A 100 g ball moving to the right at 4.0 m/s catches up and collides with a 400 g ball that is moving to the right at 1.0 m/s. If the collision is perfectly elastic, what are the speed and direction of each ball after the collision?

Mass of the first ball m₁= 100g = 0.1 kg

Initial Velocity of the first ball v_i1 = 4 m/s

Mass of the second ball m₂= 400g = 0.4 kg

Initial Velocity of the second ball v_i2 = 1 m/s

Collision is elastic therefore momentum and energy is conserved.

Using Law of conservation of Momentum,

$$\begin{gathered} m_1{v}_{i1}+m_2{v}_{i2}=m_1{v}_{f1}+m_2{v}_{f2} \\ (0.1kg)(4m/s)+(0.4kg)(1m/s)=(0.1kg)v_{f1}+(0.4kg)v_{f2} \\ 0.4kgm/s+0.4kgm/s=(0.1kg)v_{f1}+(0.4kg)v_{f2} \\ 0.8kgm/s=(0.1kg)v_{f1}+(0.4kg)v_{f2} \\ {v}_{f1}=8-4v_{f2} \end{gathered}$$

Using Law of conservation of Energy,

$$\begin{gathered} \frac{1}{2}{m}_1{v_{i1}}^2+\frac{1}{2}{m}_2{v_{i2}}^2=\frac{1}{2}{m}_1{v_{f1}}^2+\frac{1}{2}{m}_2{v_{f2}}^2 \\ {m}_1{v_{i1}}^2+m_2{v_{i2}}^2=m_1{v_{f1}}^2+m_2{v_{f2}}^2 \\ (0.1kg){(4m/s)}^2+(0.4kg){(1m/s)}^2=(0.1kg){v_{f1}}^2+(0.4kg){v_{f2}}^2 \\ (1.6+0.4)kgm/s=(0.1kg){v_{f1}}^2+(0.4kg){v_{f2}}^2 \\ 2kgm/s=(0.1kg){(8-4[v_{f2}\left]\right)}^2+(0.4kg){v_{f2}}^2 \\ 2kgm/s=6.4+1.6{v_{f2}}^2-6.4v_{f2}+0.4{v_{f2}}^2 \\ 2kgm/s=2{v_{f2}}^2-6.4v_{f2}+6.4 \\ 2{v_{f2}}^2-6.4v_{f2}+4.4=0 \\ {v_{f2}}^2-3.2v_{f2}+2.2=0 \\ Onsolving, \\ {v}_{f2}=2.2m/sor1m/s(1m/sisneglectedas{v}_{i2}=1m/s) \\ {v}_{f2}=2.2m/s \\ and{v}_{f1}=8-4v_{f2}=8-4(2.2)m/s=-0.8m/s \end{gathered}$$

Negative sign indicates that the ball moves in backward direction or Rebounds.