Chapter 7, Problem 2SP

A bullet is fired into a block of wood sitting on a block of ice. The bullet has an initial velocity of 800 m/s and a mass of 0.007 kg. The wooden block has a mass of 1.3 kg and is initially at rest. The bullet remains embedded in the block of wood afterward.

1. a. Assuming that momentum is conserved, find the velocity of the block of wood and bullet after the collision.
2. b. What is the magnitude of the impulse that acts on the block of wood in this process?
3. c. Does the change in momentum of the bullet equal that of the block of wood? Explain.

To determine

The velocity of the block of wood and bullet after collision.

Answer to Problem 2SP

The velocity of the block of wood and bullet after collision is $4.28\text{m}/\text{s}$.

Explanation of Solution

Given info: The initial velocity of the bullet is $800\text{m}/\text{s}$ and the mass of the bullet is $0.007\text{kg}$. The mass of the wooden block is $1.3\text{kg}$ and the initial velocity of the wooden block is $0\text{m}/\text{s}$.

Write the expression for conservation of momentum.

$p_i=p_f$

Here,

$p_i$ is the initial momentum

$p_f$ is the final momentum

Write the expression to find the initial momentum of the bullet.

$p_i=m_b{v}_b$

Here,

$p_i$ is the initial momentum of the bullet

$m_b$ is the mass of the bullet

$v_b$ is the velocity of the bullet

Substitute $0.007\text{kg}$ for $m_b$ and $800\text{m}/\text{s}$ for $v_b$ in the above equation to find the initial momentum of the bullet.

$\begin{array}{c}{p}_i=\left(0.007\text{kg}\right)\left(800\text{m}/\text{s}\right)\\ =5.6\text{kg}\cdot \text{m}/\text{s}\end{array}$

Write the expression to find the final momentum of the bullet and the wooden block.

$p_f=\left(m_b+m_w\right)v_f$

Here,

$p_i$ is the final momentum of the bullet and the wooden block

$m_w$ is the mass of the wooden block

$v_f$ is the final velocity of the bullet and the wooden block

Substitute $0.007\text{kg}$ for $m_b$ and $1.3\text{kg}$ for $m_w$ in the above equation to find the final momentum of the bullet and the wooden block.

$p_f=\left(0.007\text{kg}+1.3\text{kg}\right)v_f$

Write the expression for conservation of momentum.

$p_i=p_f$

Substitute $5.6\text{kg}\cdot \text{m}/\text{s}$ for $p_i$ and $\left(0.007\text{kg}+1.3\text{kg}\right)v_f$ in the above equation to find the final velocity of the bullet and the wooden block.

$\begin{array}{c}\left(0.007\text{kg}+1.3\text{kg}\right)v_f=5.6\text{kg}\cdot \text{m}/\text{s}\\ v_f=\frac{5.6\text{kg}\cdot \text{m}/\text{s}}{\left(0.007\text{kg}+1.3\text{kg}\right)}\\ =4.28\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the velocity of the block of wood and bullet after collision is $4.28\text{m}/\text{s}$.

To determine

The magnitude of the impulse that acts on the block of the wood.

Answer to Problem 2SP

The magnitude of the impulse that acts on the block of the wood is $5.6\text{Ns}$.

Explanation of Solution

Write the expression for the final momentum of the wooden block.

$p_w=m_w{v}_f$

Substitute $1.3\text{kg}$ for $m_w$ and $4.28\text{m}/\text{s}$ for $v_f$ in the above equation.

$\begin{array}{c}{p}_w=\left(1.3\text{kg}\right)\left(4.28\text{m}/\text{s}\right)\\ =5.6\text{kg}\cdot \text{m}/\text{s}\end{array}$

Write the expression for change in momentum of the wooden block.

$\Delta p=p_{\text{wi}}-p_{\text{wf}}$

Substitute $5.6\text{kg}\cdot \text{m}/\text{s}$ for $p_{\text{wf}}$ and $0\text{m}/\text{s}$ for $p_{\text{wi}}$ to find the change in momentum.

$\begin{array}{c}\Delta p=5.6\text{kg}\cdot \text{m}/\text{s}-\left(0\text{m}/\text{s}\right)\\ =5.6\text{kg}\cdot \text{m}/\text{s}\end{array}$

Write the expression of the impulse associated with the change in momentum.

$\text{impulse}=\Delta p$

Here,

$\Delta p$ is the change in momentum

Substitute $5.6\text{Ns}$ for $\Delta p$ in the above equation.

$\text{impulse}=5.6\text{Ns}$

Conclusion:

Therefore, the magnitude of the impulse that acts on the block of the wood is $5.6\text{Ns}$.

To determine

Whether the change in momentum of the bullet is equal to that of the block of the wood.

Answer to Problem 2SP

Yes, the magnitude of change in momentum of the bullet is equal to that of the block of the wood.

Explanation of Solution

Since the motion of the wooden block is due to the momentum imparted by the bullet, the magnitude of the change in momentum of the bullet is same as the change in momentum associated with the block of wood. The direction is opposite which facilitates the momentum conservation before and after the collision.

Conclusion:

Therefore, the magnitude of change in momentum of the bullet is equal to that of the block of the wood.