Chapter 8, Problem 47SP

Two balls of equal mass, moving with speeds of 3 m/s, collide head-on. Find the speed of each after impact if (a) they stick together, (b) the collision is perfectly elastic, (c) the coefficient of restitution is 1/3.

To determine

The speed of two balls of equal masses after collision if they stick together, given that both the balls move with the speed of $3\text{ m/s}$ before collision.

Answer to Problem 47SP

Solution: $0\text{ m/s}$

Explanation of Solution

Given data:

Both the balls have equal masses.

The balls collide head on.

Both the balls before collision were moving at a speed of $3\text{ m/s}$.

Formula used:

The momentum of a body is expressed as,

$p=mv$

Here, $v$ is the velocity of the body, $m$ is the mass of the body, and $p$ is the momentum of the body.

Understand that during a collision between bodies A and B, when the bodies after the collision combine to form a single body, the momentum before and after the collision is expressed as,

$p_{1i}+p_{2i}=p_f$

Here, $p_{1i}$ is the momentum of body A before collision, $p_{2i}$ is the momentum of body B before collision, and $p_f$ is the momentum of the combined system after collision.

Explanation:

Understand that both the balls are moving with the same speed and they collide head on. Therefore, both must be moving toward each other with velocities of equal magnitude from opposite directions. Therefore, the velocity of one ball is negative of the other. Consider the ball moving with positive velocity as body A and the ball moving with negative velocity as body B.

Consider the expression for initial momentum of body A before collision in horizontal direction

$p_{1i}=m_1{v}_{1i}$

Here, $v_{1i}$ is the velocity of body A before collision and $m_1$ is the mass of body A.

Substitute $3\text{ m/s}$ for $v_{1i}$

$\begin{array}{c}{p}_{1i}=m_1\left(3\text{ m/s}\right)\\ =3m_1\end{array}$

Consider the expression for initial momentum of body B before collision

$p_{2i}=m_1{v}_{2i}$

Here, $v_{2i}$ is the horizontal speed or the magnitude of velocity of body B before collision and $m_2$ is the mass of body B.

Understand that body B has negative velocity. Therefore, substitute $-3\text{ m/s}$ for $v_{2i}$.

$\begin{array}{c}{p}_{2i}=m_1\left(-3\text{ m/s}\right)\\ =-3m_1\end{array}$

Consider the mass of the final combined body after the collision

$\begin{array}{c}{m}_3=m_1+m_1\\ =2m_1\end{array}$

Here, $m_3$ is the mass of the combined system formed by sticking of bodies A and B after collision.

Understand that both the balls stick together after the collision. Consider the expression for the momentum of the combined system after collision

$p_f=m_{3}v$

Here, $v$ is the final speed of the combined system.

Substitute $\text{2}{m}_1$ for $m_3$

$\begin{array}{c}{p}_f=\left(\text{2}{m}_1\right)v\\ =2m_{1}v\end{array}$

Consider the expression for conservation of momentum for the collision

$p_{1i}+p_{2i}=p_f$

Substitute $2m_{1}v$ for $p_f$, $-3m_1$ for $p_{2i}$, and $3m_1$ for $p_{1i}$

$\begin{array}{c}3m_1+\left(-3m_1\right)=2m_{1}v\\ 3m_1-3m_1=2m_{1}v\\ 0=2m_{1}v\\ v=0\text{m/s}\end{array}$

The final speed of the combined system is $0\text{ m/s}$. Understand that both the balls stick together as a unit after the collision.

Conclusion:

Therefore, the speed of both the balls after impact is $0\text{ m/s}$.

To determine

The speed of two balls of equal masses after the perfectly elastic collision, given that both the balls move with the speed of $3\text{ m/s}$ before collision.

Answer to Problem 47SP

Solution: Both rebound at $3\text{ m/s}$

Explanation of Solution

Given data:

Both the balls have equal masses.

The balls collide head on.

Both the balls before collision were moving at a speed of $3\text{ m/s}$.

Collision is elastic.

Formula used:

The momentum of a body is expressed as,

$p=mv$

Here, $v$ is the velocity of the body, $m$ is the mass of the body, and $p$ is the momentum of the body.

The expression for kinetic energy of the body is written as,

$KE=\frac{1}{2}m{v}^2$

Here, $KE$ is the kinetic energy of the body.

Understand that during a collision between bodies A and B, the conservation of momentum for the collision is expressed as,

$p_{1i}+p_{2i}=p_{1f}+p_{2f}$

Here, $p_{1i}$ is the momentum of body A before collision, $p_{2i}$ is the momentum of body B before collision, $p_{1f}$ is the momentum of body A after collision, and $p_{2f}$ is the momentum of body B after collision.

