, and it equals the change of momentum, $\Delta p$ ⁠.
3. Substitute: $J = 3 \times 0.2 = 0.6$ ⁠.
4. The momentum changes by 0.6 N·s, that is 0.6 kg·m/s.
If you got $15$ ⁠, you divided the force by the time. A force that acts for longer changes the momentum more, so you multiply.
Step 2 of 7
Open this step
What B received
Practice · Calculate
Question
Cart B is twice as heavy as cart A: 2 kg. The carts stick, and after the hit they roll on together at 0.2 m/s. What impulse did cart A give cart B?

Correct answer
0.4 N·s ± 0.01 N·s
Setup, choices and feedback
The Scene at this step
Mass ratio B : A
2 ×fixed
Bounciness
0fixed
Show momentum and energy
Offfixed
Accepted within
± 0.01 N·s
Computed from the Scene at this step’s setting
The impulse cart A gives cart B, its change of momentum (N·s)/nodes/carts/outputs/impulse
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 0.2 N·s ± 0.01 N·s
  Misconception: Gives the change in velocity as the impulse, leaving out the mass.
  Feedback from the catalogue
  Impulse is the change in momentum, $\Delta p = m\,\Delta v$ ⁠, not the change in velocity. The same change in velocity takes twice the impulse for a cart of twice the mass. Multiply the change in velocity by the mass of the cart in question.
Feedback for a right answer
$J = m_B\,\Delta v_B = 2 \times (0.2 - 0) = 0.4$ ⁠, that is 0.4 N·s, to the right: all the momentum B leaves the hit with.

Hint
The impulse on B is the change in B’s momentum, and B started at rest.

Worked solution
1. The givens: $m_B = 2\ \mathrm{kg}$ ⁠; B starts at rest, $v_B = 0$ ⁠, and ends at $v_B' = 0.2\ \mathrm{m/s}$ ⁠.
2. The impulse is the change of momentum: $J = m_B (v_B' - v_B)$ ⁠.
3. Substitute: $J = 2 \times (0.2 - 0) = 0.4$ ⁠.
4. The impulse: 0.4 N·s, to the right.
If you answered $0.2$ ⁠, you gave the change in velocity. Impulse is a change in momentum, so multiply by the mass.
Step 3 of 7
Open this step
Equal and opposite pushes
Practice · Choose
Question
Cart B is four times as heavy as cart A. While the bumpers touch, which cart pushes harder on the other?

Correct answer
- Correct They push on each other with forces equal in size and opposite in direction
Setup, choices and feedback
The Scene at this step
Mass ratio B : A
4 ×
Bounciness
0.5
Show momentum and energy
On
Choices
Shown in a random order.
- Correct They push on each other with forces equal in size and opposite in direction
- Wrong Cart B, because it is heavier
  Misconception: Thinks the heavier cart pushes harder on the lighter one during a collision.
  Feedback from the catalogue
  By Newton’s third law, the force A exerts on B and the force B exerts on A are equal in size and opposite in direction, always, even when one cart is four times as heavy. What differs is the result: the same force changes the heavy cart’s velocity less, because $a = \dfrac{F}{m}$ ⁠.
- Wrong Cart A, because it is the one doing the hitting
  Misconception: Thinks the cart that does the hitting pushes harder than the cart it hits.
  Feedback from the catalogue
  It feels that way, because A is the one coming in with a run-up, but while the bumpers touch each cart pushes on the other with a force equal in size and opposite in direction: Newton’s third law. The two forces start and stop at the same moment, so the impulses are equal and opposite too, and that is what keeps the total.
- Wrong It depends on the bounciness
  Feedback
  The bounciness changes how hard and how long the bumpers press, but at every moment of the contact the force of A on B is the same size as the force of B on A.
Feedback for a right answer
That is Newton’s third law. The forces are equal and last the same time, so the impulses are equal in size and opposite in direction too: whatever momentum B gains, A loses. What differs is the result: the same force changes the heavy cart’s velocity less.

Hint
Think of Newton’s third law.

Worked solution
1. During the contact A pushes on B and B pushes on A: a pair of action and reaction forces.
2. By Newton’s third law, $F_{AB} = -F_{BA}$ ⁠: equal in size and opposite in direction, at every moment of the contact.
3. The masses differ, so the accelerations differ: $a = \dfrac{F}{m}$ ⁠, and B, four times as heavy, accelerates four times less.
If you chose A or B, you mixed up the force with what it does: more happens to the lighter cart, but it is not pushed harder.
Step 4 of 7
Open this step
The impulse on A
Practice · Calculate
Question
Cart B is twice as heavy, and the collision is perfectly bouncy. Cart A, 1 kg, arrives at 0.6 m/s and comes back at 0.2 m/s. Right is positive. What impulse did cart B give cart A?

