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

The correct option is v²/2g.

Let us understand why this is the correct answer step-by-step.

Step 1: Analyzing the Collision
Let the mass of the moving ball be $m$ and its initial velocity be $v$ (moving horizontally).
The bob of the simple pendulum is initially at rest, so its initial velocity is $0$.
The mass of the bob is also given to be equal to the mass of the ball, which is $m$.
Since the collision is perfectly elastic and occurs between two bodies of equal mass, they exchange their velocities after the collision. Therefore, immediately after the collision:
1. The incoming ball comes to rest (velocity = $0$).
2. The bob of the pendulum acquires the horizontal velocity of the ball, which is $v$.

Step 2: Applying the Conservation of Mechanical Energy
As the bob swings upwards, its kinetic energy is converted into gravitational potential energy. Let $h$ be the maximum height to which the bob rises.
According to the law of conservation of mechanical energy:
Kinetic Energy at the lowest point = Potential Energy at the highest point

$\frac{1}{2}m{v}^2=mgh$

where:
- $m$ is the mass of the bob,
- $v$ is the velocity acquired by the bob,
- $g$ is the acceleration due to gravity,
- $h$ is the maximum height reached by the bob.

Step 3: Solving for Height (h)
Dividing both sides of the equation by mass ($m$), we get:

$\frac{1}{2}{v}^2=gh$

Rearranging the equation to solve for $h$:

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

Thus, the bob will rise to a height of v²/2g.