Question

A cue ball of mass 0.16 kg, rolling at 4.0 m/s, hits a stationary ball (numbered 8) of the same mass. If the cue ball travels ${45}^{\circ }$ with respect to its original path, and the eight number ball also at ${45}^{\circ }$ with respect to the horizontal then what is the velocity of each ball after the collision if they travel with the same velocity.

• A. 3.2 m/s
• B. 7.9 m/s
• C. 5.6 m/s
• D. 2.8 m/s

Answer 1

The correct option is D

2.8 m/s

$P_{before}=\left(0.16kg\right)\left(4.0m/s\right)=0.64kgm/s$

$P_g=P_c$

$2\times 0.16\times cos{45}^{\circ }\times v=P_{final}$

Applying linear momentum conservation, we get

$0.64kgm/s=2\times 0.16\times cos{45}^{\circ }\times v$

$V=2.8m/s$