Question

A ball moving horizontally with speed v strikes the bob of a simple pendulum at rest. The mass of the bob is equal to that of the ball. If the collision is elastic the bob will rise to a height

• A. $\frac{v^{2}}{g}$
• B. $\frac{v^{2}}{2g}$
• C. $\frac{v^{2}}{4g}$
• D. $\frac{v^{2}}{8g}$

Answer 1

Answer: (B)

$\frac{v^{2}}{2g}$

Answer 2

Total kinetic energy of the ball will transfer to the bob of simple pendulum. Let it rises to height ‘h’ by the law of conservation of energy.

$\frac{1}{2}mv^{2} = mgh$

∴ $h = \frac{v^{2}}{2g}$