Chapter 7, Problem 30P

a.

To determine

To show: Final velocities are

  $v_A{}^{\text{'}}=\left(\frac{m_A-m_B}{m_A+m_B}\right)v_A$

  $v_B{}^{\text{'}}=\left(\frac{2m_A}{m_A+m_B}\right)v_A$

Explanation of Solution

Given:

An object of mass m_A and velocity v_A elastically striking an stationary object ( v_B =0) mass m_B

Formula used:

According to energy conservation,

  $\frac{1}{2}{m}_A{v}_A{}^2+0=\frac{1}{2}{m}_A{v}_A{{}^{\text{'}}}^2+\frac{1}{2}{m}_B{v}_B{{}^{\text{'}}}^2$

  $m_A{v}_A+0=m_A{v}^{\text{'}}{}_A+m_B{v}^{\text{'}}{}_B$

Calculation:

By the law of conservation of energy

  $\frac{1}{2}{m}_A{v}_A{}^2+0=\frac{1}{2}{m}_A{v}_A{{}^{\text{'}}}^2+\frac{1}{2}{m}_B{v}_B{{}^{\text{'}}}^2$

  $m_A{v}_A{}^2+0=m_A{v}_A{{}^{\text{'}}}^2+m_B{v}_B{{}^{\text{'}}}^2$

  $m_A(v_A{}^2-v^{\text{'}}{{}_A}^2)=m_B{v}_B{{}^{\text{'}}}^2$

  $m_A(v_A-v^{\text{'}}{}_A)(v_A+v^{\text{'}}{}_A)=m_B{v}_B{{}^{\text{'}}}^2$

(1)

Now, applying law of conservation of linear momentum,

  $m_A{v}_A+0=m_A{v}^{\text{'}}{}_A+m_B{v}^{\text{'}}{}_B$

  $m_A(v_A-v^{\text{'}}{}_A)=m_B{v}^{\text{'}}{}_B$

(2)

Divide equation (1) by (2),

  $(v_A+v^{\text{'}}{}_A)=v^{\text{'}}{}_B$

(3)

Substituting the value of $v^{\text{'}}{}_B$ from equation (3) to equation (2)

  $m_A(v_A-v^{\text{'}}{}_A)=m_B(v_A+v^{\text{'}}{}_A)$

Rearranging this equation

  $v_A(m_A-m_B)=v^{\text{'}}{}_A(m_B+m_A)$

  $v^{\text{'}}{}_A=\left(\frac{m_A-m_B}{m_B+m_A}\right)v_A$

Substituting the value of $v^{\text{'}}{}_A=(v^{\text{'}}{}_B-v_A)$ from equation (3) to equation(2)

  $m_A(v_A-v_B+v_A)=m_B{v}_B{}^{\text{'}}$

  $2m_A{v}_A-m_A{v}_B=m_B{v}_B{}^{\text{'}}$

  $2m_A{v}_A=v_B{}^{\text{'}}(m_B+m_A)$

  $v_B{}^{\text{'}}=\frac{2m_A{v}_A}{m_B+m_A}$

b.

To determine

To explain: The effect on the final velocities for the given condition.

Also, state an example.

Answer to Problem 30P

  $v_A{}^{\text{'}}=-v_A$ and

  $v_B{}^{\text{'}}=0$ (almost zero)

Explanation of Solution

Given:

An object of mass m_A and velocity v_A elastically striking an stationary object ( v_B =0) mass m_B

Formula used:

According to energy conservation,

  $\frac{1}{2}{m}_A{v}_A{}^2+0=\frac{1}{2}{m}_A{v}_A{{}^{\text{'}}}^2+\frac{1}{2}{m}_B{v}_B{{}^{\text{'}}}^2$

  $m_A{v}_A+0=m_A{v}^{\text{'}}{}_A+m_B{v}^{\text{'}}{}_B$

Calculation:

From part (a),

  $v^{\text{'}}{}_A=\left(\frac{m_A-m_B}{m_B+m_A}\right)v_A$

  $v_B{}^{\text{'}}=\frac{2m_A{v}_A}{m_B+m_A}$

When m_A is much smaller than m_B , then m_A can be neglected. Substituting the values in final velocities expression,

