Question

A ball of mass m moving horizontally at a speed v collides with the bob of a simple pendulum at rest. The mass of the bob is also m. If the collision is perfectly inelastic, the ratio of the kinetic energy of the system immediately after the collision to that before the collision will be

• A. 1 : 1
• B. 1 : 2
• C. 1 : 3
• D. 1 : 4

Answer 1

The correct option is B 1 : 2

Mass of the ball and the bob sticking together is $m^{{}^{\prime }}=2m$.
KE after collision $=\frac{1}{2}{m}^{{}^{\prime }}{v}^{{}^{\prime }2}=\frac{1}{2}\times 2m\times {\left(\frac{v}{2}\right)}^2=\frac{1}{4}m{v}^2$.
KE before collision = $\frac{1}{2}m{v}^2$. Therefore, their ratio is 1:2. Hence the correct choice is (b).