# A bullet of mass 50g is shot at the block of a ballistic pendulum whose mass is 2 kg.However ,the block is too small ,and the bullet passes through the block.if the initial speed of the bullet is 425m//sand the block rises a vertical distance of 45 cm,how fast is the bullet goinfg when it leaves the block?

1
by kaushikravikant

2015-11-12T22:04:51+05:30

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Mass of block = 2kg
height reached = 45cm = 0.45m
final PE = mgh = 2*10*0.45 = 9 J
Initial KE = 1/2 mv² = 9
⇒ 1/2 * 2 * v² = 9
⇒ v = √9 = 3m/s

Before strike, speed of bullet = 425 m/s
speed of block = 0
momentum = 0.05*425 + 2*0 = 21.25 kgm/s

After strike,
speed ofbullet = V
speed of block =  3 m/s
momentum = 0.05*V + 2*3 = 0.05V + 6 kgm/s

Conserving momentum,
21.25 = 6 + 0.05V
⇒ 0.05V = 21.25 - 6  =15.25
⇒ V = 15.25/0.05 = 305 m/s

When the bullet leaves the block, its velocity was 305 m/s
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