  $\begin{array}{c}{v}^{\text{'}}{}_A=\left(\frac{m_A-m_B}{m_B+m_A}\right)v_A\\ =\left(\frac{0-m_B}{m_B+0}\right)v_A\\ =\left(-1\right)v_A\\ =-v_A\end{array}$

Now,

  $\begin{array}{c}{v}_B{}^{\text{'}}=\frac{2m_A{v}_A}{m_B+m_A}\\ =\frac{2m_A{v}_A}{m_B+m_A}\\ \approx 0\end{array}$

Example for such a case is a rubber ball thrown against a wall.

c.

To determine

To explain: The effect on the final velocities for the given condition along with an example.

Answer to Problem 30P

Object A will move with the same velocity and object B will move double the initial velocity of object A.

Explanation of Solution

Given:

An object of mass m_A and velocity v_A elastically striking an stationary object ( v_B =0) mass m_B .

Formula used:

According to energy conservation,

  $\frac{1}{2}{m}_A{v}_A{}^2+0=\frac{1}{2}{m}_A{v}_A{{}^{\text{'}}}^2+\frac{1}{2}{m}_B{v}_B{{}^{\text{'}}}^2$

  $m_A{v}_A+0=m_A{v}^{\text{'}}{}_A+m_B{v}^{\text{'}}{}_B$

Calculation:

From part (a),

  $v^{\text{'}}{}_A=\left(\frac{m_A-m_B}{m_B+m_A}\right)v_A$

  $v_B{}^{\text{'}}=\frac{2m_A{v}_A}{m_B+m_A}$

When m_A is much larger than m_B , then it can be neglected. Substituting the values,

  $\begin{array}{c}{v}^{\text{'}}{}_A=\left(\frac{m_A-m_B}{m_B+m_A}\right)v_A\\ =\left(\frac{m_A-0}{m_A+0}\right)v_A\\ =v_A\end{array}$

Now,

  $\begin{array}{c}{v}_B{}^{\text{'}}=\frac{2m_A{v}_A}{m_B+m_A}\\ =\frac{2m_A{v}_A}{m_A}\\ =2v_A\end{array}$

This means object A will move with the same velocity but in opposite direction while object B will move with double the initial velocity of an object A

Example for such a case is collision between a fast moving truck and a stationary drum.

d.

To determine

To explain: The effect on the final velocities for the given condition along with an example.

Answer to Problem 30P

This means object A will come to rest and object B will move with the same velocity as the initial velocity of object A.

Explanation of Solution

Given:

An object of mass m_A and velocity v_A elastically striking an stationary object ( v_B =0) mass m_B .

Formula used:

According to energy conservation,

  $\frac{1}{2}{m}_A{v}_A{}^2+0=\frac{1}{2}{m}_A{v}_A{{}^{\text{'}}}^2+\frac{1}{2}{m}_B{v}_B{{}^{\text{'}}}^2$

  $m_A{v}_A+0=m_A{v}^{\text{'}}{}_A+m_B{v}^{\text{'}}{}_B$

Calculation:

From part (a),

  $v^{\text{'}}{}_A=\left(\frac{m_A-m_B}{m_B+m_A}\right)v_A$

  $v_B{}^{\text{'}}=\frac{2m_A{v}_A}{m_B+m_A}$

When m_A is equal to m_B , substituting the values,

  $\begin{array}{c}{v}^{\text{'}}{}_A=\left(\frac{m_A-m_B}{m_B+m_A}\right)v_A\\ =\left(\frac{m_B-m_B}{m_B+m_B}\right)v_A\\ =0\end{array}$

Now,

  $\begin{array}{c}{v}_B{}^{\text{'}}=\frac{2m_A{v}_A}{m_B+m_A}\\ =\frac{2m_B{v}_A}{m_B+m_B}\\ =\frac{2v_A}{2}\\ =v_A\end{array}$

This means object A will come to rest and object B will move with the same velocity as the initial velocity of object A.

Example: When a ball at a billiard table hits another ball, the first ball stops, and second ball moves with the speed of the first ball.