Correct answer
-0.8 N·s ± 0.01 N·s
Setup, choices and feedback
The Scene at this step
Mass ratio B : A
2 ×fixed
Bounciness
1fixed
Show momentum and energy
Offfixed
Accepted within
± 0.01 N·s
Computed from the Scene at this step’s setting
The impulse cart B gives cart A, its change of momentum (N·s): minus impulse/nodes/carts/outputs/impulseOnA
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- -0.2 N·s ± 0.01 N·s
  Misconception: Takes the momentum after the collision as the impulse, forgetting to subtract the momentum before.
  Feedback from the catalogue
  Impulse is the change in momentum: the momentum at the end minus the momentum at the start, $\Delta p = p' - p$ ⁠. Only when the object starts at rest does the final momentum equal the change. A cart that arrived with 0.6 kg·m/s and is left with −0.2 kg·m/s received an impulse of $-0.2 - 0.6 = -0.8$ ⁠, that is −0.8 N·s.
- -0.4 N·s ± 0.01 N·s
  Misconception: Drops the sign of a velocity or momentum that points left, as if every motion were to the right.
  Feedback from the catalogue
  Momentum and velocity have a direction. On the track right is taken as positive, so a cart rolling left, for instance after bouncing back, gets a minus sign: 0.2 m/s to the left is −0.2 m/s. Without the sign the books do not balance: the momentum of a cart coming back takes away from the total.
- 0.8 N·s ± 0.01 N·s
  Misconception: Drops the sign of a velocity or momentum that points left, as if every motion were to the right.
  Feedback
  0.8 N·s is the right size, but the impulse on A points left: B pushes A back. Right is positive, so the impulse is −0.8 N·s, exactly the opposite of the impulse A gave B.
Feedback for a right answer
$J = m_A (v_A' - v_A) = 1 \times (-0.2 - 0.6) = -0.8$ ⁠, that is −0.8 N·s. And B received +0.8 N·s from A: equal in size and opposite in direction.

Hint
The change of momentum is the final momentum minus the starting one, and the final velocity is negative.

Worked solution
1. The givens: $m_A = 1\ \mathrm{kg}$ ⁠, $v_A = 0.6\ \mathrm{m/s}$ ⁠, and after the hit $v_A' = -0.2\ \mathrm{m/s}$ ⁠, because A moves left.
2. The impulse on A is the change in its momentum: $J = m_A (v_A' - v_A)$ ⁠.
3. Substitute: $J = 1 \times (-0.2 - 0.6) = -0.8$ ⁠.
4. The impulse: −0.8 N·s, to the left.
If you answered $-0.4$ ⁠, you subtracted the speeds without the sign, $0.2 - 0.6$ ⁠. A cart that comes back has to lose all its forward momentum and then gain some backward momentum, so the change is bigger than the momentum it had. If you answered $-0.2$ ⁠, you gave the final momentum, not the change.
Step 5 of 7
Open this step
Bounce or stick
Practice · Predict, then look
Question
Equal carts. When they stick, A gives B an impulse of 0.3 N·s. What happens to the impulse when the collision is perfectly bouncy?

Correct answer
- Correct It gets bigger
Setup, choices and feedback
The Scene at this step
Mass ratio B : A
1 ×fixed
Bounciness
0fixed
Show momentum and energy
Offhidden
Choices
- Correct It gets bigger
- Wrong It stays at 0.3 N·s
  Feedback
  The total momentum does stay the same, but the question is how much of it goes to B. In a bounce A stops instead of rolling on with B, so B gets more.
- Wrong It gets smaller, because a bounce is a gentler contact
  Misconception: Thinks a bounce takes a smaller push than stopping or sticking.
  Feedback from the catalogue
  A bounce takes more impulse. To stop, the momentum only has to come down to zero. To come back, it has to come down to zero and then build up again the other way. So a ball bouncing off a wall receives a bigger impulse from the wall than one that sticks to it, and gives the wall a bigger one.
Once the prediction is locked in, the Scene changes
Bounciness
1fixed
Show momentum and energy
Onhidden
The Scene then shows: 0.6 N·s · The impulse cart A gives cart B, its change of momentum (N·s)/nodes/carts/outputs/impulse
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
It doubles, to 0.6 N·s: A stops dead, and B leaves at 0.6 m/s instead of 0.3 m/s. A bounce pushes harder than a stick: the bumpers do not only stop the carts closing in, they also push them apart.

Worked solution
1. When the carts stick, they roll on together at 0.3 m/s, so B received $1 \times 0.3 = 0.3\ \mathrm{N \cdot s}$ ⁠.
2. In a perfectly bouncy hit between equal carts, A stops and B leaves at A’s speed, 0.6 m/s.
3. The impulse on B: $1 \times 0.6 = 0.6\ \mathrm{N \cdot s}$ ⁠, twice as much.
If you predicted a smaller impulse, you thought of a bounce as a gentle contact. But in a bounce the bumpers do two jobs: they stop the carts closing in, then push them apart.
Step 6 of 7
Open this step
How hard
Practice · Calculate
Question
In the sticky collision of the equal carts, A gives B an impulse of 0.3 N·s. Suppose the bumpers touch for 0.05 s. What average force does A exert on B?

