Momentum, impulse and collisions · Momentum, projectiles, work & energy

The momentum of a body is its mass times its velocity, $p = mv$ (units kg m/s, or N s). Because velocity has a direction, so does momentum: choose one direction as positive, and a body moving the other way has a negative velocity.

Impulse and change of momentum

A force $F$ acting for a time $t$ gives an impulse $Ft$ , equal to the change of momentum (from F = ma and $v = u + at$ ):

Ft = mv - mu

When a body turns back, its velocity changes sign. The change of velocity is then the sum of the two speeds:

Bouncing backAway from the wall positive: v − (−u) = u + v

Worked example · WAEC 2019

WAEC 2019 · Paper 2 · Q7

A ball of mass $0.1\text{ kg}$ approaching a tennis player with a velocity of $10\text{ m s}^{-1}$ is hit in the opposite direction with a velocity of $15\text{ m s}^{-1}$ . If the time of impact between the racket and the ball is $0.01$ second, calculate the magnitude of the force with which the ball was hit.

Signs
- Before: ${u = -10}$ m/s. After: ${v = +15}$ m/s.
Think first. Take the direction the ball is hit as positive. What is the velocity before?
Change of momentum
- ${m(v - u) = 0.1(15 - (-10)) = 0.1 \times 25 = 2.5}$ N s.
The force
- ${F = \frac{2.5}{0.01} = 250}$ N.

Momentum in two dimensions is a vector: subtract the velocity vectors, then find the length.

More: impulse and change of momentum

Conservation of momentum

When two bodies collide, the forces between them are equal and opposite, so the total momentum is the same before and after:

Two bodies joinm₁u₁ + m₂u₂ = (m₁ + m₂)v

If they bounce apart instead, each has its own velocity afterwards: ${m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2}$ .

Momentum in a collisionSet the masses and velocities (right is +)

18 kg m/smomentum before and after3 m/scommon velocity

m₁ (kg) = 4u₁ (m/s) = 6m₂ (kg) = 2u₂ (m/s) = −3

Before: 4(6) + 2(−3) = 18. After: (4 + 2)v. So v = 18 ÷ 6 = 3 m/s, to the right.

Worked example · WAEC 2018

WAEC 2018 · Paper 2 · Q8

A body of mass $20\text{ kg}$ moving with a velocity of $80\text{ m s}^{-1}$ collides with another body of mass $30\text{ kg}$ moving with a velocity of $50\text{ m s}^{-1}$ . If the two bodies moved together after collision, find their common velocity if they moved in the:

same direction before collision;

opposite directions before collision.

Same direction
- ${20(80) + 30(50) = 50v}$ .
- ${3100 = 50v}$ , so ${v = 62}$ m/s.
Think first. Both velocities are positive.
Opposite directions
- ${20(80) + 30(-50) = 50v}$ .
- ${100 = 50v}$ , so ${v = 2}$ m/s, in the direction the 20 kg body was moving.
Think first. Take the 20 kg body's direction as positive. What is the 30 kg body's velocity?

Worked example · WAEC 2022

WAEC 2022 · Paper 2 · Q15 (a)

Two particles, $P$ and $Q$ , of masses $3\text{ kg}$ and $1.5\text{ kg}$ respectively moved in opposite directions. $P$ moved with a velocity of $5\text{ m s}^{-1}$ while $Q$ moved with a velocity of $7\text{ m s}^{-1}$ . The particles collided head-on and moved in the same direction after collision. The difference in the velocities after collision is $\frac34\text{ m s}^{-1}$ , where the final velocity of $P$ ( $V_P$ ) is greater than the final velocity of $Q$ ( $V_Q$ ). Find the velocities of $P$ and $Q$ after collision.

Momentum before
- ${3(5) + 1.5(-7) = 15 - 10.5 = 4.5}$ .
Think first. Take P's direction as positive. Q moves the other way.
Two equations
- ${3V_P + 1.5V_Q = 4.5}$ , so ${2V_P + V_Q = 3}$ .
- ${V_P - V_Q = \frac34}$ .
Think first. Momentum is conserved, and you know the difference of the velocities.
Solve
- Add: ${3V_P = \frac{15}{4}}$ , so ${V_P = \frac54}$ m/s.
- Then ${V_Q = \frac54 - \frac34 = \frac12}$ m/s.

More: conservation of momentum

Your turn

WAEC 2016 · Paper 2 · Q7

Momentum, projectiles, work & energy

(a)
A body of mass $3\text{ kg}$ moves with a velocity of $8\text{ m s}^{-1}$ . It collides with a second body moving in the same direction with a velocity of $5\text{ m s}^{-1}$ . After collision, the bodies move together with a velocity of $6\text{ m s}^{-1}$ . Find the mass of the second body.
(b)
If the second body in (a) moves with a velocity of $5\text{ m s}^{-1}$ in the opposite direction to the $3\text{ kg}$ body moving at $8\text{ m s}^{-1}$ , find, correct to two decimal places, the common velocity of the two bodies if they move together after collision.

Worked solution (try it first)

(a)

Momentum before: $3(8) + 5m = 24 + 5m$ .
After, moving together: $6(3 + m) = 18 + 6m$ .
Momentum is conserved: $24 + 5m = 18 + 6m$ , so $m = 6\text{ kg}$ .

(b)

Now the $6\text{ kg}$ body moves the other way: momentum before $= 3(8) + 6(-5) = -6$ .
After: $(3 + 6)v = 9v$ , so $v = -\dfrac69 = -\dfrac23$ .
The common velocity is $0.67\text{ m s}^{-1}$ , in the direction the $6\text{ kg}$ body was moving.

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