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

Sparsh Jain , 6 Years ago
Grade 12th pass
anser 1 Answers
Arun

Last Activity: 6 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= m1 v1ð 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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