Correct answer
6 N ± 0.05 N
Setup, choices and feedback
The Scene at this step
Mass ratio B : A
1 ×fixed
Bounciness
0fixed
Show momentum and energy
On
Accepted within
± 0.05 N
A fixed value in the question
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 0.015 N ± 0.001 N
  Misconception: Turns the time upside down: divides the force by the time to get the impulse, or multiplies the impulse by the time to get the force.
  Feedback from the catalogue
  Impulse is force times time, $J = F\,\Delta t$ ⁠, so its unit is the N·s. From it, the average force is impulse divided by time, $F = \dfrac{\Delta p}{\Delta t}$ ⁠. Check with the units: $\dfrac{\mathrm{N \cdot s}}{\mathrm{s}} = \mathrm{N}$ ⁠. And by common sense: a force acting for longer changes the momentum more, and the same change of momentum in a shorter contact takes a bigger force.
Feedback for a right answer
$F = \dfrac{\Delta p}{\Delta t} = \dfrac{0.3}{0.05} = 6$ ⁠, that is 6 N. With softer bumpers and a contact of 0.1 s, the same impulse would take only 3 N.

Hint
$F\,\Delta t = \Delta p$ ⁠. Solve for $F$ ⁠.

Worked solution
1. The givens: $\Delta p = 0.3\ \mathrm{N \cdot s}$ ⁠, $\Delta t = 0.05\ \mathrm{s}$ ⁠.
2. $F\,\Delta t = \Delta p$ ⁠, so $F = \dfrac{\Delta p}{\Delta t}$ ⁠.
3. Substitute: $F = \dfrac{0.3}{0.05} = 6$ ⁠.
4. The average force: 6 N, to the right, and B pushes A with 6 N to the left.
If you got $0.015$ ⁠, you multiplied instead of dividing. Check the units: N·s × s is not a newton.
Step 7 of 7
Open this step
A soft landing
Practice · Choose
Question
In a crash, an airbag stops the passenger. Why does it protect them?

Correct answer
- Correct It makes the stop last longer, so the same change of momentum takes a smaller force
Setup, choices and feedback
The Scene at this step
Mass ratio B : A
1 ×
Bounciness
0.5
Show momentum and energy
On
Choices
Shown in a random order.
- Correct It makes the stop last longer, so the same change of momentum takes a smaller force
- Wrong It makes the passenger’s change of momentum smaller
  Misconception: Thinks an airbag or a soft landing makes the change of momentum smaller.
  Feedback from the catalogue
  An airbag does not change the change of momentum: the passenger is moving before the crash and at rest after it, either way. What it changes is the time the change takes. From the same impulse, $F\,\Delta t = \Delta p$ ⁠, ten times the time means a tenth of the average force.
- Wrong It soaks up the momentum, so the momentum is not conserved
  Feedback
  The passenger’s momentum passes to the airbag, the car and in the end the Earth; it does not vanish. And what hurts the passenger is the force, not where the momentum goes.
- Wrong It stops them sooner
  Feedback
  The opposite: a quicker stop means a shorter time, and so a bigger force.
Feedback for a right answer
$F\,\Delta t = \Delta p$ ⁠, and $\Delta p$ ⁠ is fixed. When the airbag makes $\Delta t$ ⁠ ten times longer, the average force on the passenger is ten times smaller. For the same reason you bend your knees when you land, and draw your hands back when you catch a ball.

Hint
The passenger’s change of momentum is fixed in advance: from the car’s speed to a stop.

Worked solution
1. Before the crash the passenger has momentum $mv$ ⁠, and afterwards they are at rest: $\Delta p = -mv$ ⁠, with an airbag or without one.
2. $F\,\Delta t = \Delta p$ ⁠, so the average force is $F = \dfrac{\Delta p}{\Delta t}$ ⁠.
3. The airbag makes $\Delta t$ ⁠ longer, and the same $\Delta p$ ⁠ gives a smaller force.
If you chose the airbag making the change of momentum smaller, notice that the passenger stops either way. The only difference is how long the stop takes.

Lesson 3 of 4 · About 30 minutes

Conservation of momentum

The total no collision can change: the total momentum before and after, carts that stick, momentum handed over, a cart that bounces back, and carts heading towards each other.

Open the lesson

The 6 steps and their answers

Step 1 of 6
Open this step
What stays the same
Practice · Predict, then look
Question
Cart A arrives carrying 0.6 kg·m/s into cart B, three times as heavy and standing still. The collision is half bouncy, and some of the kinetic energy becomes heat and sound. What will the total momentum of the two carts be after the hit?

Correct answer
- Correct Exactly 0.6 kg·m/s
Setup, choices and feedback
The Scene at this step
Mass ratio B : A
3 ×fixed
Bounciness
0.5fixed
Carts / Momentum and energy
Cartshidden
Choices
- Correct Exactly 0.6 kg·m/s
- Wrong Less than 0.6 kg·m/s, because some went into heat and sound
  Misconception: Thinks some of the total momentum is lost in a collision, as heat or in the crash.
  Feedback from the catalogue
  The total momentum is not lost in a collision. Whatever impulse A gives B, B gives back to A in the opposite direction, so the total does not change, even when the carts squash and stick. What becomes heat and sound is part of the kinetic energy, a different quantity. The total momentum changes only when an outside force, such as friction, pushes on the carts.
- Wrong More than 0.6 kg·m/s, because B is heavy
  Feedback
  A large mass does not add momentum by itself: B gets only what A gives it, and A loses exactly that much.
Once the prediction is locked in, the Scene changes
Carts / Momentum and energy
Momentum and energyhidden
The Scene then shows: 0.6 kg·m/s · Total momentum just after the collision (equal to momentumBefore)/nodes/carts/outputs/momentumAfter
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The total stays at 0.6 kg·m/s. A bounces back a little with −0.075 kg·m/s, and B moves off with 0.675 kg·m/s: the total bar does not leave its dashed line, even while energy is lost.

