Chapter 13.4, Problem 13.161P

Three steel spheres of equal mass are suspended from the ceiling by cords of equal length that are spaced at a distance slightly greater than the diameter of the spheres. After being pulled back and released, sphere A hits sphere B, which then hits sphere C. Denoting the coefficient of restitution between the spheres by e and the velocity of A just before it hits B by v₀, determine (a) the velocities of A and B immediately after the first collision, (b) the velocities of B and C immediately after the second collision. (c) Assuming now that n spheres are suspended from the ceiling and that the first sphere is pulled back and released as described here, determine the velocity of the last sphere after it is hit for the first time. (d) Use the result of part c to obtain the velocity of the last sphere when $n=8$ and $e=0.9$ .

    

To determine

(a)

The velocities of sphere A and B immediately after the first collision.

Answer to Problem 13.161P

$\begin{array}{l}v{\text{'}}_A=\frac{v_0(1-e)}{2}\\ v{\text{'}}_B=\frac{v_0(1+e)}{2}\end{array}$

Explanation of Solution

First collision between the sphere A and B :

The total momentum is conserved:

$\begin{array}{l}{\text{mv}}_{\text{A}}{\text{+ mv}}_{\text{B}}{\text{= mv'}}_{\text{A}}{\text{+ mv'}}_{\text{B}}\\ v_0+0={\text{v'}}_{\text{A}}{\text{+ v'}}_{\text{B}}\end{array}$

$v_0=v{\text{'}}_A+v{\text{'}}_B$ ------------- (1)

From the relation of coefficient of restitution of Relative velocities:

$\begin{array}{l}(v_A-v_B)e=(v{\text{'}}_B-v{\text{'}}_A)\\ \left(v_0-0\right)e=v{\text{'}}_B-v{\text{'}}_A\end{array}$

$v_{0}e=v{\text{'}}_B-v{\text{'}}_A$ ----------------(2)

Solving Equations (1) and (2) simultaneously,

$\begin{array}{l}{v}_{0}e=v_B^{\text{'}}-v_A^{\text{'}}\Rightarrow v_B^{\text{'}}=v_{0}e+v_A^{\text{'}}\\ v_0=v_A^{\text{'}}+v_B^{\text{'}}\\ \Rightarrow v_0=v_A^{\text{'}}+v_{0}e+v_A^{\text{'}}\\ \Rightarrow v_0=2v_A^{\text{'}}+v_{0}e\\ \Rightarrow v_0-v_{0}e=2v_A^{\text{'}}\\ \Rightarrow 2v_A^{\text{'}}=v_0-v_{0}e\\ v{\text{'}}_A=\frac{v_0(1-e)}{2}\end{array}$

And, from equation (2)

$\begin{array}{l}{v}_0=v_A^{\text{'}}+v_B^{\text{'}}\\ v_{0}e=v_B^{\text{'}}-v_A^{\text{'}}\\ \Rightarrow \left(v_A^{\text{'}}+v_B^{\text{'}}\right)e=v_B^{\text{'}}-v_A^{\text{'}}\\ \Rightarrow v_A^{\text{'}}e+v_B^{\text{'}}e=v_B^{\text{'}}-v_A^{\text{'}}\\ \Rightarrow v_B^{\text{'}}-v_B^{\text{'}}e=v_A^{\text{'}}e+v_A^{\text{'}}\\ \Rightarrow v_B^{\text{'}}\left(1-e\right)=v_A^{\text{'}}\left(1+e\right)\\ \Rightarrow v_B^{\text{'}}=v_A^{\text{'}}\times \frac{\left(1+e\right)}{\left(1-e\right)}\\ \Rightarrow v_B^{\text{'}}=\frac{v_0(1-e)}{2}\times \frac{\left(1+e\right)}{\left(1-e\right)}\\ v{\text{'}}_B=\frac{v_0(1+e)}{2}\end{array}$

Conclusion:

The velocities of sphere A and B immediately after collision is $v{\text{'}}_A=\frac{v_0(1-e)}{2}$ and $v{\text{'}}_B=\frac{v_0(1+e)}{2}$.

To determine

(b)

The velocities of B and C, after the second collision.

Answer to Problem 13.161P

$\begin{array}{l}v{\text{'}}_C=\frac{v_0{(1+e)}^2}{4}\\ v\text{'}{\text{'}}_B=\frac{v_0{(1-e)}^2}{4}\end{array}$

Explanation of Solution

Second Collision (between sphere B and C ):

The total momentum is conserved:

$mv{\text{'}}_B+m{v}_C=mv\text{'}{\text{'}}_B+mv{\text{'}}_C$

Using the result from part (a) for $v{\text{'}}_B$ ,

$\frac{v_0(1+e)}{2}+0=v\text{'}{\text{'}}_B+v{\text{'}}_C$ ------------------(3)

From the relation of coefficient of restitution of Relative velocities:

