← Impulse and momentum change

Find the momentum change

Esther's 2 kg skateboard rolls at 3 m/s. She gives it one strong boost, and it rolls away at 7 m/s in the same direction.

By how much did the skateboard's momentum change?

Momentum change compares two moments: before and after.

Both momenta carry signs along one line, so name the positive direction first.

Take the skateboard's direction of travel as positive.

Before the boost: p_initial = 2 × 3 = 6 kg·m/s.

After the boost: p_final = 2 × 7 = 14 kg·m/s.

The change is final minus initial: Δp = p_final − p_initial.

Δp = 14 − 6 = 8 kg·m/s.

The boosted skateboardmomentvelocitymomentumbefore3 m/s6 kg·m/safter7 m/s14 kg·m/schangeΔp = 14 − 6 = 8 kg·m/s
Before, after, subtract: Δp = 8 kg·m/s.

The boost added 8 kg·m/s of momentum.

A 3 kg cart speeds up from 2 m/s to 6 m/s. Take its direction of travel as positive. Enter its momentum change in kg·m/s.

A velocity that points the negative way takes a negative sign, and so does its momentum.

Always subtract in the same order: final minus initial.

A negative Δp means the object lost forward momentum, or swung over to the negative direction.

Momentum change is final minus initial, Δp = p_final − p_initial, with every momentum signed under the stated positive direction.

A 20 kg canoe glides at 4.5 m/s. Backpaddling slows it to 1.5 m/s, still in the same direction. Take its direction of travel as positive. Calculate its momentum change.

Write down the values:

m = 20 kg, v_initial = 4.5 m/s, v_final = 1.5 m/s, Δp = ?

Write down the equation:

Δp = p_final − p_initial

Substitute in the values, and calculate:

p_initial = 20 × 4.5 = 90

p_final = 20 × 1.5 = 30

Δp = 30 − 90 = −60

The canoe's momentum change is −60 kg·m/s: it lost 60 kg·m/s.

A 60 kg skater slows from 5 m/s to 1 m/s, still moving the same way. Take her direction of travel as positive. Enter her momentum change in kg·m/s (keep the sign).

A 0.5 kg ball hits a wall at 6 m/s and rebounds along the same line at 4 m/s. Take the direction toward the wall as positive. Calculate the ball's momentum change.

The rebounding ball (toward the wall is positive)momentvelocitymomentumbefore6 m/s3 kg·m/safter−4 m/s−2 kg·m/schangeΔp = −2 − 3 = −5 kg·m/s
The rebound velocity is negative, so the after momentum is negative too.

Write down the values:

m = 0.5 kg, v_initial = 6 m/s, v_final = −4 m/s (it rebounds), Δp = ?

Write down the equation:

Δp = p_final − p_initial

Substitute in the values, and calculate:

p_initial = 0.5 × 6 = 3

p_final is negative: p_final = 0.5 × (−4) = −2

Δp = −2 − 3 = −5

The ball's momentum change is −5 kg·m/s. The reversal makes the change bigger than a plain stop would be.

+−3−2−10+1+2+3+4beforeafterΔp = −5 kg·m/smomentum of the ball (kg·m/s)
The ball's momentum slides from +3 kg·m/s through zero to −2 kg·m/s: a change of −5 kg·m/s.

A 0.4 kg handball hits a wall at 8 m/s and rebounds along the same line at 3 m/s. Take the direction toward the wall as positive. Enter the ball's momentum change in kg·m/s (keep the sign).

Your turn

A 4 kg scooter speeds up from 2.5 m/s to 6.5 m/s. Take its direction of travel as positive. Enter its momentum change in kg·m/s.

A 0.6 kg basketball hits a backboard at 8 m/s and rebounds along the same line at 5 m/s. Take the direction toward the backboard as positive. Enter the ball's momentum change in kg·m/s (keep the sign).

A 1 kg ball moves at 5 m/s. Case one: it hits soft clay and stops dead. Case two: it hits a springy net and bounces straight back at 5 m/s. Which statement about the size of the momentum change is correct?