4.1 Linear Momentum

Unit 4: Linear Momentum. Unit overview · Games · Practice set

Six little picture stories

Press Next (or Play) to walk through each story one small step at a time. The numbers come last. Tap a round play button to hear a step read aloud. After every three stories there is one quick check.

1. A tennis ball and a bowling ball

Read the story as text

You are pushing a loaded shopping cart across a parking lot at a slow walk. A friend tosses you a tennis ball, fast. You catch the ball with one hand and it barely moves you. Then the cart starts to roll down a gentle slope toward a parked car, and you have to dig your heels in to stop it, even though it is moving slowly.

The ball is fast but light. The cart is slow but heavy. So which one has "more motion", and how do we put a number on it?

2. Mass and speed both count

3. Momentum has a direction

Quick check. One question before the next stories.

A 1 kg ball rolls left at 3 m/s. Right is positive. What is its momentum?

4. A parked truck has no momentum

5. Two brothers, one total

6. Twice as fast, twice the momentum

Quick check. One question before the next stories.

Ella rides at twice her old speed on the same bike. What happens to her momentum?

7. Check yourself

Think of your answer first, then tap to see it.

a) A bowling ball and a ping-pong ball roll at the same speed. Which one has more momentum? Why?

Show answer

The bowling ball. Momentum is mass × velocity. Same velocity, much more mass, so much more momentum.

b) A 0.15 kg baseball flies at 40 m/s. What is its momentum?

Show answer

p = m v = 0.15 kg × 40 m/s = 6 kg·m/s, in the direction the ball flies.

c) A 2 kg cart rolls right at 3 m/s. A 3 kg cart rolls left at 2 m/s. Right is positive. What is their total momentum?

Show answer

+2 × 3 = +6 kg·m/s and 3 × (−2) = −6 kg·m/s. Total = +6 + (−6) = 0. Both move, but the momenta cancel.

Already know this?

Three quick questions. Get all three right on the first try and you can skip ahead to the simulation.

A 2000 kg car is parked. A 0.5 kg ball rolls past at 4 m/s. Which has more momentum?

Two carts roll toward each other. One has momentum +4 kg·m/s, the other −4 kg·m/s. What is the total momentum?

A 3 kg cart and a 1 kg cart have the same momentum. How do their speeds compare?

The idea

Picture a shopping cart rolling toward you in a parking lot. A cart full of groceries is hard to stop. An empty cart at the same speed stops easily.

Now picture the empty cart rolling much faster. It gets harder to stop too. So two things count: how much mass it has, and how fast it moves.

full 30 kg 1 m/s 10 kg empty, 3 m/s equally hard to stop

Momentum is mass times velocity. It measures how much motion an object carries, and how hard it is to stop.

p = m v

Units: kilograms times metres per second, written kg·m/s. (This is the same as N·s.)

Momentum is a vector. It points the same way as the velocity. On a line, we pick a positive direction (say right is +) and give momentum a sign. A cart moving left has negative momentum.

A heavy object that is not moving has zero momentum. A light object can have a lot of momentum if it is fast enough. Mass and speed both count.

A system of objects has a total momentum: add the momenta of the parts as vectors (with signs on a line).

psystem = m₁v₁ + m₂v₂ + … = Mtotal vcm

The second form says the total momentum equals the total mass times the velocity of the centre of mass. So if you know the total momentum, you know how the centre of mass moves: vcm = psystem / Mtotal.

A 2 kg B 3 kg +3 m/s −1.5 m/s right is +

The same idea with numbers

Read the steps as text

Cart A (2.0 kg) moves right at 3.0 m/s. Cart B (3.0 kg) moves left at 1.5 m/s. Right is positive. Find each momentum, the total momentum, and the velocity of the centre of mass.

  1. Write each velocity with its sign: vA = +3.0 m/s, vB = −1.5 m/s. Why: momentum is a vector. On a line, the sign carries the direction.
  2. pA = mAvA = (2.0 kg)(+3.0 m/s) = +6.0 kg·m/s. Why: p = mv for one object.
  3. pB = (3.0 kg)(−1.5 m/s) = −4.5 kg·m/s. Why: B moves left, so its momentum is negative.
  4. psystem = +6.0 + (−4.5) = +1.5 kg·m/s (to the right). Why: total momentum is the vector sum. We do NOT add 6.0 and 4.5 to get 10.5.
  5. vcm = psystem / M = 1.5 kg·m/s ÷ 5.0 kg = +0.30 m/s. Why: psystem = M vcm. The centre of mass drifts slowly right.

