Answer to Question #132941 in Physics for Iliya

Question

Object A travels in a straight line and has 100.0J of kinetic energy. It collides with a stationary object B of mass 3.0kg. After the collision assumed to be elastic, object A has 4.0J of kinetic energy. Determine the speed of object B after the collision.

Answer 1

In elastic collisions the kinetic energy conserves. This means than energies before and after the collision are equal:

$$K_A +K_B = K'_A + K'_B$$

where $K_A = 100J$ is energy of the object before the collision, $K_B = 0$ is energy of the object B before the collision (because it is stationary), $K_A' = 4J$ is the energy of the object A after the collision and $K_B'$ is the energy of the object B after the collision. Thus, obtain:

$$K_B' = K_A + K_B - K_A' = 100J + 0J - 4J = 96J$$

On the other hand, by definition, the kinetic energy is:

$$K_B' = \dfrac{m_Bv_B'^2}{2}$$

where $m_B = 3kg$ is the mass of the object B, $v_B'$ is the speed of the object B after the collision. Expressing $v_B'$, obtain:

$$v_B' = \sqrt{\dfrac{2K_B'}{m_B}}=\sqrt{\dfrac{2\cdot 96}{3}}= 8 m/s$$

Answer. 8 m/s.