Jake is bowling with an 8 kg bowling ball roll the ball at 7 Ms and it hits 1 stationary pain with a mass of 2kg the pain goes flying forward at 12 Ms how fast is the ball now traveling

Solution

In this question we have given

mass of bowling ball=8kg

Velocity of bowling ball = 7m/s

mass of pain=2kg

velocity of pain=12m/s

In this case, law of conservation of momentum will be applied. According to law of conservation of momentum,

initial momentum of system = final momentum of the system

Total initial momentum of the system = mass of bowling ball * Velocity of bowling ball

=8 kg x 7 m/s

= 56 kg m/s.

momentum of the pain after getting hit is = mass of pain * velocity of pain

=2kg*12m/s

=24 kg m/s.

So the momentum of the bowling ball after hitting the pain = 56 kg m/s - 24 kg m/s

= 32 kg m/s

velocity of the bowling ball after hitting the pain = momentum of the bowling ball after hitting the pain/mass of bowling ball

=32 kg m/s / 8 kg

= 4 m/s

Therfore, velocity of the bowling ball after hitting the pain is 4m/s