Question

A particle of mass m moving with velocity ${V_0}$ strikes a simple pendulum of mass m and sticks to it. The maximum height attained by the pendulum will be
(A) $\dfrac{{{v_0}^2}}{{8g}}$
(B) $\sqrt {{v_0}g}$
(C) $2\sqrt {\dfrac{{{v_0}}}{g}}$
(D) $\dfrac{{{v_0}^2}}{{4g}}$

Answer 1

We are looking for the ratio of kinetic energies before and after contraction. We can put the values in the formula of angular momentum.

Formula used: To calculate the momentum:
$p = mv$
Here, $p$ is the momentum of the body,
$m$ is the mass of the body
$v$ is the velocity of the body,

Complete step by step answer:
We are already given
Initial momentum of particle $= m{v_0}$​
Final momentum of system (particle + pendulum) $= 2mv$
By the law of conservation of momentum
So,
$m{v_0} = 2mv$
$\Rightarrow v = {v_0}/2$
Initial K.E. of the system = $\dfrac{1}{2}2m{v^2} = \dfrac{1}{2}2m{\left( {\dfrac{{{v_0}}}{2}} \right)^2}$
If the system rises up to height h then P.E.= 2mgh
By the law of conservation of energy
$\dfrac{1}{2}2m\left( {\dfrac{{{v_0}}}{2}} \right) = 2mgh$
On calculating further,
$h = \dfrac{{{v_0}^2}}{{2g}}$
So, we need to see from the above options, and select the correct value.

Thus, the correct answer is option A.

Additional Information: A property of a moving body that the body has by virtue of its mass and motion and that is equal to the product of the body's mass and velocity. In everyday life momentum is used many times. This is because the momentum of vehicles running at high speeds is very high and causes a lot of damage to the vehicles and injuries to passengers during the collision. A bullet, although small in mass, has a large momentum because of an extremely large velocity

Note: Some examples of change in kinetic energy from daily life are:
Any object in motion is using kinetic energy: a person walking, a thrown baseball, a crumb falling from a table, and a charged particle in an electric field are all examples of kinetic energy at work.