Question

The three balls each weigh 0.5 lb and have a coefficient of restitution of e = 0.85. If ball A is released from rest and strikes ball B and then ball B strikes ball C, determine the velocity of each ball after the second collision has occurred. The balls slide without friction.

 

Answer 1

Concepts and reason

Collision:

The process in which two bodies of defined mass collide with each other and transfers energy with each other is known as a collision. Velocity:

Velocity is a vector quantity that is calculated as the rate of deviation in position with respect to time. It is denoted by \(V\) and its unit is \(\mathrm{m} / \mathrm{s}\) Acceleration:

Acceleration is a vector quantity that is calculated as the rate of deviation in velocity with respect to time. It is denoted by \(a\) and its unit is \(\mathrm{m} / \mathrm{s}^{2}\) Linear momentum:

The linear momentum of a particle can be calculated as the product of its mass and the velocity it travels. It is denoted by \(\vec{P}\) and unit is \(\mathrm{kg} \cdot \mathrm{m} / \mathrm{s}\)

Law of conservation of momentum:

When two objects collide in an isolated system, the total momentum of the two objects will be equal before and after the collision.

Coefficient of restitution:

The coefficient of restitution is the ratio of the relative speed after the collision to before the collision. It is denoted by the symbol e. Firstly, calculate the velocity of ball A before the collision. Then, calculate the velocity of ball A after collision using the equations of coefficient of restitution and conservation of momentum. Then, calculate the velocities of ball B and ball C after the collision. using the equations of coefficient of restitution and conservation of momentum.

Fundamentals

The expression for the velocity of the body falling from a height is given as follows:

\(V=\sqrt{2 g h}\)

Here, the height at which the body falls is \(h\), and acceleration due to gravity is \(g\). The expression for the conservation of linear momentum is given as follows:

\(\sum m v=\sum m v^{\prime}\)

Here, mass is \(m\), velocity is \(v\), velocity after impact is \(v^{\prime}\), the sum of products of masses and velocities before impact is \(\sum m v\)

and the sum of products of masses and velocities after impact is \(\sum m v^{\prime}\)

\( \)

The expression to calculate the coefficient of restitution (e) is given as follows:

\(e=\frac{v_{2}^{\prime}-v_{1}^{\prime}}{v_{1}-v_{2}}\)

Here, velocities of two objects before impact are \(v_{1}\) and \(v_{2}\) respectively and velocities of the objects after impact are \(v_{1}^{\prime}\) and \(v_{2}^{\prime}\) respectively.

Determine the initial velocity of ball \(\mathrm{A}\) before collision \(\left(V_{A_{1}}\right)\).

\(V_{A_{1}}=\sqrt{2 g h}\)

Substitute \(32.2 \mathrm{ft} / \mathrm{s}^{2}\) for \(g\) and \(3 \mathrm{ft}\) for \(\mathrm{h}\).

\(\begin{aligned} V_{A_{1}} &=\sqrt{2 \times 32.2 \mathrm{ft} / \mathrm{s}^{2} \times 3 \mathrm{ft}} \\ &=\sqrt{193.2} \\ &=13.9 \mathrm{ft} / \mathrm{s} \end{aligned}\)

Write the equation of coefficient of restitution \(\left(e_{1}\right)\) for a collision of ball \(\mathrm{A}\) and \(\mathrm{B}\).

\(e_{1}=\frac{V_{B_{2}}-V_{A_{2}}}{V_{A_{1}}-V_{B_{1}}}\)

Here, velocities of ball \(A\) and \(B\) before impact are \(V_{A_{1}}\) and \(V_{B_{1}}\) respectively and the velocities of the balls \(A\) and \(B\) after

impacts are \(V_{A_{2}}\) and \(V_{B_{2}}\) respectively.

Substitute \(0.85\) for \(e_{1}, 0\) for \(V_{B_{1}}\), and \(13.9 \mathrm{ft} / \mathrm{s}\) for \(V_{A_{1}}\) \(0.85=\frac{V_{B_{2}}-V_{A_{2}}}{13.9 \mathrm{ft} / \mathrm{s}-0}\)

\(V_{B_{2}}=V_{A_{2}}+11.815\)

Write the equation of conservation of momentum.

\(m_{A} V_{A_{1}}+m_{B} V_{B_{1}}=m_{A} V_{A_{2}}+m_{B} V_{B_{2}}\)

Here, the mass of the ball \(\mathrm{A}\) is \(m_{A}\), and mass of the ball \(\mathrm{B}\) is \(m_{B}\).

