Question

A ball of mass 'm' moving with a horizontal velocity 'v' strikes the bob of a pendulum at rest. Mass of the bob of the pendulum is also 'm''. 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 )

Answer 1

it is the shortest method

Answer 2

Answer:

Height of the combined masses = $h\ =\ \dfrac{v^2}{8g}$

Explanation:

Given,

• Mass of the ball = m
• Horizontal initial speed of the ball = v
• Mass of the bob of the pendulum = m

Let h be the maximum height attained by the bob of the pendulum after the collision with the ball.

Now from the conservation of linear momentum, the total kinetic energy of the ball and bob after the collision is fully converted into the potential energy of the bob and ball (they stick together) because at the maximum height the kinetic energy of the bob and ball becomes zero due to the velocity at the top of the height is zero.

Let u be the final velocity of the combined body after the collision,

From the conservation of linear momentum,

$mu=v\ +\ 0\ =\ (m\ +\ m)u\\\Rigtharrow u\ =\ \dfrac{v}{2}$

Therefore Kinetic energy of the ball and bob after the collision = potential energy of the bob and ball after the collision,

$\therefore K.E.\ =\ P.E.\\\Rightarrow \dfrac{1}{2}(m\ +\ m)\dfrac{v^2}{4}\ =\ (m\ +\ m)gh\\\Rightarrow h\ =\ \dfrac{mv^2}{8mg}\\\Rightarrow h\ =\ \dfrac{v^2}{8g}$

Hence the ball and the bob attains a height of $\dfrac{v^2}{8g}$ after the collision.