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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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Physics
6 months
2021-07-26T22:51:35+00:00
2021-07-26T22:51:35+00:00 1 Answers
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Answers ( )
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