Question

A bullet of mass m1 = 5 g is moving with speed v1. It strikes a block of mass m2 = 6 kg that is hanging at rest from a cord. The bullet embeds itself into the block and the bullet and block swing up on the cord, coming to rest at a height h = 2 cm above the original position of the block. What is the correct expression for momentum conservation when the bullet strikes the block?

Answer 1

The initial velocity of bullet is 7518.26 m/s.

Step-by-step explanation:

Given that,

Mass of bullet = 5 g

Mass of block = 6 kg

Height = 2 cm

We need to calculate the velocity

Using conservation of energy

`(m_(bu)+m_(bl))gh=(1)/(2)(m_(1)+m_(2))v^2`

`v=√(2* g* h)`

`v=√(2*9.8*2)`

`v=6.26\ m/s`

We need to calculate the velocity of bullet

Using conservation of momentum

`m_(bu)v'=(m_(bu)+m_(bl))v`

Put the value into the formula

`v'=(0.005+6)/(0.005)*6.26`

`v'=7518.26\ m/s`

Hence, The initial velocity of bullet is 7518.26 m/s.