The expression for conservation of kinetic energy for the perfectlyelastic collision of bodies A and B is written as,

$K{E}_{1i}+K{E}_{2i}=K{E}_{1f}+K{E}_{2f}$

Here, $K{E}_{1i}$ is the kinetic energy of body A before collision, $K{E}_{2i}$ is the kinetic energy of body B before collision, $K{E}_{1f}$ is the kinetic energy of body A after collision, and $K{E}_{1f}$ is the kinetic energy of body B after collision.

Explanation:

Understand that both the balls are moving with the same speed and they collide head on. Therefore, both must be moving towards each other with velocities of equal magnitude from opposite directions. Therefore, the velocity of one ball is negative of the other. Consider the ball moving with positive velocity as body A and the ball moving with negative velocity as body B.

Consider the expression for final momentum of body A after collision

$p_{1f}=m_1{v}_{1f}$

Here, $v_{1f}$ is the velocity of body A after collision.

Consider the expression for the momentum of body B after collision

$p_{2f}=m_1{v}_{2f}$

Here, $v_{2f}$ is the final speed of body B after collision.

Consider the expression for conservation of momentum for the collision

$p_{1i}+p_{2i}=p_{1f}+p_{2f}$

Substitute $m_1{v}_{1f}$ for $p_{1f}$, $m_1{v}_{2f}$ for $p_{2f}$, $-3m_1$ for $p_{2i}$, and $3m_1$ for $p_{1i}$.

$\begin{array}{c}3m_1+\left(-3m_1\right)=m_1{v}_{1f}+m_1{v}_{2f}\\ 3m_1-3m_1=m_1\left(v_{1f}+v_{2f}\right)\\ \frac{\left(3-3\right)m_1}{m_1}=\left(v_{1f}+v_{2f}\right)\\ 0=\left(v_{1f}+v_{2f}\right)\end{array}$

Further, solve as,

$\begin{array}{c}{v}_{1f}+v_{2f}=0\\ v_{2f}=-v_{1f}\mathrm{...................................}(1)\end{array}$

Understand that the collision is elastic. Therefore, consider the expression for kinetic energy of body A before collision

$K{E}_{1i}=\frac{1}{2}{m}_1{v}_{1i}^2$

Substitute $3\text{ m/s}$ for $v_{1i}$

$\begin{array}{c}K{E}_{1i}=\frac{1}{2}{m}_1{\left(3\text{ m/s}\right)}^2\\ =\frac{1}{2}\left(9m_1\right)\\ =4.5m_1\end{array}$

Consider the expression for kinetic energy of body B before collision

$K{E}_{2i}=\frac{1}{2}{m}_1{v}_{2i}^2$

Substitute $-3\text{ m/s}$ for $v_{2i}$.

$\begin{array}{c}K{E}_{2i}=\frac{1}{2}{m}_1{\left(-3\text{ m/s}\right)}^2\\ =\frac{1}{2}\left(9m_1\right)\\ =4.5m_1\end{array}$

Consider the expression for kinetic energy of body A after collision

$K{E}_{1f}=\frac{1}{2}{m}_1{v}_{1f}^2$

Consider the expression for kinetic energy of body B after collision

$K{E}_{2f}=\frac{1}{2}{m}_1{v}_{2f}^2$

Using equation 1, substitute $-v_{1f}$ for $v_{2f}$.

$\begin{array}{c}K{E}_{2f}=\frac{1}{2}{m}_1{\left(-v_{1f}\right)}^2\\ =\frac{1}{2}{m}_1{v}_{1f}^2\end{array}$

Understand that the kinetic energy of the system is conserved during an elastic collision. Therefore, write the expression for conservation of kinetic energy for the elastic collision of bodies A and B.

$K{E}_{1i}+K{E}_{2i}=K{E}_{1f}+K{E}_{2f}$

Substitute $\frac{1}{2}{m}_1{v}_{1f}^2$ for $K{E}_{1f}$, $\frac{1}{2}{m}_1{v}_{1f}^2$ for $K{E}_{2f}$, $4.5m_1$ for $K{E}_{2i}$, and $4.5m_1$ for $K{E}_{1i}$.

$\begin{array}{c}4.5m_1+4.5m_1=\frac{1}{2}{m}_1{v}_{1f}^2+\frac{1}{2}{m}_1{v}_{1f}^2\\ 9m_1=m_1{v}_{1f}^2\\ v_{1f}^2=\frac{9m_1}{m_1}\\ v_{1f}^2=9\end{array}$

Further, solve as,

$\begin{array}{c}{v}_{f1}=\sqrt{9}\\ =\pm 3\text{ m/s}\end{array}$

Considering negative sign of the velocity since on a head-on collision, if two objects of equal masses collide with equal velocity then they will rebound. Hence,

$v_{f1}=-3\text{ m/s}$

Consider the expression for final velocity of body B after collision

$v_{2f}=-v_{1f}$

Substitute $-3\text{ m/s}$ for $v_{f1}$

$v_{2f}=3\text{ m/s}$

Conclusion:

Therefore, the speed of both body A and body B, that is the speed of both the balls after the collision impact, is $3\text{ m/s}$.And, both the balls rebound.

