Question

A bullet of mass 30 gm is fired horizontally
with a velocity 100 ms 7 into a block of
mass 3 kg at rest. The bullet is embedded
into the block. The velocity with which the
block moving is:
30
38
ms-1
100
101
ms-1
muo
ms-7
ſms-1​

Answer 1

Solution :-

As per the given data ,

• Mass of the bullet(m) = 30 g = 0.03 kg
• Initial velocity of the bullet (u) = 100 m/s
• Final velocity of the bullet (v) = 0 m/s
• Mass of the block (M) = 3 kg
• Initially the block was at rest

As no external force is acting on the system ( block + bullet ) we can say that the momentum of the system is conserved .

As per law of conservation of momentum ,

Initial momentum of the bullet + Initial momentum of the block = Final momentum of the system

As finally , the bullet get's imbedded inside the block and both are moving with a common velocity  

➜ mu + MU = (m+ M )V

Now let's substitute the given values in the above equation ,

➜ 0.03 x 100 + 3 x 0 =( 0.03 + 3 )x V

➜ 0.03 x 100 = 3.03x V

➜ V = 3 / 3 .03

➜ V = 0.9  m/s

The block is moving with the velocity of 0.9 m/s