Worked solution
1. Before the hit only A moves: $p = 1 \times 0.6 = 0.6\ \mathrm{kg \cdot m/s}$ ⁠.
2. During the contact A and B receive equal and opposite impulses, and nothing else pushes along the track.
3. So the total momentum after the hit equals the total before: $p' = p = 0.6\ \mathrm{kg \cdot m/s}$ ⁠.
If you predicted less momentum, you mixed up momentum with kinetic energy. Some of the kinetic energy becomes heat and sound; the total momentum does not change.
Step 2 of 6
Open this step
Sticking together
Practice · Calculate
Question
Cart A, 1 kg, arrives at 0.6 m/s at cart B, 3 kg, and sticks to it. At what speed do the two roll on together?

Correct answer
0.15 m/s ± 0.005 m/s
Setup, choices and feedback
The Scene at this step
Mass ratio B : A
3 ×fixed
Bounciness
0fixed
Carts / Momentum and energy
Cartsfixed
Accepted within
± 0.005 m/s
Computed from the Scene at this step’s setting
Cart B's velocity just after the collision/nodes/carts/outputs/velocityAfterB
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 0.3 m/s ± 0.005 m/s
  Misconception: Splits the speed evenly between the carts (the average of their velocities), whatever their masses and bounciness.
  Feedback from the catalogue
  The speed is not shared out evenly. What is kept is the momentum, so the masses have to be weighed in: when the carts stick, $v' = \dfrac{m_A v_A + m_B v_B}{m_A + m_B}$ ⁠. A cart three times as heavy drags the common speed down to 0.15 m/s, not 0.3 m/s. And when the collision is bouncy, the carts do not end up at the same speed at all.
- 0.2 m/s ± 0.005 m/s
  Misconception: Leaves one cart out after the collision: divides the momentum by one cart's mass, or counts only one cart's kinetic energy.
  Feedback from the catalogue
  After the collision there are two carts, and you need both. When they stick, the momentum is spread over the mass of both, $m_A + m_B$ ⁠. And the kinetic energy after the collision is the sum for both, $\tfrac12 m_A v_A'^2 + \tfrac12 m_B v_B'^2$ ⁠.
Feedback for a right answer
$v' = \dfrac{1 \times 0.6}{1 + 3} = 0.15$ ⁠, that is 0.15 m/s. The same momentum, 0.6 kg·m/s, is now carried by four times the mass, so the speed is four times smaller.

Hint
The momentum before equals the momentum after, and afterwards there is one body of mass $m_A + m_B$ ⁠.

Worked solution
1. The givens: $m_A = 1\ \mathrm{kg}$ ⁠, $v_A = 0.6\ \mathrm{m/s}$ ⁠, $m_B = 3\ \mathrm{kg}$ ⁠, $v_B = 0$ ⁠.
2. Conservation of momentum: $m_A v_A + m_B v_B = (m_A + m_B)\,v'$ ⁠.
3. Substitute: $1 \times 0.6 + 3 \times 0 = 4\,v'$ ⁠, so $v' = \dfrac{0.6}{4} = 0.15$ ⁠.
4. The carts roll on together at 0.15 m/s, to the right.
If you got $0.3$ ⁠, you took the average of the speeds as if the carts were equal. If you got $0.2$ ⁠, you divided by B’s mass alone.
Step 3 of 6
Open this step
Momentum handed over
Practice · Predict, then look
Question
Equal carts, and they stick: after the hit both roll on at 0.3 m/s. What will happen when the ends are perfectly bouncy?

Correct answer
- Correct A will stop dead, and B will move off at 0.6 m/s
Setup, choices and feedback
The Scene at this step
Mass ratio B : A
1 ×fixed
Bounciness
0fixed
Carts / Momentum and energy
Cartshidden
Choices
- Correct A will stop dead, and B will move off at 0.6 m/s
- Wrong Both will roll on together at 0.3 m/s, as before
  Misconception: Splits the speed evenly between the carts (the average of their velocities), whatever their masses and bounciness.
  Feedback from the catalogue
  The speed is not shared out evenly. What is kept is the momentum, so the masses have to be weighed in: when the carts stick, $v' = \dfrac{m_A v_A + m_B v_B}{m_A + m_B}$ ⁠. A cart three times as heavy drags the common speed down to 0.15 m/s, not 0.3 m/s. And when the collision is bouncy, the carts do not end up at the same speed at all.
- Wrong A will bounce back at 0.6 m/s, and B will stay put
  Feedback
  Then the total momentum would become −0.6 kg·m/s, and no outside force pushes the carts to do that. That would happen only if B were a wall that does not move.
Once the prediction is locked in, the Scene changes
Bounciness
1fixed
Carts / Momentum and energy
Momentum and energyhidden
The Scene then shows: 0.6 m/s · Cart B's velocity just after the collision/nodes/carts/outputs/velocityAfterB
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
A stops, and B moves off at 0.6 m/s: all the momentum is handed over, as with billiard balls. Only this way are both the momentum, 0.6 kg·m/s, and the kinetic energy, 0.18 J, the same before and after.

