# A-level Physics (Advancing Physics)/Conservation of Momentum/Worked Solutions

1. A ball of mass 0.5 kg collides with a stationary ball of 0.6 kg at a velocity of 3ms−1. If the stationary ball moves off at a speed of 2ms−1, what is the new velocity of the first ball?

$0.5\times 3=0.5v+(0.6\times 2)$ $1.5=0.5v+1.2$ $0.5v=0.3$ $v=0.6{\mbox{ ms}}^{-1}$ 2. Two balls are moving in opposite directions with velocities 5ms−1 and 10ms−1. They collide, and move off in opposite directions with new velocities of 7.5ms−1 each. If the mass of the first ball was 1.25 kg, what is the mass of the second ball?

$(5\times 1.25)+(-10m)=(-7.5\times 1.25)+(7.5m)$ $6.25-10m=7.5m-9.375$ $17.5m=15.625$ $m=0.893{\mbox{ kg}}$ 3. A totally elastic collision occurs between two balls of equal mass. One of the balls is stationary. What happens?

Since momentum must be conserved:

$mu=mv_{1}+mv_{2}$ $u=v_{1}+v_{2}$ (1)

Since kinetic energy must be conserved:

${\frac {mu^{2}}{2}}={\frac {m{v_{1}}^{2}}{2}}+{\frac {m{v_{2}}^{2}}{2}}$ $u^{2}={v_{1}}^{2}+{v_{2}}^{2}$ Substitute in the value of u from (1):

$(v_{1}+v_{2})^{2}={v_{1}}^{2}+{v_{2}}^{2}$ ${v_{1}}^{2}+2v_{1}v_{2}+{v_{2}}^{2}={v_{1}}^{2}+{v_{2}}^{2}$ $2v_{1}v_{2}=0$ Therefore, either v1 or v2 is 0. Using equation (1), if v1 is zero, then v2 = u, and vice versa. However, v1 cannot be the same as u, as this would mean that the first ball had to move through the second ball! So, the only physical solution is that the first ball stops, and the second ball continues moving with the first ball's original velocity.

4. A particle explodes to become two particles with masses 1 kg and 2 kg. The 1 kg particle moves with velocity 45ms−1. With what velocity does the other particle move?

$0=(1\times 45)+2v$ $2v=-45$ $v=-22.5{\mbox{ ms}}^{-1}$ i.e. in the opposite direction to the motion of the 1 kg particle.

5. A 3 kg ball moving at 3ms−1 collides with a 5 kg ball moving at -5ms−1. The collision is perfectly elastic. What are the new velocities of the balls?

Since momentum must be conserved:

$(3\times 3)+(5\times -5)=3v_{1}+5v_{2}=-16{\mbox{ kgms}}^{-1}$ (1)

Since kinetic energy must be conserved:

${\frac {3\times 3^{2}}{2}}+{\frac {5\times (-5)^{2}}{2}}={\frac {3\times {v_{1}}^{2}}{2}}+{\frac {5\times {v_{2}}^{2}}{2}}$ $3{v_{1}}^{2}+5{v_{2}}^{2}=152$ From (1):

$v_{2}={\frac {-16-3v_{1}}{5}}$ $3{v_{1}}^{2}+5\left({\frac {-16-3v_{1}}{5}}\right)^{2}=152$ $3{v_{1}}^{2}+5\left({\frac {9{v_{1}}^{2}+96v_{1}+256}{25}}\right)=152$ $3{v_{1}}^{2}+{\frac {9{v_{1}}^{2}+96v_{1}+256}{5}}=152$ $15{v_{1}}^{2}+9{v_{1}}^{2}+96v_{1}+256=760$ $24{v_{1}}^{2}+96v_{1}-504=0$ ${v_{1}}^{2}+4v_{1}-21=0$ $(v_{1}+7)(v_{1}-3)=0$ So, v1 is either -7 or 3 ms−1.

If v1 = -7ms−1, by (1):

$-21+5v_{2}=-16$ $5v_{2}=5$ $v_{2}=1{\mbox{ ms}}^{-1}$ If v1 = 3ms−1:

$9+5v_{2}=-16$ $5v_{2}=-25$ $v_{2}=-5{\mbox{ ms}}^{-1}$ This last solution is non-physical since it requires the balls to move through each other. So, v1 = -7ms−1 and v2 = 1ms−1

6. A ball collides with a wall, and rebounds at the same velocity. Why doesn't the wall move?

Let the mass of the wall be M, and the mass of the ball be m:

$mu=Mv_{wall}-mu$ $2mu=Mv_{wall}$ $v_{wall}={\frac {2mu}{M}}$ The mass of the wall is large. As M tends to infinity, therefore, vwall tends to 0.