Question

A bullet of mass 0.012 kg and horizontal speed $70m{s}^{-1}$ strikes a block of wood of mass 0.4 kg and instantly comes to rest with respect to the block. The block is suspended from the ceiling by means of thin wires. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block.

Answer 1

Mass of the bullet, m = 0.012 kg
Initial speed of the bullet, $u_b$ = 70 m/s
Mass of the wooden block, M = 0.4 kg
Initial speed of the wooden block, $u_B=0$
Final speed of the system of the bullet and the block = ν
Applying the law of conservation of momentum:
$m{u}_b+m{u}_B=(m+M)v0.012\times 70+0.4\times 0=(0.012+0.4)v\therefore v=\frac{0.84}{0.412}=2.04m/s$
For the system of the bullet and the wooden block:
Mass of the system, m' = 0.412 kg
Velocity of the system = 2.04 m/s

Height up to which the system rises = h
Applying the law of conservation of energy to this system:
Potential energy at the highest point = Kinetic energy at the lowest point
$m^{\prime }gh=\frac{1}{2}{m}^{\prime }{v}^2\therefore h=\frac{1}{2}\left(\frac{v^2}{g}\right)=\frac{1}{2}\times \frac{(2.04{)}^2}{9.8}=0.2123m$
The wooden block will rise to a height of 0.2123 m.
Heat produced = Kinetic energy of the bullet - Kinetic energy of the system
$=\frac{1}{2}m{u}^2-\frac{1}{2}{m}^{\prime }{v}^2=\frac{1}{2}\times 0.012\times (70{)}^2-\frac{1}{2}\times 0.412\times (2.04{)}^2=29.4-0.857=28.54J$