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A bullet of mass 20g moving horizontally at a speed of 300m/s is fired into a wooden block of mass 500g suspended by a long string. the bullet crosses the block and emerges on the other side. if the centre of the mass of the block rises by 20cm, find the speed of the bullet as it emerges from the block.
(while conserving energy in the last step r we supposed to equate the potential energey possessed by the block and the K.E. imparted to it. bcoz i think both the energies should be equated with the KE. of the bullet but i didn't get the answer through that method. plz help by providing the sol. , if poss, and with suitable explanation)
The potential energy of the block is equal to the difference between the initial and final kinetic energies of the bullet..Solve and get the answer..
Mass of bullet = M = 20gm = 0.02kg.
Mass of wooden block M = 500gm = 0.5kg
Velocity of the bullet with which it strikes u = 300 m/sec.
Let the bullet emerges out with velocity V and the velocity of block = V'
As per law of conservation of momentum.
mu = Mv' +mu .........(1)
Again applying work – energy principle for the block after the collision,
0 – (1/2) M × v'2 =-
Mgh (where h = 0.2m)
v'2 = 2gh
V' = (2gh)1/2 = 2m/s
Substituting the value of V' in the equation (1), we get\
0.02 × 300 = 0.5 × 2 + 0.2 × v
v= 250m/sec.
approve if u like my ans
equate the kinetic energy of the bullet after emerging from the block and potential energy of the block at 20 cm to initial kinetic energy.........
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