Question

A ball A, moving with kinetic energy E, makes a head on collision with a stationary ball, with mass n times that of A. The maximum potential energy stored in the system during the collision is?
(A) $ \dfrac{{nE}}{{(n + 1)}} $
(B) $ \dfrac{{(n + 1)E}}{n} $
(C) $ \dfrac{{(n - 1)E}}{n} $
(D) $ \dfrac{E}{n} $

Answer 1

Hint: This question is based on the concept of elastic collision.In this question firstly we will calculate an expression for velocity and then we will apply the law of conservation of linear momentum. At last, we will apply the concept of elastic collision and reach the conclusion.

Complete answer:
We know that kinetic energy $ E = \dfrac{1}{2}m{v^2} $
So we can rewrite the above equation as,
 $ 2E = m{v^2} $
 $ {v^2} = \dfrac{{2E}}{m} $
Now by taking square root on both the sides, we get,
 $ v = \sqrt {\dfrac{{2E}}{m}} $
According to the law of conservation of linear momentum,
Total momentum before collision = Total momentum after collision
So, $ mv = {m_1}{v_1} + {m_2}{v_2} $
We also know that the mass of the stationary ball is n times the mass of ball A
So, $ mv = m{v_1} + nm{v_2} $
On cancelling ‘m’ on both the sides, we get,
 $ v = {v_1} + n{v_2}........(1) $
We also know that,
 $ v = e({v_2} - {v_1}) $
Also, in elastic collision $ e = 1 $
So, $ v = {v_2} - {v_1}........(2) $
Adding (1) and (2),
 $ v + v = {v_1} + n{v_2} + {v_2} - {v_1} $
On simplifying the above equation, we get,
 $ 2v = n{v_2} + {v_2} $
 $ 2v = {v_2}(n + 1) $
Now, on taking (n+1) on the other side,
 $ {v_2} = \dfrac{{2v}}{{(n + 1)}} $
Now, due to deformation the second mass starts moving because its potential energy converts into kinetic energy
So, $ K{E_{max }} = \dfrac{1}{2} \times nm \times {\left( {\dfrac{{2v}}{{1 + n}}} \right)^2} $
 $ K{E_{max }} = \dfrac{1}{2} \times nm \times \dfrac{{4{v^2}}}{{{{(1 + n)}^2}}} $
On, further simplification, we get
 $ K{E_{max }} = \dfrac{{nE}}{{(n + 1)}} $
Hence, the correct answer is (A) $ \dfrac{{nE}}{{(n + 1)}} $ .

Note:
In elastic collision, there is no loss in the kinetic energy of the system when the collision takes place. Both kinetic energy and potential energy is conserved in the case of elastic collision. Thus, we can reach the conclusion that when two bodies make an elastic collision, there is no loss in speed.