Question:

In the figure, pendulum bob on the left side is pulled aside to a height $ h $ from its initial position. After it is released, it collides with the right pendulum bob at rest, which is of the same mass. After the collision, the two bobs stick together and rise to a height

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For perfectly inelastic collisions where two objects stick together, the total kinetic energy after the collision is less than the total energy before the collision. The new height reached by the bobs can be calculated using the principle of conservation of mechanical energy.

Updated On: May 3, 2025

$ \frac{3h}{4} $
$ \frac{2h}{3} $
$ \frac{h}{2} $
$ \frac{h}{4} $

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The Correct Option is D

Approach Solution - 1

To solve the problem of the pendulum bob, we'll use the principles of conservation of energy and conservation of momentum.

Step 1: Calculate the velocity of the left bob before collision.

When the left pendulum bob is released from a height $ h $, it converts potential energy into kinetic energy at the lowest point (just before collision). The potential energy at height $ h $ is given by $ mgh $, where $ m $ is the mass and $ g $ is the acceleration due to gravity. At the lowest point, kinetic energy $ \frac{1}{2}mv^2 $ is equal to potential energy:

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

Simplifying, we get:

$ v^2 = 2gh $

$ v = \sqrt{2gh} $

Step 2: Use conservation of momentum for the collision.

The left bob (mass $ m $) collides with the right bob (of same mass $ m $) which is initially at rest. The initial momentum before collision is:

$ mv = m\sqrt{2gh} $

After the collision, they stick together, so the combined mass is $ 2m $ and let the velocity be $ V $. By conservation of momentum:

$ m\sqrt{2gh} = 2mV $

Solving for $ V $, we get:

$ V = \frac{\sqrt{2gh}}{2} $

Step 3: Calculate the height the two bobs rise to after collision.

After collision, the kinetic energy of the combined mass is converted to potential energy at the highest point. Thus, we have:

$ \frac{1}{2}(2m)V^2 = 2mgh' $, where $ h' $ is the maximum height reached after collision.

Substituting for $ V $:

$ \frac{1}{2}\times 2m \left(\frac{\sqrt{2gh}}{2}\right)^2 = 2mg h' $

$ \frac{1}{2}\times 2m \times \frac{2gh}{4} = 2mg h' $

$ \frac{mgh}{2} = 2mg h' $

$ h' = \frac{h}{4} $

Thus, the two bobs rise to a height of $ \frac{h}{4} $ after collision, which matches the correct option.

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Approach Solution -2

To solve this problem, we apply the principles of conservation of energy and conservation of momentum in a perfectly inelastic collision.

Step 1: Calculate the initial velocity of the first pendulum bob.

Initially, the pendulum is at height $ h $, so all the energy is gravitational potential energy. When released from height $ h $, this energy is converted to kinetic energy just before the collision.

Potential energy at height $ h $: $ U_i = mgh $

Kinetic energy just before collision: $ K = \frac{1}{2}mv^2 $

By conservation of energy, $ U_i = K $:

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

Solving for $ v $, the velocity, gives:

$ v = \sqrt{2gh} $

Step 2: Use conservation of momentum during collision.

Before the collision, only the moving bob has momentum:

$ p_{initial} = mv $

After collision, the two bobs stick together and have a combined mass of $ 2m $ and move with velocity $ v_f $:

$ p_{final} = 2mv_f $

By conservation of momentum:

$ mv = 2mv_f $

Simplifying, we find:

$ v_f = \frac{v}{2} = \frac{\sqrt{2gh}}{2} $

Step 3: Find the maximum height after the collision.

After the collision, all kinetic energy is converted into potential energy as the combined bob system rises to a new height $ h_f $.

Kinetic energy just after collision:

$ K_f = \frac{1}{2}(2m)v_f^2 = mv_f^2 $

Potential energy at maximum height $ h_f $:

$ U_f = 2mgh_f $

By conservation of energy:

$ mv_f^2 = 2mgh_f $

Simplifying gives:

$ v_f^2 = 2gh_f $

Substitute $ v_f = \frac{\sqrt{2gh}}{2} $ into the equation:

$ \left(\frac{\sqrt{2gh}}{2}\right)^2 = 2gh_f $

$ \frac{2gh}{4} = 2gh_f $

$ \frac{gh}{2} = 2gh_f $

$ h_f = \frac{h}{4} $

Thus, the final height to which the bobs rise is $ \frac{h}{4} $