Question

A ball of mass ${}^{\prime }{m}^{\prime }$ moving with a horizontal velocity ${}^{\prime }{v}^{\prime }$ strikes the bob of a pendulum at rest. Mass of the bob of the pendulum is also ${}^{\prime }{m}^{\prime }$. During this collision the ball sticks with the bob of the pendulum. The height to which the combined mass rises: (g=acceleration due to gravity)

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

Answer 1

The correct option is A $\frac{v^2}{8g}$

From conservation of linear momentum before and after the collision,

$mv=(m+m)v^{\prime }$

$⟹v^{\prime }=v/2$

All the initial kinetic energy gets converted to gravitational potential energy at the maximum height.

Thus, $\frac{1}{2}\left(2m\right)v^{\prime 2}=\left(2m\right)gh$

$⟹h=\frac{v^2}{8g}$