Worked solution
1. Momentum is conserved: $1 \times v_A' + 1 \times v_B' = 0.6$ ⁠.
2. When perfectly bouncy, the carts move apart as fast as they came together: $v_B' - v_A' = 0.6$ ⁠.
3. Add the two equations: $2\,v_B' = 1.2$ ⁠, so $v_B' = 0.6$ ⁠ and $v_A' = 0$ ⁠.
If you chose rolling on together, that is the sticky collision. Its total momentum is the same, but perfectly bouncy bumpers push the carts apart.
Step 4 of 6
Open this step
Bouncing back
Practice · Calculate
Question
Cart B is four times as heavy, 4 kg, and the collision is perfectly bouncy. After the hit B moves off to the right at 0.24 m/s. What is the velocity of cart A, 1 kg, after the hit? Right is positive.

Correct answer
-0.36 m/s ± 0.005 m/s
Setup, choices and feedback
The Scene at this step
Mass ratio B : A
4 ×fixed
Bounciness
1fixed
Carts / Momentum and energy
Cartsfixed
Accepted within
± 0.005 m/s
Computed from the Scene at this step’s setting
Cart A's velocity just after the collision (negative: rolling back)/nodes/carts/outputs/velocityAfterA
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 0.36 m/s ± 0.005 m/s
  Misconception: Drops the sign of a velocity or momentum that points left, as if every motion were to the right.
  Feedback from the catalogue
  Momentum and velocity have a direction. On the track right is taken as positive, so a cart rolling left, for instance after bouncing back, gets a minus sign: 0.2 m/s to the left is −0.2 m/s. Without the sign the books do not balance: the momentum of a cart coming back takes away from the total.
Feedback for a right answer
$v_A' = 0.6 - 4 \times 0.24 = -0.36$ ⁠, that is −0.36 m/s: A bounces back. B took 0.96 kg·m/s, more than all the momentum A brought, so A must carry −0.36 kg·m/s for the total to stay at 0.6 kg·m/s.

Hint
Write the total momentum before and after, and put in what you know about B.

Worked solution
1. The givens: $m_A = 1\ \mathrm{kg}$ ⁠, $v_A = 0.6\ \mathrm{m/s}$ ⁠, $m_B = 4\ \mathrm{kg}$ ⁠, $v_B' = 0.24\ \mathrm{m/s}$ ⁠.
2. Conservation of momentum: $m_A v_A = m_A v_A' + m_B v_B'$ ⁠.
3. Substitute: $0.6 = v_A' + 4 \times 0.24 = v_A' + 0.96$ ⁠, so $v_A' = -0.36$ ⁠.
4. A comes back: $v_A' = -0.36\ \mathrm{m/s}$ ⁠.
If you answered $0.36$ ⁠, you lost the sign. A cannot carry on forward: the total momentum would then be 1.32 kg·m/s, more than at the start.
Step 5 of 6
Open this step
Stopping A dead
Practice · Reach the target
Question
Cart B is twice as heavy. Change Bounciness until cart A stops dead after the hit and hands all its momentum to B, then press Check.

Goal
- 0 m/s ± 0.005 m/s
  At the start of the step: 0.12 m/s
Setup, choices and feedback
The Scene at this step
Mass ratio B : A
2 ×fixed
Bounciness
0.2
Carts / Momentum and energy
Momentum and energy
Read from the Scene at
Cart A's velocity just after the collision (negative: rolling back)/nodes/carts/outputs/velocityAfterA
One setting that reaches it
Bounciness
0.5
Feedback for a right answer
At half bouncy, 0.5. A stops, and B, twice as heavy, carries the whole 0.6 kg·m/s at 0.3 m/s.

Feedback for a wrong answer
Not yet. If A rolls on forward after the hit, raise the bounciness; if it comes back, lower it.

Hint
Too little bounce and A rolls on forward; too much and it bounces back.

Worked solution
1. When A stops, all the momentum goes to B: $2\,v_B' = 0.6$ ⁠, so $v_B' = 0.3\ \mathrm{m/s}$ ⁠.
2. The bounciness is the speed of moving apart divided by the speed of closing in: $e = \dfrac{v_B' - v_A'}{v_A}$ ⁠.
3. Substitute: $e = \dfrac{0.3 - 0}{0.6} = 0.5$ ⁠.
4. Set Bounciness to 0.5.
Perfectly bouncy, A would bounce back, because B is heavier. A perfectly bouncy hit stops A dead only when the carts are equal.
Step 6 of 6
Open this step
Towards each other
Practice · Calculate
Question
Two carts roll towards each other: a 2 kg cart to the right at 0.5 m/s, and a 1 kg cart to the left at 0.4 m/s. They stick. What is their common velocity after the hit? Right is positive.

