Chapter 7, Problem 2SP

(a)

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 $2.9\text{m}/\text{s}$.

Explanation of Solution

Given info: The initial velocity of the bullet is $700\text{m}/\text{s}$ and the mass of the bullet is $0.005\text{kg}$. The mass of the wooden block is $1.2\text{kg}$ and the initial velocity of the wooden block is $0\text{m}/\text{s}$.

Write the expression for conservation of momentum.

$p_i=p_f$ (1)

Here,

$p_i$ is the initial momentum

$p_f$ is the final momentum

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

$p_{i_b}=m_b{v}_b$

Here,

$p_{i_b}$ is the initial momentum of the bullet

$m_b$ is the mass of the bullet

$v_b$ is the velocity of the bullet

Substitute $0.005\text{kg}$ for $m_b$ and $700\text{m}/\text{s}$ for $v_b$ in the above equation to find the initial momentum of the bullet.

$\begin{array}{c}{p}_{i_b}=\left(0.005\text{kg}\right)\left(700\text{m}/\text{s}\right)\\ =3.5\text{kg}\cdot \text{m}/\text{s}\end{array}$

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,

$m_w$ is the mass of the wooden block

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

Substitute $0.005\text{kg}$ for $m_b$ and $1.2\text{kg}$ for $m_w$ in the above equation to find the final momentum of the bullet and the wooden block.

$p_f=\left(0.005\text{kg}+1.2\text{kg}\right)v_f$

Initial momentum of the bullet is the initial momentum itself.

Substitute $3.5\text{kg}\cdot \text{m}/\text{s}$ for $p_i$ and $\left(0.005\text{kg}+1.2\text{kg}\right)v_f$ for $p_f$ in equation (1) to find the final velocity of the bullet and the wooden block.

$\begin{array}{c}\left(0.005\text{kg}+1.2\text{kg}\right)v_f=3.5\text{kg}\cdot \text{m}/\text{s}\\ v_f=\frac{3.5\text{kg}\cdot \text{m}/\text{s}}{\left(0.005\text{kg}+1.2\text{kg}\right)}\\ =2.9\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the velocity of the block of wood and bullet after collision is $2.9\text{m}/\text{s}$.

(b)

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 $3.48\text{Ns}$.

Explanation of Solution

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

$p_{w_f}=m_w{v}_f$

Here,

$p_{w_f}$ is the final momentum of the wooden block

Substitute $1.2\text{kg}$ for $m_w$ and $2.9\text{m}/\text{s}$ for $v_f$ in the above equation to find $p_{w_f}$.

$\begin{array}{c}{p}_{w_f}=\left(1.2\text{kg}\right)\left(2.9\text{m}/\text{s}\right)\\ =3.48\text{kg}\cdot \text{m}/\text{s}\end{array}$

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

$\Delta p=p_{{\text{w}}_i}-p_{{\text{w}}_f}$

Substitute $3.48\text{kg}\cdot \text{m}/\text{s}$ for $p_{w_f}$ and $0\text{m}/\text{s}$ for $p_{{\text{w}}_i}$ in the above equation to find the change in momentum.

$\begin{array}{c}\Delta p=3.48\text{kg}\cdot \text{m}/\text{s}-\left(0\text{m}/\text{s}\right)\\ =3.48\text{kg}\cdot \text{m}/\text{s}\end{array}$

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

$\text{impulse}=\Delta p$

Substitute $3.48\text{Ns}$ for $\Delta p$ in the above equation.

$\text{impulse}=3.48\text{Ns}$

Conclusion:

Therefore the magnitude of the impulse that acts on the block of the wood is $3.48\text{Ns}$.

(c)

To determine

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

Answer to Problem 2SP

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

Explanation of Solution

Linear momentum is the product of mass of an object and its velocity. It was called as the change in quantity of the motion. It is a vector and it has the same direction as that of velocity.

Here the motion of the wooden block is due to the momentum imparted by the bullet. So that 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 so that it 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.