Conservation of Momentum (AQA A-Level Further Maths): Revision Notes

Worked Example 1: Direct collision with body at rest

Problem: A body $P$ of mass 1 kg moving with velocity $5 \text{ m s}^{-1}$ collides directly with another body $Q$ of mass 2 kg moving towards $P$ with velocity $4 \text{ m s}^{-1}$. After the collision, $Q$ is at rest. Find the final velocity of $P$.

Solution:

Choose the direction of $P$'s initial motion as positive.

Initial velocities: $u_P = 5 \text{ m s}^{-1}$, $u_Q = -4 \text{ m s}^{-1}$ (opposite direction)

Final velocities: $v_P = v$ (unknown), $v_Q = 0$ (at rest)

There are no external forces, so momentum is conserved.

Total initial momentum: $= 1 \times 5 + 2 \times (-4) = 5 - 8 = -3 \text{ N s}$

Total final momentum: $= 1 \times v + 2 \times 0 = v \text{ N s}$

The momentum equation is: $5 - 8 = v + 0$

Solving: $v = -3 \text{ m s}^{-1}$

The final velocity is $3 \text{ m s}^{-1}$ in the opposite direction to that shown on the diagram.

Worked Example 2: Coalescence with vector components

Problem: A particle $P$ of mass 1 kg with velocity $(2\mathbf{i} - 3\mathbf{j}) \text{ m s}^{-1}$ collides and coalesces with a particle $Q$ of mass 4 kg and velocity $(\mathbf{i} + 2\mathbf{j}) \text{ m s}^{-1}$. They move off together with a common velocity $\mathbf{v}$. Calculate the speed $|\mathbf{v}|$ and the angle $\mathbf{v}$ makes with the $\mathbf{i}$-direction.

Solution:

The combined mass after impact is: $1 + 4 = 5 \text{ kg}$

There are no external forces, so momentum is conserved.

The momentum equation is: $1 \times (2\mathbf{i} - 3\mathbf{j}) + 4 \times (\mathbf{i} + 2\mathbf{j}) = 5 \times \mathbf{v}$

Expanding: $2\mathbf{i} - 3\mathbf{j} + 4\mathbf{i} + 8\mathbf{j} = 5\mathbf{v}$

$6\mathbf{i} + 5\mathbf{j} = 5\mathbf{v}$

Final velocity: $\mathbf{v} = 1.2\mathbf{i} + \mathbf{j} \text{ m s}^{-1}$

Speed: $|\mathbf{v}| = \sqrt{1.2^2 + 1^2} = \sqrt{2.44} = 1.56 \text{ m s}^{-1}$

Angle with positive $\mathbf{i}$-direction: $\tan^{-1}\left(\frac{1}{1.2}\right) = 39.8°$

Note: Use Pythagoras and trigonometry to calculate the speed and angle when working with vector components.

Worked Example 3: Skateboard problem

Problem: A child of mass 30 kg travels in a horizontal straight line on a skateboard of mass 5 kg with a velocity of $1.4 \text{ m s}^{-1}$. The child jumps off the skateboard with an initial horizontal backwards velocity of $0.2 \text{ m s}^{-1}$ while the skateboard continues forwards. All motion is in the same straight line. Calculate the final velocity, $v$, of the skateboard.

Solution:

Total mass initially: $30 + 5 = 35 \text{ kg}$

Total initial momentum: $= (30 + 5) \times 1.4 = 49 \text{ N s}$

After the child jumps, taking the forward direction as positive:

• Child's velocity: $-0.2 \text{ m s}^{-1}$ (backwards)
• Skateboard's velocity: $v$ (forwards, unknown)

Total final momentum: $= (30 \times (-0.2)) + (5 \times v) = -6 + 5v \text{ N s}$

There are no external forces, so momentum is conserved.

The momentum equation is: $49 = 5v - 6$

Solving: $5v = 55$ $v = 11 \text{ m s}^{-1}$

The final velocity of the skateboard is $11 \text{ m s}^{-1}$ in the forward direction.

Worked Example 4: Sequential collisions

Problem: Three particles $A$, $B$ and $C$, with masses 1 kg, 4 kg and 12 kg respectively, are positioned in a straight line. Particles $B$ and $C$ are at rest and particle $A$ is moving towards $B$ with a speed of $10 \text{ m s}^{-1}$.

After $A$ and $B$ collide, particle $A$ rebounds backwards and $B$ moves towards $C$ with twice the speed of $A$.

After $B$ and $C$ collide, they move in opposite directions with the same speed.

Show that there are no more collisions between the particles, assuming no other forces act on them.

Solution:

Let $A$ rebound with speed $u$ (backwards) after the first collision. Then $B$ moves with speed $2u$ (forwards).

First collision (A strikes B):

Taking the forward direction (towards $C$) as positive:

Initial momentum: $1 \times 10 + 4 \times 0 = 10 \text{ N s}$

Final momentum: $1 \times (-u) + 4 \times 2u = -u + 8u = 7u \text{ N s}$

There are no external forces, so momentum is conserved: $10 = 7u$

Therefore: $u = \frac{10}{7} \text{ m s}^{-1}$

After the first collision:

• $A$ moves at $\frac{10}{7} \text{ m s}^{-1}$ backwards
• $B$ moves at $\frac{20}{7} \text{ m s}^{-1}$ forwards

Second collision (B strikes C):

Let $B$ and $C$ move with speeds $v$ after the collision (in opposite directions).

Initial momentum: $4 \times \frac{20}{7} + 12 \times 0 = \frac{80}{7} \text{ N s}$

Final momentum (taking $B$ moving backwards as negative, $C$ forwards as positive): $= 4 \times (-v) + 12 \times v = -4v + 12v = 8v \text{ N s}$

Conserving momentum: $\frac{80}{7} = 8v$

Therefore: $v = \frac{10}{7} \text{ m s}^{-1}$

After both collisions:

• $A$ moves at $\frac{10}{7} \text{ m s}^{-1}$ backwards (towards the left)
• $B$ moves at $\frac{10}{7} \text{ m s}^{-1}$ backwards (towards the left, same direction as $A$)
• $C$ moves at $\frac{10}{7} \text{ m s}^{-1}$ forwards (towards the right, opposite to $A$ and $B$)

Since $A$ and $B$ are moving in the same direction with the same speed, and in the opposite direction to $C$, there are no further collisions. Hence, there are no more collisions between the particles.