Question

A block a of mass ma = 1 kg is kept on a smooth horizontal surface and attached by a light thread to another block b of mass mb=2 kg .block b is resting on ground and thread and pulley are massless and frictionless.a bullet of mass m 0.25 kg moving horizontally with velocity of u=200m/s penetrates through the block and comes out with a velocity of 100m/s.find the height through which the block b will rise.

Answer 1

Answer:see it's easy first calculate velocity of the system:

200×0.25=3v+100×0.25

V=25/3

By energy conservation u get

1/2×3×25/3^2=2×10×h

h=5.21m

Here we have to take mass of the whole system and in PE mass of the 2kg block as 1kg block's PE will remain same. Energy will be conserved by system as whole