Lab: momentum of two carts

Set the mass and velocity of each cart (right is +). Predict the direction of the total momentum, then play. Watch the momentum graph: each cart's momentum changes when they bump, but the dashed total line stays flat. (That is a preview of 4.3.) The centre of mass marker moves at a steady vcm the whole time.

Try this: make cart 1 heavy and slow (4 kg at 0.5 m/s) and cart 2 light and fast (0.5 kg at −3 m/s). Which momentum is bigger in size? (Answer: 2.0 vs 1.5 kg·m/s, so cart 1 wins even though it is slow.) Now find settings where the total momentum is exactly zero. What does the centre of mass do then?

Examples

Example 1: a kicked ball basic

A 0.40 kg soccer ball flies toward the goal at 12 m/s. What is its momentum?

Show solution
Read the steps as text
  1. p = mv = (0.40 kg)(12 m/s) = 4.8 kg·m/s, toward the goal. Why: one object, so just mass times velocity, with its direction.

Example 2: two skaters medium

A 60 kg skater glides east at 4.0 m/s. A 50 kg skater glides west at 5.0 m/s. Find the total momentum of the two-skater system and the velocity of its centre of mass. Take east as positive.

Show solution
Read the steps as text
  1. p₁ = (60 kg)(+4.0 m/s) = +240 kg·m/s. Why: east is +.
  2. p₂ = (50 kg)(−5.0 m/s) = −250 kg·m/s. Why: west is −.
  3. ptotal = 240 − 250 = −10 kg·m/s, so 10 kg·m/s west. Why: add with signs. Notice the lighter skater has the bigger momentum because she is faster.
  4. vcm = −10 ÷ 110 = −0.091 m/s, about 0.09 m/s west. Why: vcm = ptotal / Mtotal, with M = 60 + 50 = 110 kg.
Trap: saying the heavier object always has more momentum. Instead: p = mv, so speed counts too: the 50 kg skater at 5.0 m/s (250 kg·m/s) beats the 60 kg skater at 4.0 m/s (240 kg·m/s).

Example 3: two pucks at right angles AP

On an air table, a 3.0 kg puck moves east at 4.0 m/s and a 2.0 kg puck moves north at 3.0 m/s. Find the size and direction of the total momentum.

Show solution
Read the steps as text
  1. East part: px = (3.0)(4.0) = 12 kg·m/s. North part: py = (2.0)(3.0) = 6.0 kg·m/s. Why: momentum is a vector, so work in components.
  2. Size: |p| = √(12² + 6.0²) = √180 = 13.4 kg·m/s. Why: the components are at right angles, so use Pythagoras. Adding 12 + 6 = 18 is wrong.
  3. Direction: θ = tan⁻¹(6.0 / 12) = 26.6° north of east. Why: the angle from the east axis has tangent py/px.

Example 4: same kinetic energy, different momentum AP

Object X has mass m. Object Y has mass 4m. Both have the same kinetic energy K. Find the ratio pY / pX.

Show solution
Read the steps as text
  1. From K = ½mv², v = √(2K/m). Then p = mv = √(2mK). Why: we want p in terms of m and K only, so we remove v.
  2. pY / pX = √(2·4m·K) / √(2mK) = √4 = 2. Why: K is the same, so only the mass changes under the square root.
  3. Check with numbers: m = 1 kg at 2 m/s has K = 2 J and p = 2 kg·m/s. 4 kg with K = 2 J has v = 1 m/s and p = 4 kg·m/s. Ratio 2. Why: a quick number check catches algebra slips.

Practice (AP style)

1. A 2000 kg truck moves at 10 m/s. A 1000 kg car moves at 20 m/s in the same direction. How do their momenta compare?

Show answer

Truck: (2000)(10) = 20 000 kg·m/s. Car: (1000)(20) = 20 000 kg·m/s. Same. (Their kinetic energies are different: the car has twice the KE. Momentum and KE are not the same thing.)

2. Cart P (1.0 kg) moves at +2.0 m/s. Cart Q (2.0 kg) moves at −1.0 m/s. What is the velocity of the centre of mass of the two-cart system?

