A 4.25g bullet traveling horizontally with a velocity ofmagnitude 375 m/s is fir

Question

A 4.25g bullet traveling horizontally with a velocity ofmagnitude 375 m/s is fired into a wooden block with mass1.10 kg, initially at rest on a level frictionless surface.The bullet passes through the block and emerges with itsspeed reduced to 120 m/s. a)How fast is the block moving just after the bullet emergesfrom it?                       v=              m/s A 4.25g bullet traveling horizontally with a velocity ofmagnitude 375 m/s is fired into a wooden block with mass1.10 kg, initially at rest on a level frictionless surface.The bullet passes through the block and emerges with itsspeed reduced to 120 m/s. a)How fast is the block moving just after the bullet emergesfrom it?                       v=              m/s

Explanation / Answer

Applying the law of conservation of momentum we have as, m1u1+m2u2=m1v1+m2v2 4.25*10-3*375+1.10*0=4.25*10-3*120+1.10*v From this we get the velocity of it. Hence we get by it.

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