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 re

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?

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Trúc Chi 6 months 2021-07-26T22:51:35+00:00 1 Answers 99 views 0

Answers ( )

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    2021-07-26T22:53:24+00:00

    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

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