Show answer

ptotal = (1.0)(2.0) + (2.0)(−1.0) = 0. So vcm = 0 / 3.0 kg = 0. The carts move, but the centre of mass stays put.

3. A runner's kinetic energy doubles while her mass stays the same. By what factor does the size of her momentum change?

Show answer

p = √(2mK). If K doubles with m fixed, p grows by √2 ≈ 1.41.

4. A student measures the momentum p of a cart at several velocities v and plots p against v. The points lie on a straight line through the origin with slope 0.80 kg. What does the slope tell the student?

Show answer

p = mv has the form y = (slope)x with slope m. Units check: (kg·m/s) ÷ (m/s) = kg.

5. Two 1.0 kg pucks slide on ice: one east at 3.0 m/s, one north at 4.0 m/s. What is the size of the total momentum?

Show answer

Components 3.0 east and 4.0 north (kg·m/s). Size √(3² + 4²) = 5.0 kg·m/s. Adding to 7.0 ignores direction.

6. Which of these has the greatest size of momentum?

Show answer

Ball: 6.0. Bus: 0 (not moving). Person: 105. Bullet: 3.0 (all in kg·m/s). The walking person wins.

7. (Short answer) A 3.0 kg cart and a 1.0 kg cart are on a track. The 3.0 kg cart moves at +2.0 m/s. The centre of mass of the pair moves at +1.0 m/s. Find the velocity of the 1.0 kg cart.

Show answer
  1. ptotal = M vcm = (4.0 kg)(1.0 m/s) = 4.0 kg·m/s.
  2. 4.0 = (3.0)(2.0) + (1.0)v → v = 4.0 − 6.0 = −2.0 m/s.
  3. So the 1.0 kg cart moves left at 2.0 m/s.

8. (Short answer, reasoning) A classmate says: "A bowling ball always has more momentum than a tennis ball, because it is heavier." Is the claim right? Use p = mv in your answer and give a number example.

Show answer

No. Momentum depends on mass and velocity. A 6.0 kg bowling ball rolling at 0.50 m/s has p = 3.0 kg·m/s. A 0.058 kg tennis ball served at 60 m/s has p = 3.5 kg·m/s, which is more. And a bowling ball at rest has zero momentum. A good answer names both quantities and shows a case where the light object wins.

9. (Short free response: translation between representations) A 2.0 kg cart moves along a track. Its velocity-time graph (right positive) is: flat at +3.0 m/s from 0 to 2.0 s, a very quick drop at t = 2.0 s (it bounces off a bumper), then flat at −1.0 m/s from 2.0 s to 4.0 s. (a) Describe the momentum-time graph, with values. (b) Which bar chart shows the cart's momentum before and after the bounce? (A) +6.0 and +2.0; (B) +6.0 and −2.0; (C) +3.0 and −1.0; (D) +6.0 and 0. (c) Calculate the change in momentum.

Show answer
  1. (a) p = mv, so the graph has the same shape as the velocity graph, scaled by 2.0 kg: flat at +6.0 kg·m/s from 0 to 2.0 s, a quick drop, then flat at −2.0 kg·m/s from 2.0 s to 4.0 s.
  2. (b) (B). (C) gives velocities, not momenta; (A) loses the sign of the bounce.
  3. (c) Δp = −2.0 − (+6.0) = −8.0 kg·m/s (8.0 kg·m/s to the left).

Point guide (4 points):

  • 1 point: p-t graph has the shape of v-t scaled by m.
  • 1 point: values +6.0 and −2.0 kg·m/s with signs.
  • 1 point: picks (B).
  • 1 point: Δp = −8.0 kg·m/s.

10. (Short free response: experimental design and analysis) A student wants to find which has more momentum: a light cart pushed hard or a heavy cart pushed gently. She has a balance, a photogate and a card 0.10 m long that sits on top of each cart. (a) Describe how she finds each cart's momentum. (b) Her data: cart A, 0.50 kg, card blocks the gate for 0.050 s; cart B, 1.5 kg, blocks it for 0.20 s. Calculate each momentum and say which is bigger. (c) The timer is accurate to ±0.002 s. Estimate the percent uncertainty in cart A's speed, and say whether it could change the answer to (b).

