Chapter 15, Problem 94A

(a)

To determine

To sketch: The given system, showing the velocities and momenta before and after the collision.

Answer to Problem 94A

Explanation of Solution

Given:

Mass of ball A = $1.0\text{ kg}$

Mass of ball B = $2.0\text{ kg}$

Before collision:

Velocity of ball A ( ${\text{v}}_{\text{A}1}$ )= $\text{3}\text{.0 m/s}$ (moving west)

Velocity of ball B ( ${\text{v}}_{\text{B1}}$ ) = $0\text{ m/s}$ (at rest)

After collision:

Velocity of ball A ( ${\text{v}}_{\text{A2}}$ ) = $\text{2}\text{.0 m/s}$ (moves south)

Formula used:

Momentum of a body:

  $p=mv$

Where $p$ is the momentum of the body, $m$ is the mas and $v$ is the velocity of the body.

Calculation:

As per problem:

  $\begin{array}{l}{m}_A=1.0\text{ kg}\\ m_B=2.0\text{ kg}\end{array}$

Before collision

  $\begin{array}{l}{\text{v}}_{\text{A}1}\text{=3}\text{.0 m/s (moving west)}\\ {\text{v}}_{\text{B1}}=0\text{ m/s}\end{array}$

After collision

  ${\text{v}}_{\text{A2}}=\text{2}\text{.0 m/s (moving south)}$

The momentum of the balls is calculated by the formula

  $p=mv$

The momentum of the ball A

Before collision, ${\text{p}}_{\text{A1}}=3.0\text{ kg}\cdot \text{m/s}$ [ $p=(1.0\text{ kg)(3}\text{.0 m/s)}$ ]

After collision, ${\text{p}}_{\text{A2}}=2.0\text{ kg}\cdot \text{m/s}$ [ $p=(1.0\text{ kg)(2}\text{.0 m/s)}$ ]

The momentum of ball B

Before collision, ${\text{p}}_{\text{B1}}=0\text{ kg}\cdot \text{m/s}$ [ $p=(1.0\text{ kg)(0 m/s)}$ ]

Sketch the system with the given data

The ball B moves at an angle $\theta$ north-westwards after collision.

(b)

To determine

To calculate: The velocity and momentum of ball B.

Answer to Problem 94A

The momentum of ball B is $3.6\text{ kg}\cdot \text{m/s}$ at ${34}^o$ north-west and velocity is $1.8\text{ m/s}$ at ${34}^o$ north-west.

Explanation of Solution

Given:

Mass of ball A = $1.0\text{ kg}$

Mass of ball B = $2.0\text{ kg}$

Before collision:

Velocity of ball A ( ${\text{v}}_{\text{A}1}$ ) = $\text{3}\text{.0 m/s}$ (moving west)

Momentum of ball A ${\text{p}}_{\text{A1}}=3.0\text{ kg}\cdot \text{m/s}$

Velocity of ball B ( ${\text{v}}_{\text{B1}}$ ) = $0\text{ m/s}$ (at rest)

Momentum of ball B ${\text{p}}_{\text{B1}}=0\text{ kg}\cdot \text{m/s}$

After collision:

Velocity of ball A ( ${\text{v}}_{\text{A2}}$ ) = $\text{2}\text{.0 m/s}$ (moves south)

Momentum of ball A ${\text{p}}_{\text{A2}}=2.0\text{ kg}\cdot \text{m/s}$

Formula used:

Conservation of momentum:

  $m_{\text{A}}{v}_{\text{A}}=m_{\text{B}}{v}_{\text{B}}$

Where $m_{\text{A}}$ and $m_{\text{B}}$ are the masses of body A and B respectively, $v_{\text{A}}$ is the velocity of the body A and $v_{\text{B}}$ is the velocity of body B.

Calculation:

As per problem,

  $\begin{array}{l}{m}_{\text{A}}=1.0\text{ kg}\\ v_{\text{A1}}=3.0\text{ m/s}\\ v_{\text{A2}}=-2.0\text{ m/s [ the ball moves south]}\\ m_{\text{B}}=2.0\text{ kg}\end{array}$

To calculate horizontal momentum of ball B after collision rearrange the formula and substitute the values

  $\begin{array}{l}{m}_{\text{A}}{v}_{\text{A1}}=m_{\text{B}}{v}_{\text{B2}}\\ (1.0\text{ kg)(3}\text{.0 m/s)=}{m}_{\text{B}}{v}_{\text{B2}}\\ m_{\text{B}}{v}_{\text{B2}}=3.0\text{ kg}\cdot \text{m/s}\end{array}$

To calculate horizontal momentum of ball B after collision rearrange the formula and substitute the values

  $\begin{array}{l}{m}_{\text{A}}{v}_{\text{A2}}=m_{\text{B}}{v}_{\text{B2}}\\ (1.0\text{ kg)(-2}\text{.0 m/s)=}{m}_{\text{B}}{v}_{\text{B2}}\\ m_{\text{B}}{v}_{\text{B2}}=-2.0\text{ kg}\cdot \text{m/s}\end{array}$

The vector sum of momentum ball B

  $\begin{array}{l}{m}_{\text{B}}{v}_{\text{B2}}=\sqrt{{(3.0\text{ kg}\cdot \text{m/s)}}^2\text{ +(-2}\text{.0 kg}\cdot {\text{m/s)}}^2}\\ m_{\text{B}}{v}_{\text{B2}}=3.6\text{ kgm/s and tan}\theta \text{=}\frac{\text{2}\text{.0 kg}\cdot \text{m/s}}{3.0\text{ kg}\cdot \text{m/s}}\\ \text{ so }\theta \text{=}{34}^o\end{array}$

Therefore, momentum of ball B is $3.6\text{ kg}\cdot \text{m/s}$ at ${34}^o$ north-west.

The velocity of ball B

  $\begin{array}{l}{v}_{\text{B2}}=\frac{m_{\text{B}}{v}_{\text{B2}}}{m_{\text{B}}}\\ v_{\text{B2}}=\frac{3.6\text{ kg}\cdot \text{m/s}}{2.0\text{ kg}}\\ v_{\text{B2}}=1.8\text{ m/s}\end{array}$

Conclusion:

The momentum of ball B is $3.6\text{ kg}\cdot \text{m/s}$ at ${34}^o$ north-west and velocity is $1.8\text{ m/s}$ at ${34}^o$ north-west.