#### Answer

Choice C.

#### Work Step by Step

The system’s momentum is zero before the collision because the identical balls are moving at the same speed and have oppositely directed velocities.
Small masses, system momentum before the collision is zero:
$$(m_{1})(v_{before})+(m_{2})(-v_{before})$$
The system’s momentum is still zero after the collision by conservation of momentum. The identical balls are still moving at the same speed as each other, with oppositely directed velocities.
Small masses, system momentum after the collision is zero:
$$(m_{1})(-v_{after})+(m_{2})(v_{after})$$
Now let the masses be doubled. A ball’s momentum and its kinetic energy are proportional to its mass. Doubling each ball’s mass means that each momentum term in the conservation equation is doubled. However, this factor of two can be canceled out in all the equations, bringing us back to the original equations of motion.
Large masses, system momentum before the collision is zero:
$$(2 m_{1})(v_{before})+(2 m_{2})(-v_{before})$$
$$(m_{1})(v_{before})+(m_{2})(-v_{before})$$
Large masses, system momentum after the collision is zero:
$$(2 m_{1})(-v_{after})+(2 m_{2})(v_{after})$$
$$(m_{1})(-v_{after})+(m_{2})(v_{after})$$
Doubling the masses doesn’t affect the final velocities.