Chapter 3, Problem 2C

A bullet with a mass of 0.01 kg is tired horizontally into a block of wood hanging on a string. The bullet sticks in the wood and causes it to swing upward to a height of 0.1 m. If the mass of the wood block is 2 kg, what was the initial speed of the bullet?

To determine

The initial speed of the bullet.

Answer to Problem 2C

The initial speed of the bullet is $281.4m/s$.

Explanation of Solution

Given:

Given that a bullet with a mass of $0.01kg$ is fired horizontally into a block of wood hanging on a string. The bullet sticks in the wood and causes it to swing upward to a height of $0.1m$ .Also, the mass of the wood block is $2kg$

Formula used:

According to the law of conservation of momentum, when the two objects collide in an isolated system, the total momentum before and after the collision is same.

Calculation:

Given that the bullet sticks in the wood and swing upward to a height of $0.1m$

Thus, kinetic energy of bullet is converted into the potential energy

Now, kinetic energy of combined mass just after the impact is calculated as

$KE=\frac{1}{2}\left(m_{Bullet}+m_{Wood}\right)v_{Bullet}{}^2$

We have $m_{Bullet}=0.01kg$ and $m_{Wood}=2kg$

Substituting the value in above formula, we get

$KE=\frac{1}{2}\left(0.01kg+2kg\right)v_{Bullet}{}^2$ ------ Eq(1)

Also, potential energy of wood and bullet together is calculated as

$PE=\left(m_{Bullet}+m_{Wood}\right)gh$

We have $h=0.1m$, $m_{Bullet}=0.01kg$, $m_{Wood}=2kg$ and $g=9.8m/s^2$

Substituting the value in above formula, we get

$PE=\left(0.01kg+2kg\right)\left(9.8m/s^2\right)\left(0.1m\right)$ ------- Eq(2)

Equating Eq(1) and Eq(2), we get

$\frac{1}{2}\left(0.01kg+2kg\right)v_{Bullet}{}^2=\left(0.01kg+2kg\right)\left(9.8m/s^2\right)\left(0.1m\right)$

$\frac{1}{2}\left(2.01kg\right)v_{Bullet}{}^2=\left(2.01kg\right)\left(9.8m/s^2\right)\left(0.1m\right)$

$v_{Bullet}{}^2=\frac{\left(2.01kg\right)\left(9.8m/s^2\right)\left(0.1m\right)\times 2}{\left(2.01kg\right)}$

$v_{Bullet}{}^2=1.96m^2/s^2$

$v_{Bullet}=1.4m/s$

The impact of bullet on wooden block is an inelastic collision, thus momentum is conserved.

Using the conservation of momentum, we have

$m_{Bullet}{v}_{Initial}+m_{Wood}{v}_{Wood}=\left(m_{Bullet}+m_{Wood}\right)v_{Final}$

We have $v_{Bullet}=v_{Final}=1.4m/s$, $m_{Bullet}=0.01kg$, $m_{Wood}=2kg$ and $v_{Wood}=0m/s$

Substituting the values, we have

$\left(0.01kg\right)v_{Initial}+\left(2kg\right)\left(0m/s\right)=\left(0.01kg+2kg\right)\left(1.4m/s\right)$

$\left(0.01kg\right)v_{Initial}=\left(2.01kg\right)\left(1.4m/s\right)$

$v_{Initial}=\frac{\left(2.01kg\right)\left(1.4m/s\right)}{\left(0.01kg\right)}$

$v_{Initial}=281.4m/s$.

Conclusion:

Hence, the initial speed of the bullet is $281.4m/s$.