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 us analyze the given physical situation step-by-step:
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 $m_2=m$, which is initially stationary, so its initial velocity is $u_2=0$.

Let the velocities of the balls after the perfectly elastic direct central impact be $v_1$ and $v_2$, respectively.

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

Substituting the given values:
$6=\frac{1}{2}\left(3\right)v_1^2$
$6=1.5v_1^2$
$v_1^2=4$
Taking the square root, we get two possible final velocities for the first ball:
$v_1=2\text{m/s}$ or $v_1=-2\text{m/s}$

For a perfectly elastic collision ($e=1$) of a moving mass with a stationary mass, the final velocity of the first mass is given by the standard formula derived from the conservation of momentum and conservation of kinetic energy:
$v_1=\left(\frac{m_1-m_2}{m_1+m_2}\right)u_1$

Substituting $m_1=3$, $m_2=m$, and $u_1=4$ into this formula:
$v_1=\left(\frac{3-m}{3+m}\right)\times 4$

Now, we analyze the two possible values of $v_1$:

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

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

Thus, the possible values for the mass $m$ are 1 kg and 9 kg.