Show answer
  1. (a) Measure each cart's mass on the balance. Push the cart through the photogate on a level track. Speed v = card length / blocking time. Momentum p = mv, in the direction of motion. Repeat several runs and average.
  2. (b) A: v = 0.10 / 0.050 = 2.0 m/s, p = 0.50 × 2.0 = 1.0 kg·m/s. B: v = 0.10 / 0.20 = 0.50 m/s, p = 1.5 × 0.50 = 0.75 kg·m/s. The lighter cart A has more momentum, because it is much faster.
  3. (c) 0.002 / 0.050 = 4%, so pA = 1.0 ± 0.04 kg·m/s. That cannot bring it down to 0.75, so the answer stands.

Point guide (4 points):

  • 1 point: measures mass and speed (card length / time).
  • 1 point: p = mv for each cart, 1.0 and 0.75 kg·m/s.
  • 1 point: concludes the lighter cart has more momentum.
  • 1 point: 4% uncertainty and a correct judgement.

11. (Short free response: mathematical routines) A 1500 kg car moves east at 20 m/s. A 9000 kg truck moves west at 4.0 m/s. Take east as positive. (a) Calculate the momentum of each vehicle. (b) Calculate the total momentum of the car + truck system. (c) Calculate the speed the car would need for its momentum to have the same size as the truck's.

Show answer
  1. (a) Car: p = 1500 × 20 = +30 000 kg·m/s. Truck: p = 9000 × (−4.0) = −36 000 kg·m/s.
  2. (b) ptotal = 30 000 + (−36 000) = −6000 kg·m/s (6000 kg·m/s west).
  3. (c) v = 36 000 / 1500 = 24 m/s.

Point guide (3 points):

  • 1 point: +30 000 and −36 000 kg·m/s.
  • 1 point: −6000 kg·m/s (west).
  • 1 point: 24 m/s.

12. (Short free response: qualitative/quantitative translation) Cart A has mass m and cart B has mass 4m. Both start at rest on a level frictionless track, and each is pushed by the same constant force F for the same time Δt. (a) Without equations, compare the final momenta of the carts and explain. (b) Derive expressions for each cart's final speed in terms of F, Δt and m. (c) Calculate both momenta and both speeds for m = 0.50 kg, F = 2.0 N and Δt = 1.5 s. (d) Are your numbers consistent with (a)?

Show answer
  1. (a) The same force for the same time gives the same change in momentum, so the final momenta are equal; the heavier cart just moves more slowly.
  2. (b) FΔt = Δp = mv, so vA = FΔt / m and vB = FΔt / (4m).
  3. (c) pA = pB = 2.0 × 1.5 = 3.0 kg·m/s. vA = 3.0 / 0.50 = 6.0 m/s; vB = 3.0 / 2.0 = 1.5 m/s.
  4. (d) Yes: equal momenta as predicted in (a), and vA is 4 times vB, matching (b).

Point guide (4 points):

  • 1 point: equal momenta, same F and Δt.
  • 1 point: vA = FΔt/m, vB = FΔt/(4m).
  • 1 point: 3.0 kg·m/s, 6.0 m/s, 1.5 m/s.
  • 1 point: consistency with (a) and (b).

Common mistakes

The mistake: adding sizes of momentum when objects move in opposite directions (6 + 4.5 = 10.5).

Why it is wrong: momentum is a vector. Opposite directions partly cancel.

How to spot it: did you write a sign on every velocity before multiplying? If two objects move toward each other, one momentum must be negative.

The mistake: "heavier means more momentum" (or "faster means more momentum").

Why it is wrong: it is the product mv. A slow heavy object and a fast light one can have the same momentum.

How to spot it: always compute both products before comparing.

The mistake: mixing up momentum and kinetic energy.

Why it is wrong: p = mv is a vector with units kg·m/s. K = ½mv² is a scalar in joules and is never negative. Two objects can have equal momentum and different KE.

How to spot it: check units. If your answer is in J, it is not momentum.

The mistake: in two dimensions, adding the sizes (3 + 4 = 7).

Why it is wrong: perpendicular components combine with Pythagoras.

How to spot it: if the objects move in different directions that are not opposite, split into x and y parts.

The mistake: thinking the centre of mass must be at rest, or must move with the heavier object.

Why it is wrong: vcm = ptotal / M. It is set by the total momentum, which can point either way.

How to spot it: compute ptotal first. If it is zero, the centre of mass is at rest; otherwise it moves in that direction.