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

in simple terms, the principle states that the buoyancy force on an object is equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g. this is also known as upthrust.

beta is electron from nucleus

something from nucleus has changed, but charge must be same for cons of charge