$(v{\text{'}}_B-0)e=v{\text{'}}_C-v\text{'}{\text{'}}_B$

Substituting again for $v{\text{'}}_B$ from (a),

$\frac{v_0(1+e)}{2}(e)=v{\text{'}}_C-v\text{'}{\text{'}}_B$ -------------(4)

Solving equations (3) and (4) simultaneously,

$\begin{array}{l}\frac{v_0(1+e)}{2}(e)=v{\text{'}}_C-v\text{'}{\text{'}}_B\\ \frac{v_{0}e(1+e)}{2}+v\text{'}{\text{'}}_B=v{\text{'}}_C\_\_\_\_\_\_\_\_(5)\end{array}$

Substitute the value of $v{\text{'}}_C$ in equation (3)

$\begin{array}{l}\frac{v_0(1+e)}{2}=v\text{'}{\text{'}}_B+v{\text{'}}_C\\ \frac{v_0(1+e)}{2}=v\text{'}{\text{'}}_B+\frac{v_{0}e(1+e)}{2}+v\text{'}{\text{'}}_B\\ \frac{v_0(1+e)}{2}=2v\text{'}{\text{'}}_B+\frac{v_{0}e(1+e)}{2}\\ \frac{v_0(1+e)}{2}=\frac{4v\text{'}{\text{'}}_B+v_{0}e(1+e)}{2}\\ v_0(1+e)=4v\text{'}{\text{'}}_B+v_{0}e(1+e)\\ v_0+v_{0}e=4v\text{'}{\text{'}}_B+v_{0}e+v_0{e}^2\\ v_0\left(1-e^2\right)=4v\text{'}{\text{'}}_B\\ v\text{'}{\text{'}}_B=\frac{v_0\left(1-e^2\right)}{4}\end{array}$

Again, put the value of $v\text{'}{\text{'}}_B$ in equation (5) and calculate $v{\text{'}}_C.$

$\begin{array}{l}\frac{v_{0}e(1+e)}{2}+v\text{'}{\text{'}}_B=v{\text{'}}_C\\ v{\text{'}}_C=\frac{v_{0}e(1+e)}{2}+\frac{v_0(1-e^2)}{4}\\ v{\text{'}}_C=\frac{2v_{0}e+2v_0{e}^2+v_0-v_0{e}^2}{4}\\ v{\text{'}}_C=\frac{2v_{0}e+v_0{e}^2+v_0}{4}\\ v{\text{'}}_C=\frac{v_0\left(2e+e^2+1\right)}{4}\\ v{\text{'}}_C=\frac{v_0{\left(1+e\right)}^2}{4}\end{array}$

Conclusion:

The velocities of sphere B and C, after the second collision is $v\text{'}{\text{'}}_B=\frac{v_0{(1-e)}^2}{4}andv{\text{'}}_C=\frac{v_0{(1+e)}^2}{4}.$

To determine

(c)

The velocity of the last sphere after it is hit for the first time if there is n number of spheres.

Answer to Problem 13.161P

$v{\text{'}}_n=\frac{v{\text{'}}_0{(1+e)}^{n-1}}{2^{n-1}}$

Explanation of Solution

Given:

There are $n$ number of spheres.

Calculation:

For n spheres there are (n) number of balls and $(n-1)th$ collisions.

Sphere C is the 3 number of sphere and we can take the velocity of sphere C from the above part that is part (b). Thus,

$v{\text{'}}_C=\frac{v{\text{'}}_0{(1+e)}^2}{4}$

Put n = 3 for $(n-1)th$ collision then

$\begin{array}{l}v{\text{'}}_n=v{\text{'}}_3=v{\text{'}}_C=\frac{v{\text{'}}_0{(1+e)}^2}{4}\\ \Rightarrow v{\text{'}}_3=\frac{v{\text{'}}_0{(1+e)}^{3-1}}{2^{3-1}}\end{array}$

Thus, for (n) number of balls

$\Rightarrow v{\text{'}}_n=\frac{v{\text{'}}_0{(1+e)}^{n-1}}{2^{n-1}}$

Conclusion:

The velocity of the (n) number of sphere after it is hit for the first time is $v{\text{'}}_n=\frac{v{\text{'}}_0{(1+e)}^{n-1}}{2^{n-1}}.$

To determine

(d)

The velocity of last sphere.

Answer to Problem 13.161P

$v{\text{'}}_n=0.815v_0$

Explanation of Solution

Given:

$n=5,e=0.9$

Calculation:

For $n=5,e=0.9$

From the answer of part (c) with n=5

$\begin{array}{l}v{\text{'}}_n=\frac{v{\text{'}}_0{(1+0.9)}^{5-1}}{2^{5-1}}\\ \Rightarrow v{\text{'}}_n=\frac{v{\text{'}}_0{(1.9)}^4}{2^4}\\ \Rightarrow v{\text{'}}_n=0.815v_0\end{array}$

Conclusion:

The velocity of last sphere is $v{\text{'}}_n=0.815v_0.$