Question

A 4.25 g bullet traveling horizontally with a velocity of magnitude 375 m/s is fired into a wooden block with mass 1.10 kg , initially at rest on a level frictionless surface. The bullet passes through the block and emerges with its speed reduced to 122 m/s.
How fast is the block moving just after the bullet emerges from it?

Answer 1

Answer:

block is moving at 0.97m/s after the bullet emerges from it

Explanation:

The problem is based on elastic collision

Given data

Mass of bullet m1= 4.25g to kg we have 4.25/1000= 0.00425kg

Mass of block m2= 1.10kg

Initial velocity of bullet u1= 375m/s

Initial velocity of block u2= 0m/s

Final velocity of bullet v1= 122m/s

Final velocity of block v2=?

To solve for the final velocity of the block let us apply the principles of conservation of momentum for an elastic collision

m1u1+m2u2=m1v1+m2v2

Substituting our data into the

expression we have

(4.25*10^-3*375)+(1.1*0) =

(4.25*10^-3*122) + (1.1*v2)

1.59+0=0.52+1.1v2

1.59-0.52=1.1v2

1.07=1.1v2

v2=1.07/1.1

v2= 0.97m/s