Chapter 8, Problem 48SP

A 90-g ball moving at 100 cm/s collides head-on with a stationary 10-g ball. Determine the speed of each after impact if (a) they stick together, (b) the collision is perfectly elastic, (c) the coefficient of restitution is 0.90.

To determine

The speed of the two balls of masses $10\text{ g}$ and $90\text{ g}$ after collision, if they stick together, given that the $90\text{ g}$ ball strikes the $10\text{ g}$ stationary ball with a speed of $100\text{ cm/s}$.

Answer to Problem 48SP

Solution: $90\text{ cm/s}$

Explanation of Solution

Given data:

The mass of the ball moving with $100\text{ cm/s}$ is $90\text{ g}$.

The mass of the stationary ball is $10\text{ g}$.

Both the balls stick together after the collision.

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

Explanation:

Consider the ball moving with a velocity of $100\text{ cm/s}$ as Body A and the stationary ball as Body B.

Consider the expression for initial momentum of Body A before the collision

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

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

Substitute $100\text{ cm/s}$ for $v_{1i}$ and $90\text{ g}$ for $m_1$

$\begin{array}{c}{p}_{1i}=\left(90\text{ g}\right)\left(100\text{ cm/s}\right)\\ =\left(90\text{ g}\right)\left(100\text{ cm/s}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{ cm}}\right)\left(\frac{{10}^{-3}\text{ kg}}{1\text{ g}}\right)\\ =0.09\text{ kg}\cdot \text{m/s}\end{array}$

Consider the expression for initial momentum of Body B before the collision

$p_{2i}=m_2{v}_{2i}$

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

Understand that Body B is stationary before the collision. Therefore, substitute $0\text{m/s}$ for $v_{2i}$

$\begin{array}{c}{p}_{2i}=m_2\left(0\text{m/s}\right)\\ =0\text{ kg}\cdot \text{m/s}\end{array}$

Consider the mass of the final combined body after the collision

$m_3=m_1+m_2$

Here, $m_3$ is the mass of the combined system formed as the bodies A and B stick together after the collision.

Substitute $10\text{ g}$ for $m_2$ and $90\text{ g}$ for $m_1$

$\begin{array}{c}{m}_3=90\text{ g}+10\text{ g}\\ \text{=100 g}\end{array}$

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

$p_f=m_{3}v$

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

Substitute $\text{100 g}$ for $m_3$

$\begin{array}{c}{p}_f=\left(\text{100 g}\right)v\\ =\left(\text{100 g}\right)v\left(\frac{{10}^{-3}\text{kg}}{1\text{ g}}\right)\\ =0.1v\end{array}$

Consider the expression for conservation of momentum for the collision

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

Substitute $0.1v$ for $p_f$, $0\text{kg}\cdot \text{m/s}$ for $p_{2i}$, and $0.09\text{ kg}\cdot \text{m/s}$ for $p_{1i}$

$\begin{array}{c}0.09\text{ kg}\cdot \text{m/s}+0\text{ kg}\cdot \text{m/s}=0.1v\\ 0.09=0.1v\\ \frac{0.09}{0.1}=v\\ v=0.9\text{m/s}\end{array}$

Further solve,

$\begin{array}{c}v=0.9\text{m/s}\left(\frac{100\text{ cm}}{1\text{ m}}\right)\\ =90\text{ cm/s}\end{array}$

The final speed of the combined system is $90\text{ cm/s}$. Understand that both the balls stick together as a unit after the collision.

Conclusion:

Therefore, the speed of both the balls after impact is $90\text{ cm/s}$.

To determine

The speed of the two balls of masses $10\text{ g}$ and $90\text{ g}$ after the perfectly elastic collision, given that the $90\text{ g}$ ball strikes the $10\text{ g}$ stationary ball with a speed of $100\text{ cm/s}$.

Answer to Problem 48SP

Solution: The speed of $90\text{ g}$ ball after impact is $0.80\text{ m/s}$ and the speed of $90\text{ g}$ ball after impact is $1.8\text{m/s}$.

Explanation of Solution

Given data:

The mass of the ball moving with $100\text{ cm/s}$ is $90\text{ g}$.

The mass of the stationary ball is $10\text{ g}$.

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

The expression for conservation of kinetic energy for the elastic collision of the 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 the collision, $K{E}_{2i}$ is the kinetic energy of Body B before the collision, $K{E}_{1f}$ is the kinetic energy of Body A after the collision, and $K{E}_{1f}$ is the kinetic energy of Body B after the collision.

Explanation:

Consider the ball moving with a velocity of $100\text{ cm/s}$ as Body A and the stationary ball as Body B.

