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This question is of chapter center of mass. This question belongs to Conservation of linear momentum.

This question is of chapter center of mass.
This question belongs to Conservation of linear momentum.
 

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Grade:12th pass

1 Answers

Arun
25750 Points
5 years ago
Dear student
 
Assume that v1 and v2 be the velocity of bullet after collision.As the block raise to vertical distance of 0.1 m then kinetic energy converted into potential energy.From the law of conservation of energygh2^2 = m2v2½ m =√2gh2ð v = √2*9.8*0.12ð v= 1.4 m/sec2ð vu1 = initial velocity of bulletBy the law of conservation of momentum2v2+ m1v1 = m1u1ð m2v2 - m1u1= mv1ð m1)/ m2v2 - m1u1 = ( m1ð v    =(0.01*500 – 2x1.4)/0.01 = (5-2.8)/0.01= 2.2/0.01= 220 m/sec
 
 
Regards
Arun (askIITians forum expert)

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