A 0.0140 kg bullet traveling at 205 m/s east hits a motionless 1.80 kg block and bounces off it, retracing its original path with a velocity of 103 m/s west. What is the final velocity of the block? Assume the block rests on a perfectly frictionless horizontal surface.

Final velocity of the block = 2.40 m/s east.

Explanation:

Here momentum is conserved.

Initial momentum = Final momentum

Mass of bullet = 0.0140 kg

Consider east as positive.

Initial velocity of bullet = 205 m/s

Mass of Block = 1.8 kg

Initial velocity of block = 0 m/s

Initial momentum = 0.014 x 205 + 1.8 x 0 = 2.87 kg m/s

Final velocity of bullet = - 103 m/s

We need to find final velocity of the block (u)

Final momentum = 0.014 x - 103 + 1.8 x u = - 1.442 + 1.8 u

We have

2.87 = - 1.442 + 1.8 u

u = 2.40 m/s

Final velocity of the block = 2.40 m/s east.