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

Answer:

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

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.

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

$0.012\times 250+0\times 0=(1+0.012)v$

$v=2.964m/sec$

From third equation of motion

$h=\frac{v^2}{2g}=\frac{2.964^2}{2\times .8}=0.45m$

So option (b) will be the correct answer.