Question

Two identical uniform solid spherical balls A and B of mass m each are placed on a fixed wedge as shown in figure. Ball B is kept at rest and it is released just before two balls collide. Ball A rolls down without slipping on an inclined plane and collides elastically with ball B. The kinetic energy of ball A just after the collision with ball B is:

& \text{A) }\dfrac{mgh}{7} \\\ & \text{B) }\dfrac{mgh}{2} \\\ & \text{C) }\dfrac{2mgh}{5} \\\ & \text{D) }\dfrac{7mgh}{5} \\\ \end{aligned}$$

Answer 1

We are to use the conservation laws of linear momentum and kinetic energy to find the solution to the problem. We can equate the kinetic energies of the two balls before and after collision to find the velocity of each ball, and thereby find the kinetic energy of ball A.

Complete answer:
In the situation given here, the ball is undergoing a rotational motion along with the translation motion, so we have to consider the moment of inertia element of the ball A before the collision to get the correct answer. We know that the moment of inertia of a sphere rotating along the axis that passes through the center of the sphere is given by –
$I=\dfrac{2}{5}m{{r}^{2}}$

The kinetic energy of the ball related to the rotational energy is given by –
$K{{E}_{rot}}=\dfrac{1}{2}I{{(\dfrac{{{v}_{cm}}}{r})}^{2}}=\dfrac{m{{v}_{cm}}^{2}}{5}$

The ball loses its potential as it moves down the height, which is converted to the kinetic energy. So the kinetic energy possessed by the ball A while it collides the ball B at the height $\dfrac{h}{2}$ is given by –
Potential energy lost by the ball A is equal to the kinetic energy gained buy the ball A.
i.e.,

& P{{E}_{h/2}}=K{{E}_{rot}}+K{{E}_{tran}} \\\ & mg\dfrac{h}{2}=\dfrac{m{{v}_{cm}}^{2}}{5}+\dfrac{1}{2}m{{v}_{cm}}^{2} \\\ & \dfrac{mgh}{2}=\dfrac{2m{{v}_{cm}}^{2}+5m{{v}_{cm}}^{2}}{10} \\\ & \Rightarrow \dfrac{5}{7}mgh=m{{v}_{cm}}^{2} \\\ & \therefore \dfrac{1}{5}m{{v}_{cm}}^{2}=\dfrac{mgh}{7} \\\ \end{aligned}$$ But we know that the ball loses only its translational kinetic energy when hitting the ball B. i.e., the translational energy is transferred to the ball completely by the conservation of kinetic energy. The rotational kinetic energy of the ball remains the same, i.e., $$K{{E}_{rot}}=\dfrac{mgh}{7}$$ **So, the correct answer is “Option A”.** **Note:** We can say that the total translational energy of ball A is transferred to ball B because both of them are of same masses and the ball B is initially at rest. We have to consider the masses of the balls if they had different masses which will be a different case.