Substitute \(0.5 \mathrm{lb}\) for \(m_{A}, 0\) for \(V_{B_{1}}, 13.9 \mathrm{ft} / \mathrm{s}\) for \(V_{A_{1}}\) and \(0.5 \mathrm{lb}\) for \(m_{B}\)

\(0.5 \mathrm{lb} \times 13.9 \mathrm{ft} / \mathrm{s}=0.5 \mathrm{lb} \times V_{A_{2}}+0.5 \mathrm{lb} \times V_{B_{2}}\)

\(V_{A_{2}}+V_{B_{2}}=13.9\)...... (2)

Substitute Equation (1) in Equation (2). \(V_{A_{2}}+\left(V_{A_{2}}+11.815\right)=13.9\)

\(2 V_{A_{2}}=2.085\)

\(V_{A_{2}}=1.0425 \mathrm{ft} / \mathrm{s}\)

The velocity of ball A before the collision is calculated by using the values of acceleration due to gravity and height in the velocity equation. The velocity of ball A after the collision is calculated by using the equations of coefficient of restitution and conservation of momentum.

From Equation (1), calculate the velocity of ball B before the collision.

\(V_{B_{2}}=V_{A_{2}}+11.815\)

Substitute \(1.0425 \mathrm{ft} / \mathrm{s}\) for \(V_{A_{2}}\)

\(\begin{aligned} V_{B_{2}} &=(1.0425 \mathrm{ft} / \mathrm{s})+11.815 \\ &=12.8575 \mathrm{ft} / \mathrm{s} \end{aligned}\)

Write the equation of coefficient of restitution \(\left(e_{2}\right)\) for a collision of ball \(\mathrm{B}\) and \(\mathrm{C}\).

\(e_{2}=\frac{V_{C_{3}}-V_{B_{3}}}{V_{B_{2}}-V_{C_{2}}}\)

Here, velocities of ball \(\mathrm{B}\) and \(\mathrm{C}\) before impact are \(V_{B_{2}}\) and \(V_{C_{2}}\) respectively and the velocities of the balls \(\mathrm{B}\) and \(\mathrm{C}\) after

impacts are \(V_{B_{3}}\) and \(V_{c_{3}}\) respectively.

Substitute \(0.85\) for \(e_{2}, 0\) for \(V_{C_{2}}\), and \(12.8575 \mathrm{ft} / \mathrm{s}\) for \(V_{B_{2}}\). \(0.85=\frac{V_{C_{3}}-V_{B_{3}}}{12.8575 \mathrm{ft} / \mathrm{s}-0}\)

\(V_{C_{3}}=V_{B_{3}}+10.9289_{\ldots \ldots .(3)}\)

Write the equation of conservation of momentum.

\(m_{B} V_{B_{2}}+m_{C} V_{C_{2}}=m_{B} V_{B_{3}}+m_{C} V_{C_{3}}\)

Here, the mass of the ball \(\mathrm{C}\) is \(m_{C}\).

Substitute 0 for \(V_{C_{2}}, 0.5 \mathrm{lb}\) for \(m_{B}, 12.8575 \mathrm{ft} / \mathrm{s}\) for \(V_{B_{2}}\) and \(0.5 \mathrm{lb}\) for \(m_{C}\). \(0.5 \mathrm{lb} \times 12.8575 \mathrm{ft} / \mathrm{s}=0.5 \mathrm{lb} \times V_{B_{3}}+0.5 \mathrm{lb} \times V_{C_{3}}\)

\(V_{B_{3}}+V_{C_{3}}=12.8575 \ldots \ldots\) (4)

Substitute Equation (3) in Equation (4).

\(V_{B_{3}}+\left(V_{B_{3}}+10.9289\right)=12.8575\)

\(2 V_{B_{3}}=1.9286\)

\(V_{B_{3}}=0.9643 \mathrm{ft} / \mathrm{s}\)

The velocity of ball B before the second collision is calculated by using the value of the velocity of ball A after collision in Equation (1). The velocity of ball B after the collision is calculated by using the equations of coefficient of restitution and conservation of momentum.

From Equation (3), calculate the velocity of ball C after the collision.

\(V_{C_{3}}=V_{B_{3}}+10.9289\)

Substitute \(0.9643 \mathrm{ft} / \mathrm{s}\) for \(V_{B_{3}}\).

\(\begin{aligned} V_{C_{3}} &=(0.9643 \mathrm{ft} / \mathrm{s})+10.9289 \\ &=11.8932 \mathrm{ft} / \mathrm{s} \end{aligned}\)

The velocity of ball C after the collision is calculated by using the value of the velocity of ball B after collision in Equation (3).

The velocity of ball A after collision \(\left(V_{A_{2}}\right)\) is \(1.0425 \mathrm{ft} / \mathrm{s}\)

The velocity of ball B after collision \(\left(V_{B_{3}}\right)\) is \(0.9643 \mathrm{ft} / \mathrm{s}\)

The velocity of ball C after collision \(\left(V_{c_{3}}\right)\) is \(11.8932 \mathrm{ft} / \mathrm{s}\)