Correct answer
0.2 m/s ± 0.005 m/s
Setup, choices and feedback
The Scene at this step
Mass ratio B : A
1 ×
Bounciness
0
Carts / Momentum and energy
Momentum and energy
Accepted within
± 0.005 m/s
A fixed value in the question
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 0.4667 m/s ± 0.005 m/s
  Misconception: Drops the sign of a velocity or momentum that points left, as if every motion were to the right.
  Feedback from the catalogue
  Momentum and velocity have a direction. On the track right is taken as positive, so a cart rolling left, for instance after bouncing back, gets a minus sign: 0.2 m/s to the left is −0.2 m/s. Without the sign the books do not balance: the momentum of a cart coming back takes away from the total.
- 0.05 m/s ± 0.005 m/s
  Misconception: Splits the speed evenly between the carts (the average of their velocities), whatever their masses and bounciness.
  Feedback from the catalogue
  The speed is not shared out evenly. What is kept is the momentum, so the masses have to be weighed in: when the carts stick, $v' = \dfrac{m_A v_A + m_B v_B}{m_A + m_B}$ ⁠. A cart three times as heavy drags the common speed down to 0.15 m/s, not 0.3 m/s. And when the collision is bouncy, the carts do not end up at the same speed at all.
Feedback for a right answer
$v' = \dfrac{2 \times 0.5 + 1 \times (-0.4)}{2 + 1} = \dfrac{0.6}{3} = 0.2$ ⁠, that is 0.2 m/s to the right. The heavier cart had more momentum, so the pair carries on its way.

Hint
The cart moving left has negative momentum.

Worked solution
1. Right is positive: $m_1 = 2\ \mathrm{kg}$ ⁠, $v_1 = 0.5\ \mathrm{m/s}$ ⁠; $m_2 = 1\ \mathrm{kg}$ ⁠, $v_2 = -0.4\ \mathrm{m/s}$ ⁠.
2. The total momentum before: $2 \times 0.5 + 1 \times (-0.4) = 1 - 0.4 = 0.6\ \mathrm{kg \cdot m/s}$ ⁠.
3. After the hit there is one body of mass $3\ \mathrm{kg}$ ⁠: $v' = \dfrac{0.6}{3} = 0.2$ ⁠.
4. The pair moves right at 0.2 m/s.
If you got about $0.47$ ⁠, you added the momenta without the sign, $1 + 0.4$ ⁠: the momentum of the cart moving left takes away from the total. If you got $0.05$ ⁠, you averaged the velocities without weighting them by the masses.

Lesson 4 of 4 · About 25 minutes

Elastic and inelastic collisions

What happens to kinetic energy in a collision: elastic, inelastic and perfectly inelastic collisions, how much energy is lost, finding the bounciness from the velocities, and where the energy goes.

Open the lesson

The 6 steps and their answers

Step 1 of 6
Open this step
Kinetic energy before the hit
Practice · Calculate
Question
Cart A, 1 kg, rolls at 0.6 m/s. What is its kinetic energy before the collision?

Correct answer
0.18 J ± 0.001 J
Setup, choices and feedback
The Scene at this step
Mass ratio B : A
1 ×
Bounciness
0.5
Show momentum and energy
Offfixed
Accepted within
± 0.001 J
Computed from the Scene at this step’s setting
Kinetic energy before the collision: cart A's, ½ massA speed²/nodes/carts/outputs/kineticBefore
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 0.36 J ± 0.001 J
  Misconception: Works out kinetic energy as mv², forgetting the half.
  Feedback from the catalogue
  Kinetic energy is $E_k = \tfrac12 mv^2$ ⁠: mass times speed squared, and half of that. Without the half the answer is twice too big.
- 0.3 J ± 0.001 J
  Misconception: Works out kinetic energy as ½mv, without squaring the speed.
  Feedback from the catalogue
  In kinetic energy the speed is squared: $E_k = \tfrac12 mv^2$ ⁠. So a cart at twice the speed carries four times the kinetic energy. $\tfrac12 mv$ ⁠ is half the momentum, not an energy, and its units are not joules either.
Feedback for a right answer
$E_k = \tfrac12 \times 1 \times 0.6^2 = 0.18$ ⁠, that is 0.18 J. That is all the kinetic energy the two carts have before the hit: B is standing still.

Hint
$E_k = \tfrac12 mv^2$ ⁠.

Worked solution
1. The givens: $m = 1\ \mathrm{kg}$ ⁠, $v = 0.6\ \mathrm{m/s}$ ⁠.
2. $E_k = \tfrac12 mv^2$ ⁠.
3. Substitute: $E_k = \tfrac12 \times 1 \times 0.36 = 0.18$ ⁠.
4. The kinetic energy: 0.18 J.
If you got $0.36$ ⁠, you left out the half. If you got $0.3$ ⁠, you did not square the speed.
Step 2 of 6
Open this step
When the carts stick
Practice · Calculate
Question
Equal carts, 1 kg each, stick and roll on together at 0.3 m/s. What is the kinetic energy of the two together after the hit?

