Answer to Question #112489 in Quantum Mechanics for Lizwi

Question

A mysterious crate has shown up at your place of work, Firecracker Company, and you are told to measure its inertia. It is too heavy to lift, but it rolls smoothly on casters. Getting an inspiration, you lightly tape a 0.60-kg iron block to the side of the crate, slide a firecracker between the crate and the block, and light the fuse. When the firecracker explodes, the block goes one way and the crate rolls the other way. You measure the crate's speed to be 0.052 m/s by timing how long it takes to cross floor tiles. You look up the specifications of the firecracker and find that it releases 9 J of energy. That's all you need, and you quickly calculate the inertia of the crate. What is that inertia?

Answer 1

According to the conservation of momentum, we can write

$mv-Mu=0,\\ M=m\frac{v}{u}=0.6\frac{v}{0.052}=11.5v.$

This is the mass of the crate in terms of the mass of the block.

Next, apply the conservation of energy:

$\frac{1}{2}mv+\frac{1}{2}Mu=9\text{ J},\\ \space\\ \frac{1}{2}\cdot0.6v^2+\frac{1}{2}M\cdot0.052^2=9\text{ J},\\$

substitute M that we obtained from momentum, conservation:

$\frac{1}{2}\cdot0.6v^2+\frac{1}{2}11.5v\cdot0.052^2=9\text{ J},\\ v=5.45\text{ m/s}$

(take the positive root).

Now substitute this velocity to the equation for M obtained from momentum conservation:

$M=11.5\cdot5.45=62.7\text{ kg}.$

The moment of inertia is

$I=MR^2,$

where R - the radius of the crate.