Question

A ball with a mass of `4 kg` and velocity of `2 m/s` collides with a second ball with a mass of `2 kg` and velocity of `- 4 m/s`. If `20%` of the kinetic energy is lost, what are the final velocities of the balls?

Answer 1

`vecv_{1f}=+4/\sqrt{5}ms^{-1}\qquad vecv_{2f}=-8/\sqrt{5}ms^{-1}`

Explanation:

This is inelastic collision. So only the momentum is conserved, kinetic energy is NOT conserved.

Given Quantities & Unknowns:
`m_1=4kg; \quad vec v_(1i) = +2 ms^{-1}; \qquad vec v_{1f} = ?`
`m_2=2kg; \quad vec v_(2i) = -4 ms^{-1}; \qquad vec v_{2f} = ?`

Momentum Conservation : `vec P_f = vec P_i`
`m_1.vecv_{1f} + m_2vec v_{2f} = m_1vec v_{1i} + m_2vec v_{2i}`
`m_1.vec v_{1f} + m_2vec v_{2f} = [4kg\times2ms^{-1}][2kg\times(-4ms^{-1})] =0`
`vecv_{2f} = -(m_1/m_2)vecv_{1f} = - 2vecv_{1f}`

`v_{2f}^2 = vecv_{2f}.vecv_{2f} = +4v_{1f}^2` ...... (1)

Kinetic Energy Comparison: Given that there is a `20%` loss in kinetic energy, we conclude that the post collision kinetic energy is only `80%` of the kinetic energy before collision, `K_f = 0.8K_i`.

`1/2[m_1.v_{1f}^2 + m_2.v_{2f}^2] = 0.8\times1/2[m_1v_{1i}^2 + m_2.v_{2i}^2]` ...... (2)

`m_1.v_{1f}^2 + m_2.v_{2f}^2 = 0.8\times[m_1v_{1i}^2 + m_2.v_{2i}^2]=(8/10)\times48`

Eliminate `v_{2f}` in Eqn. (2) using Eqn. (1)

`m_1v_{1f}^2 + m_2(4v_{1f}^2)=(8/10)\times48`
`v_{1f}^2 = (8/10)(48)/(m_1+4m_2)=32/10; \qquad v_{1f}=4/\sqrt{5}ms^{-1};`
`vec v_{1f} = +4/\sqrt{5}ms^{-1}`
`vecv_{2f}=-2vecv_{1f}=-8/\sqrt{5}ms^{-1}`