Correct answer
0.09 J ± 0.002 J
Setup, choices and feedback
The Scene at this step
Mass ratio B : A
1 ×fixed
Bounciness
0fixed
Show momentum and energy
Offfixed
Accepted within
± 0.002 J
Computed from the Scene at this step’s setting
Both carts' kinetic energy just after the collision/nodes/carts/outputs/kineticAfter
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 0.18 J ± 0.002 J
  Misconception: Assumes kinetic energy is conserved in every collision, as momentum is.
  Feedback from the catalogue
  Momentum is conserved in every collision, but kinetic energy is not. When the bumpers squash and do not fully spring back, part of the kinetic energy becomes heat and sound. Only in an elastic collision, perfectly bouncy, is the kinetic energy after equal to the kinetic energy before. In a perfectly inelastic collision, when the carts stick, the most is lost.
- 0.045 J ± 0.002 J
  Misconception: Leaves one cart out after the collision: divides the momentum by one cart's mass, or counts only one cart's kinetic energy.
  Feedback from the catalogue
  After the collision there are two carts, and you need both. When they stick, the momentum is spread over the mass of both, $m_A + m_B$ ⁠. And the kinetic energy after the collision is the sum for both, $\tfrac12 m_A v_A'^2 + \tfrac12 m_B v_B'^2$ ⁠.
Feedback for a right answer
$E_k' = \tfrac12 \times 2 \times 0.3^2 = 0.09$ ⁠, that is 0.09 J: half of the 0.18 J there was before. The momentum is kept, 0.6 kg·m/s before and after, but half the kinetic energy has become heat and sound.

Hint
After the hit the carts are one body of mass 2 kg.

Worked solution
1. After the hit there is one body: $m = 2\ \mathrm{kg}$ ⁠, $v' = 0.3\ \mathrm{m/s}$ ⁠.
2. $E_k' = \tfrac12 m v'^2$ ⁠.
3. Substitute: $E_k' = \tfrac12 \times 2 \times 0.09 = 0.09$ ⁠.
4. The kinetic energy after: 0.09 J, against 0.18 J before.
If you answered $0.18$ ⁠, you assumed kinetic energy is kept like momentum. If you answered $0.045$ ⁠, you counted only one cart.
Step 3 of 6
Open this step
Momentum always, energy not always
Practice · Choose
Question
In which collisions is the two carts’ total momentum conserved: when they stick, when half bouncy, or when perfectly bouncy?

Correct answer
- Correct In all of them
Setup, choices and feedback
The Scene at this step
Mass ratio B : A
1 ×
Bounciness
0.5
Show momentum and energy
On
Choices
Shown in a random order.
- Correct In all of them
- Wrong Only when perfectly bouncy
  Misconception: Thinks momentum is conserved only in elastic collisions, since energy is lost in the others.
  Feedback from the catalogue
  Conservation of momentum does not depend on the bounciness. It comes from Newton’s third law: during the contact each cart receives an impulse equal and opposite to the other’s, whatever the bumpers. The difference between an elastic and an inelastic collision is in the kinetic energy alone.
- Wrong In none: a little momentum is always lost
  Misconception: Thinks some of the total momentum is lost in a collision, as heat or in the crash.
  Feedback from the catalogue
  The total momentum is not lost in a collision. Whatever impulse A gives B, B gives back to A in the opposite direction, so the total does not change, even when the carts squash and stick. What becomes heat and sound is part of the kinetic energy, a different quantity. The total momentum changes only when an outside force, such as friction, pushes on the carts.
- Wrong Only when they stick
  Feedback
  When the carts bounce apart, each still receives an impulse equal and opposite to the other’s. Slide Bounciness and see: the total bar does not move.
Feedback for a right answer
In all of them. In each one the total momentum bar stays on its dashed line. What changes from one collision to another is only the kinetic energy.

Hint
Slide Bounciness and watch the total bar against its dashed line.

Worked solution
1. During the contact each cart receives an equal and opposite impulse from the other, by Newton’s third law, whatever the bumpers.
2. Nothing outside pushes along the track, so the total does not change.
3. So momentum is conserved in all three collisions. The bounciness decides only how much kinetic energy comes back.
If you chose “Only when perfectly bouncy”, you mixed up conservation of momentum with conservation of kinetic energy: only the kinetic energy depends on the bounciness.
Step 4 of 6
Open this step
A heavy target
Practice · Predict, then look
Question
Now B is four times as heavy, and the carts stick. When the carts were equal, half the kinetic energy was lost. How much will be lost now?

Correct answer
- Correct About 80%
Setup, choices and feedback
The Scene at this step
Mass ratio B : A
4 ×fixed
Bounciness
0fixed
Show momentum and energy
Offhidden
Choices
- Wrong Half again
  Misconception: Thinks a sticky collision always loses half the kinetic energy.
  Feedback from the catalogue
  Half is lost only when the two carts are equal. When they stick, what is left is $\dfrac{m_A}{m_A + m_B}$ ⁠ of the kinetic energy, with B starting at rest. With B four times as heavy only a fifth is left, and $80\%$ ⁠ becomes heat and sound.
- Correct About 80%
- Wrong About 20%
  Feedback
  20% is what is left. With B four times as heavy, the pair rolls on very slowly, and most of the kinetic energy is lost.
- Wrong None, because B is heavy and hardly moves
  Feedback
  Precisely because B is heavy and hardly moves, the pair rolls on slowly, and little kinetic energy is left. The missing energy has become heat and sound.
Once the prediction is locked in, the Scene changes
Show momentum and energy
Onhidden
The Scene then shows: 0.144 J · The kinetic energy the collision did not give back, as heat and sound/nodes/carts/outputs/heat
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
About 80%: 0.144 J out of 0.18 J. The pair rolls on at only 0.12 m/s, with 0.036 J. When a cart hits a heavier one and sticks to it, most of the kinetic energy becomes heat and sound.

