A ball of mass 3 kg moving with a velocity of 4 m/s undergoes a perfectly-elastic directcentral impact with a stationary ball of mass m. After the impact is over, the kinetic energy of the 3 kg ball is 6 J. The possible value(s) of m is/are

The correct option is 1 kg, 9 kg.

Let the mass of the first ball be $m_1=3\text{kg}$ and its initial velocity be $u_1=4\text{m/s}$.
Let the mass of the second ball be $m_2=m$ and its initial velocity be $u_2=0$ (stationary).
After the perfectly elastic direct central impact, let the final velocity of the 3 kg ball be $v_1$.

We are given that the final kinetic energy ($K_{\text{f}}$) of the 3 kg ball is 6 J.
Using the formula for kinetic energy:
$K_{\text{f}}=\frac{1}{2}{m}_1{v}_1^2$

Substituting the values:
$6=\frac{1}{2}\left(3\right)v_1^2$
$12=3v_1^2$
$v_1^2=4$
$v_1=\pm 2\text{m/s}$

In a perfectly elastic collision ($e=1$) between two bodies where the second body is initially stationary ($u_2=0$), the final velocity $v_1$ of the first body is given by the relation:
$v_1=\left(\frac{m_1-m_2}{m_1+m_2}\right)u_1$

Substituting the values $m_1=3$, $m_2=m$, and $u_1=4$:
$v_1=\left(\frac{3-m}{3+m}\right)\left(4\right)$

Since $v_1=\pm 2\text{m/s}$, we have two cases to consider:

Case 1: $v_1=2\text{m/s}$
$2=4\left(\frac{3-m}{3+m}\right)$
$1=2\left(\frac{3-m}{3+m}\right)$
$3+m=6-2m$
$3m=3$
$m=1\text{kg}$

Case 2: $v_1=-2\text{m/s}$
$-2=4\left(\frac{3-m}{3+m}\right)$
$-1=2\left(\frac{3-m}{3+m}\right)$
$-(3+m)=6-2m$
$-3-m=6-2m$
$m=9\text{kg}$

Therefore, the possible values of $m$ are 1 kg and 9 kg.