Chapter 11, Problem 101A

a.

To determine

Change in momentum of the bullet if it is embedded in the block.

Answer to Problem 101A

  $\Delta P=-0.499\text{ kg-m/s}$

Explanation of Solution

Given:

Mass of bullet, $m_1=5\text{ g = 0}\text{.005 kg}$

Velocity of bullet before collision, $u_1=100\text{ m/s}$

Mass of solid block, $m_2=10\text{ kg}$

Formula used:

Momentum of any object is given by,

  $P=mv$

Here, P is momentum, m is mass and v is velocity

If no external force is applied on the object hen momentum is conserved.

  $P_i=P_f$

Calculation:

Initially solid block is at rest before the collision.

Therefore, velocity of solid block before the collision, $u_2=0\text{ m/s}$

According to conservation of momentum.

  $m_1{u}_1+m_2{u}_2=\left(m_1+m_2\right)v$

Here, v is common velocity of bullet and block after collision.

Now, substitute the value of $m_1,u_1,m_2,u_2$ and solve for v.

  $\begin{array}{l}0.005\times 100+10\times 0=\left(10+0.005\right)v\\ 10.005v=0.5\\ v=0.0499\text{ m/s}\end{array}$

Therefore, velocity of bullet after collision is 0.0499 m/s

Now, change in momentum of bullet is

  $\Delta P=m\left(v-u_1\right)$

Here, m is mass, $u_1$ is velocity of bullet before the collision and v is velocity of bullet after the collision.

Now, substitute the value of m, $u_1$ and v and solve.

  $\begin{array}{l}\Delta P=0.005\left(0.0499-100\right)\\ =0.005\times -99.95\\ =-0.499\text{ kg-m/s}\end{array}$

Here, negative sign indicates that change in momentum of bullet will be in opposite direction of its initial velocity.

Conclusion:

Therefore,momentum of bullet when it embedded in the block is $0.499\text{ kg-m/s}$ in opposite direction of its initial velocity.

b.

To determine

Change in momentum of the bullet if it is embedded in the block.

Answer to Problem 101A

  $\Delta P=-0.995\text{ kg-m/s}$

Explanation of Solution

Given:

Mass of bullet, $m_1=5\text{ g = 0}\text{.005 kg}$

Velocity of bullet before collision, $u_1=100\text{ m/s}$

Mass of solid block, $m_2=10\text{ kg}$

Velocity of bullet after collision is, $v_1=-99\text{ m/s}$

Formula used:

Momentum of any object is given by,

  $P=mv$

Here, P is momentum, m is mass and v is velocity

If no external force is applied on the object hen momentum is conserved.

  $P_i=P_f$

Calculation:

Now, change in momentum of bullet is

  $\Delta P=m\left(v_1-u_1\right)$

Here, m is mass, $u_1$ is velocity of bullet before the collision and v is velocity of bullet after the collision.

Now, substitute the value of m, $u_1$andv and solve.

  $\begin{array}{l}\Delta P=0.005\left(-99-100\right)\\ =0.005\times -199\\ =-0.995\text{ kg-m/s}\end{array}$

Here, the negative sign indicates that change momentum of bullet will be in opposite direction of its initial velocity.

Conclusion:

Therefore, momentum of bullet when it embedded in the block is $0.995\text{ kg-m/s}$ in opposite direction of its initial velocity.

c.

To determine

The case in which block has greater speed.

Answer to Problem 101A

Second case

Explanation of Solution

Given:

Mass of bullet, $m_1=5\text{ g = 0}\text{.005 kg}$

Velocity of bullet before collision, $u_1=100\text{ m/s}$

Mass of solid block, $m_2=10\text{ kg}$

Velocity of bullet after collision is, $v_1=-99\text{ m/s}$

Formula used:

Momentum of any object is given by

  $P=mv$

Here, P is momentum, m is mass and v is velocity

If no external force is applied on the object,then momentum is conserved.

  $P_i=P_f$

Calculation:

From part (a), velocity of block after collision is 0.0499 m/s.

Initially solid block is at rest before the collision.

Therefore, velocity of solid block before the collision, $u_2=0\text{ m/s}$

According to conservation of momentum.

  $m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$

Now, substitute the value of $m_1,u_1,m_2,u_2,v_1$ and solve for $v_2$ .

  $\begin{array}{l}0.005\times 100+10\times 0=0.005\times -99+10\times v_2\\ 10v_2-0.495=0.5\\ 10v_2=0.5+0.495\\ 10v_2=0.995\\ v_2=0.0995\text{ m/s}\end{array}$

Conclusion:

Therefore, velocity of block after collision will be greater in second case.