Worked solution
1. Conservation of momentum: $v' = \dfrac{1 \times 0.6}{1 + 4} = 0.12\ \mathrm{m/s}$ ⁠.
2. The kinetic energy after: $\tfrac12 \times 5 \times 0.12^2 = 0.036\ \mathrm{J}$ ⁠.
3. What was lost: $0.18 - 0.036 = 0.144\ \mathrm{J}$ ⁠, that is $\dfrac{0.144}{0.18} = 80\%$ ⁠.
If you predicted half, you generalised from the equal carts. What is left is $\dfrac{m_A}{m_A + m_B}$ ⁠: a half for equal carts, and a fifth when B is four times as heavy.
Step 5 of 6
Open this step
How bouncy
Practice · Calculate
Question
Cart A arrives at 0.6 m/s at B, which stands still. After the hit A comes back at 0.1 m/s, and B moves right at 0.35 m/s. What is the bounciness of the collision?

Correct answer
0.75 ± 0.01
Setup, choices and feedback
The Scene at this step
Mass ratio B : A
2 ×fixed
Bounciness
0.75hidden
Show momentum and energy
Offfixed
Accepted within
± 0.01
Computed from the Scene at this step’s setting
Bounciness/nodes/carts/props/bounciness
Wrong answers it recognises
- 0.4167 ± 0.01
  Misconception: Drops the sign of a velocity or momentum that points left, as if every motion were to the right.
  Feedback from the catalogue
  Momentum and velocity have a direction. On the track right is taken as positive, so a cart rolling left, for instance after bouncing back, gets a minus sign: 0.2 m/s to the left is −0.2 m/s. Without the sign the books do not balance: the momentum of a cart coming back takes away from the total.
Feedback for a right answer
$e = \dfrac{0.35 - (-0.1)}{0.6} = \dfrac{0.45}{0.6} = 0.75$ ⁠. Quite bouncy, but not perfectly: of the 0.18 J only about 0.13 J comes back, and the rest becomes heat and sound.

Hint
The bounciness is the speed of moving apart divided by the speed of closing in. A and B move in opposite directions, so they move apart at the sum of their speeds.

Worked solution
1. Right is positive: $v_A = 0.6$ ⁠ and $v_B = 0$ ⁠, and after the hit $v_A' = -0.1$ ⁠ and $v_B' = 0.35$ ⁠, in metres per second.
2. The speed of closing in: $v_A - v_B = 0.6$ ⁠. The speed of moving apart: $v_B' - v_A' = 0.35 - (-0.1) = 0.45$ ⁠.
3. The bounciness: $e = \dfrac{0.45}{0.6} = 0.75$ ⁠.
If you got about $0.42$ ⁠, you worked out $0.35 - 0.1$ ⁠ without the sign. The carts move in opposite directions, so the gap between them grows at the sum of their speeds.
Step 6 of 6
Open this step
Where the energy goes
Practice · Choose
Question
When the equal carts stick, half the kinetic energy is missing after the hit. Where did it go?

Correct answer
- Correct Into heat in the bumpers and the sound of the knock
Setup, choices and feedback
The Scene at this step
Mass ratio B : A
1 ×
Bounciness
0
Show momentum and energy
On
Choices
Shown in a random order.
- Correct Into heat in the bumpers and the sound of the knock
- Wrong It vanished: a collision destroys energy
  Misconception: Thinks the kinetic energy a collision loses simply disappears.
  Feedback from the catalogue
  Energy does not disappear. The kinetic energy the collision does not give back goes into heat in the bumpers, as their material squashes and its molecules jiggle harder, and into the sound wave of the knock. It is still there, only in a form that can no longer move the carts.
- Wrong It turned into momentum
  Feedback
  Momentum and energy are two different quantities, in different units, and one does not turn into the other. Nor did the total momentum grow: it stayed at 0.6 kg·m/s.
- Wrong It went into cart B
  Feedback
  B did gain kinetic energy, but that is already counted: the 0.09 J belongs to both carts together. What is missing is beyond that.
Feedback for a right answer
The energy did not vanish. When the bumpers squash and do not spring back, their material warms a little, and the knock sends out a sound wave. On the board that is the orange bar, Heat and sound: exactly what the kinetic energy is missing.

Hint
Choose On under Show momentum and energy, and look at the orange bar.

Worked solution
1. Before the hit: 0.18 J of kinetic energy. After the equal carts stick: 0.09 J.
2. 0.09 J is missing. Energy is neither made nor destroyed, so it has taken another form.
3. The bumpers squashed and did not spring back: the energy became the random motion of their molecules, that is heat, and the sound wave of the knock.
If you chose that the energy vanished, remember: in a collision it only passes into a form that no longer moves the carts.

Across the unit · 20 misconceptions

Misconception catalogue

Each common mistake this unit recognises, in a few words. Open one for what the learner reads when an answer matches it, and where it is asked.

Index

Mass times velocity

What momentum is

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

Working out momentum

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Heavy and slow, light and fast

Question

Goal

The Scene at this step

One setting that reaches it

Hint

Worked solution

Car or bullet

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

Momentum has a direction

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Impulse

Force times time

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

What B received

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Equal and opposite pushes

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

The impulse on A

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Bounce or stick

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

How hard

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

A soft landing