To determine

The speed of two balls of equal masses after collision if the coefficient of restitution for the collision is $\frac{1}{3}$, given that both the balls move with the speed of $3\text{ m/s}$ before collision.

Answer to Problem 47SP

Solution: Both balls rebounds at $1\text{ m/s}$.

Explanation of Solution

Given data:

Both the balls have equal masses.

The balls collide head on.

Both the balls before collision were moving at a speed of $3\text{ m/s}$.

Coefficient of restitution is $\frac{1}{3}$.

Formula used:

The momentum of a body is expressed as,

$p=mv$

Here, $v$ is the velocity of the body, $m$ is the mass of the body, and $p$ is the momentum of the body.

Understand that during a collision between bodies A and B, the conservation of momentum for the collision is expressed as,

$p_{1i}+p_{2i}=p_{1f}+p_{2f}$

Here, $p_{1i}$ is the momentum of body A before collision, $p_{2i}$ is the momentum of body B before collision, $p_{1f}$ is the momentum of body A after collision, and $p_{2f}$ is the momentum of body B after collision.

The coefficient of restitution for a collision of bodies A and B is expressed as,

$e=\frac{v_{2f}-v_{1f}}{v_{1i}-v_{2i}}$

Here, $e$ is the coefficient of restitution for the collision, $v_{2f}$ is the final velocity of body B after collision, $v_{1f}$ is the final velocity of body A after collision, $v_{1i}$ is the velocity of body A before collision, and $v_{2i}$ is the velocity of body B before collision.

Explanation:

Understand that both the balls are moving with the same speed and they collide head on. Therefore, both must be moving towards each other with velocities of equal magnitude from opposite directions. Therefore, velocity of one ball is negative of the other. Consider the ball moving with positive velocity as body A and the ball moving with negative velocity as body B.

Consider the ball with positive velocity as body A and the ball with negative velocity as body B.

Consider the expression for final momentum of body A after collision.

$p_{1f}=m_1{v}_{1f}$

Consider the expression for the momentum of body B after collision.

$p_{2f}=m_1{v}_{2f}$

Consider the expression for conservation of momentum for the collision.

$p_{1i}+p_{2i}=p_{1f}+p_{2f}$

Substitute $m_1{v}_{1f}$ for $p_{1f}$, $m_1{v}_{2f}$ for $p_{2f}$, $-3m_1$ for $p_{2i}$, and $3m_1$ for $p_{1i}$.

$\begin{array}{c}3m_1+\left(-3m_1\right)=m_1{v}_{1f}+m_1{v}_{2f}\\ 3m_1-3m_1=m_1\left(v_{1f}+v_{2f}\right)\\ \frac{\left(3-3\right)m_1}{m_1}=\left(v_{1f}+v_{2f}\right)\\ 0=\left(v_{1f}+v_{2f}\right)\end{array}$

Further, solve as,

$\begin{array}{c}{v}_{1f}+v_{2f}=0\\ v_{2f}=-v_{1f}\mathrm{....................................}(1)\end{array}$

Consider the expression for coefficient of restitution for the collision.

$\begin{array}{l}e=\frac{v_{2f}-v_{1f}}{v_{1i}-v_{2i}}\\ =\frac{v_{2f}+\left(-v_{1f}\right)}{v_{1i}-v_{2i}}\end{array}$

Substitute $\frac{1}{3}$ for $e$, $3\text{ m/s}$ for $v_{1i}$, $v_{2f}$ for $-v_{1f}$, and $-3\text{ m/s}$ for $v_{2i}$.

$\begin{array}{c}\frac{1}{3}=\frac{v_{2f}+v_{2f}}{3\text{ m/s}-\left(-3\text{ m/s}\right)}\\ \frac{1}{3}=\frac{2v_{2f}}{3+3}\\ 3\left(2v_{2f}\right)=3+3\\ 6v_{2f}=6\end{array}$

Further, solve as,

$\begin{array}{c}{v}_{2f}=\frac{6}{6}\\ =1\text{ m/s}\end{array}$

Consider the expression for final velocity of body A after collision

$v_{2f}=-v_{1f}$

Substitute $1\text{ m/s}$ for $v_{2f}$

$\begin{array}{l}1\text{ m/s}=-v_{1f}\\ v_{1f}=-1\text{ m/s}\end{array}$

The negative sign indicates that body A moves in opposite direction to body B.

Conclusion:

Therefore, the speed of both body A and body B, that is the speed of both the balls after the collision impact, is $1\text{ m/s}$. And, opposite signs of velocities show that both balls rebound.