Question

A ball of mass m1 = 0.5 kg, moving along the x-axis with a speed of 5 m/s, has an elastic head-on collision with a target ball of mass m2 = 1 kg initially at rest. What is the velocity of each ball after the collision?

Answer 1

From the law of conservation of momentum we have:

$$m_1u_1=m_1v_1+m_2v_2.$$

Since collision is elastic, kinetic energy is conserved and we can write:

$$\dfrac{1}{2}m_1u_1^2=\dfrac{1}{2}m_1v_1^2+\dfrac{1}{2}m_2v_2^2.$$

This formula gives us an additional relationship between velocities. Therefore, with the help of these two formulas we can find the velocity of each ball after collision:

$$v_1=(\dfrac{m_1-m_2}{m_1+m_2})u_1,$$ $$v_1=(\dfrac{0.5\ kg-1\ kg}{0.5\ kg+1\ kg})\cdot5\ \dfrac{m}{s}=-1.67\ \dfrac{m}{s}.$$

The sign minus means that the first ball moves in the opposite direction (to the left) after the collision.

$$v_2=\dfrac{2m_1u_1}{m_1+m_2},$$ $$v_2=\dfrac{2\cdot0.5\ kg\cdot5\ \dfrac{m}{s}}{0.5\ kg+1\ kg}=3.33\ \dfrac{m}{s}.$$

The sign plus means that the second moves to the right after the collision.