Question

A 12.0 g bullet was fired horizontally into a 1 kg block of wood. The bullet initially had a speed of 250 m/s. The block of wood was hanging from a 2 m long piece of (massless) string. After the collision the block/bullet combined object swings upward on the string. Find the height the block/bullet combined object rises.

a. 0.66m
b. 0.45m
c. 0.12 m
d. 0.35 m
e. 0.27m

Answer 1

So height though which combination of block bullet rises is 0.45 m

Step-by-step explanation:

We have given mass of the bullet
`m_1=12gram=0.012kg`

Velocity of the bullet
`v_1=250m/sec`

Mass of block of wood
`m_2=1kg`

Block is at rest so
`v_2=0m/sec`

From conservation of momentum.

`m_1v_1+m_2v_2=(m_1+m_2)v`

`0.012* 250+0* 0=(1+0.012)v`

`v=2.964m/sec`

From third equation of motion

`h=(v^2)/(2g)=(2.964^2)/(2* .8)=0.45m`

So option (b) will be the correct answer.