Question

A 15gm bullet is fired horizontally into a 3kg block of wood suspended by a string.The bullet sticks in the block,and the impact vaused the block to swing 10cm above the initial level.the velocity of the bullet nearly is

Answer 1

Answer:

The answer will be 281.4 m/s

Explanation:

According to the problem the mass of the bullet is m1= 15/1000 gm and the mass of the block is m2=3 kg

let the velocity of the bullet is v1

When the bullet collides with the block the mechanical energy will be conserved.

Therefore we can say that

loss in kinetic energy = gain in potential energy

1/2 mv^2 = mgh

v = √2gh

  = √ 2 x 9.8 x 0.1

 = 1.4 m/s

Now this the speed of the block after the impact

Now from momentum conservation

m2v1+ 0 = (m1+m2) x v

=> v1 = (0.015+ 3) x 1.4/ 0.015

          = 281.4 m/s