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

Answer:

The initial velocity of bullet is 7518.26 m/s.

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

$(m_{bu}+m_{bl})gh=\dfrac{1}{2}(m_{1}+m_{2})v^2$

$v=\sqrt{2\times g\times h}$

$v=\sqrt{2\times9.8\times2}$

$v=6.26\ m/s$

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

$v'=\dfrac{0.005+6}{0.005}\times6.26$

$v'=7518.26\ m/s$

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