Question

If a 3.5 gram ping pong ball were traveling to the right horizontally at 12 m/s, and a larger 12 g super ball were thrown directly behind it (also to the right) at 15 m/s so that the super ball bumped into and elastically collided with the ping pong ball, what would be the velocities of the two balls after the collision

Answer 1

Answer:

v = 14.32 m/s

Explanation:

According to the principle of conservation of linear momentum, both the momentum and kinetic energy of the system are conserved. Since the two balls are in the same direction of motion before collision, then;

$m_{1} u_{1}$ + $m_{2} u_{2}$ = ($m_{1}$ + $m_{2}$) v

0.035 × 12 + 0.120 × 15 = (0.035 + 0.120) v

0.420 + 1.800 = (0.155) v

2.22 = 0.155 v

⇒ v = $\frac{2.22}{0.155}$

      = 14.323

The velocity of the balls after collision is 14.32 m/s.