$(i)$ $A$ light and a heavy object have the same momentum. What is the ratio of their kinetic energies? Which one has a larger kinetic energy?
$(ii)$ $A$ ball is dropped from a height of $10\, m$. If the energy of the ball reduces by $40\%$ after striking the ground,how high can the ball bounce back? $(g = 10\, m s^{-2})$

(A) $(i)$ The kinetic energy $(E)$ of a body is related to its momentum $(p)$ as $E = p^2 / 2m$. Since the momentum is the same for both objects,$E \propto 1/m$.
Therefore,the ratio of the kinetic energy of the light object to the heavy object is $\frac{E_{\text{light}}}{E_{\text{heavy}}} = \frac{m_{\text{heavy}}}{m_{\text{light}}}$.
Since $m_{\text{heavy}} > m_{\text{light}}$,the ratio is greater than $1$. Thus,the lighter object has a larger kinetic energy.
$(ii)$ Given: Initial height $h_1 = 10\, m$. The energy reduces by $40\%$,so the remaining energy is $60\%$ of the initial energy $(E_2 = 0.6 E_1)$.
Using the potential energy formula $E = mgh$,we have $mgh_2 = 0.6 mgh_1$.
$h_2 = 0.6 \times h_1 = 0.6 \times 10\, m = 6\, m$.
The ball will bounce back to a height of $6\, m$.