Consider the expression for final momentum of Body A after the collision

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

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

Substitute $90\text{g}$ for $m_1$

$\begin{array}{c}{p}_{1f}=\left(90\text{g}\right)v_{1f}\\ =\left(90\text{g}\right)v_{1f}\left(\frac{{10}^{-3}\text{ kg}}{1\text{ g}}\right)\\ =0.09v_{1f}\end{array}$

Consider the expression for the momentum of Body B after the collision

$p_{2f}=m_2{v}_{2f}$

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

Substitute $10\text{g}$ for $m_2$

$\begin{array}{c}{p}_{2f}=\left(10\text{g}\right)v_{2f}\\ =\left(10\text{g}\right)v_{2f}\left(\frac{{10}^{-3}\text{kg}}{1\text{ g}}\right)\\ =0.01v_{2f}\end{array}$

Consider the expression for conservation of momentum for the collision

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

Substitute $0.09v_{1f}$ for $p_{f1}$, $0.01v_{2f}$ for $p_{f2}$, $0\text{ kg}\cdot \text{m/s}$ for $p_{2i}$, and $0.09\text{ kg}\cdot \text{m/s}$ for $p_{1i}$

$\begin{array}{c}0.09\text{ kg}\cdot \text{m/s}+0\text{ kg}\cdot \text{m/s}=0.09v_{1f}+0.01v_{2f}\\ 0.09=0.09v_{1f}+0.01v_{2f}\\ 0.01v_{2f}=0.09-0.09v_{1f}\\ v_{2f}=9-9v_{1f}\mathrm{...........................................}(1)\end{array}$

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

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

Substitute $100\text{ cm/s}$ for $v_{1i}$ and $90\text{g}$ for $m_1$

$\begin{array}{c}K{E}_{1i}=\frac{1}{2}\left(90\text{g}\right){\left(100\text{ cm/s}\right)}^2\\ =\frac{1}{2}\left(90\text{g}\right)\left(\frac{{10}^{-3}\text{ kg}}{1\text{ g}}\right)\left(10000\frac{{\text{cm}}^{\text{2}}}{{\text{s}}^2}\right)\left(\frac{{10}^{-4}{\text{ m}}^2}{1{\text{ cm}}^2}\right)\\ =\frac{1}{2}\left(0.09\text{ kg}\right)\left(1{\text{m}}^2{\text{/s}}^{\text{2}}\right)\\ =0.045\text{J}\end{array}$

Consider the expression for kinetic energy of Body B before the collision

$K{E}_{2i}=\frac{1}{2}{m}_2{v}_{2i}^2$

Substitute $0\text{ m/s}$ for $v_{2i}$

$\begin{array}{c}K{E}_{2i}=\frac{1}{2}{m}_2{\left(0\text{ m/s}\right)}^2\\ =0\text{ J}\end{array}$

Consider the expression for kinetic energy of Body A after the collision

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

Substitute $90\text{g}$ for $m_1$

$\begin{array}{c}K{E}_{1f}=\frac{1}{2}\left(90\text{g}\right)v_{1f}^2\\ =\frac{1}{2}\left(90\text{g}\right)\left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)v_{1f}^2\\ =\frac{1}{2}\left(0.09\right)v_{1f}^2\\ =0.045v_{1f}^2\end{array}$

Consider the expression for kinetic energy of Body B after the collision

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

Substitute $90\text{g}$ for $m_1$ and $9-9v_{1f}$ for $v_{2f}$ from Equation (1)

$\begin{array}{c}K{E}_{2f}=\frac{1}{2}\left(10\text{g}\right){\left(9-9v_{1f}\right)}^2\\ =\frac{1}{2}\left(10\text{g}\right)\left(\frac{{10}^{-3}\text{kg}}{1\text{ g}}\right)\left(81+81v_{1f}^2-162v_{1f}\right)\\ =\frac{1}{2}\left(0.01\text{kg}\right)\left(81+81v_{1f}^2-162v_{1f}\right)\\ =0.005\left(81+81v_{1f}^2-162v_{1f}\right)\end{array}$

Further solve,

$\begin{array}{c}K{E}_{2f}=0.005\left(81+81v_{1f}^2-162v_{1f}\right)\\ =0.005\left(81\right)+0.005\left(81v_{1f}^2\right)-0.005\left(162v_{1f}\right)\\ =0.405+0.405v_{1f}^2-0.81v_{1f}\end{array}$

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

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

Substitute $0.045v_{1f}^2$ for $K{E}_{1f}$, $0.405+0.405v_{1f}^2-0.81v_{1f}$ for $K{E}_{2f}$, $0\text{J}$ for $K{E}_{2i}$, and $0.045\text{J}$ for $K{E}_{1i}$

$\begin{array}{c}0.045+0=0.045v_{1f}^2+0.405+0.405v_{1f}^2-0.81v_{1f}\\ 0.045=0.45v_{1f}^2-0.81v_{1f}+0.405\\ 0.45v_{1f}^2-0.81v_{1f}+0.405-0.045=0\\ 0.45v_{1f}^2-0.81v_{1f}+0.36=0\end{array}$

Further solve,

$\begin{array}{c}{v}_{1f}^2-1.8v_{1f}+0.8=0\\ v_{1f}^2-0.8v_{1f}-v_{1f}+0.8=0\\ v_{1f}\left(v_{1f}-0.8\right)-1\left(v_{1f}-0.8\right)=0\\ \left(v_{1f}-0.8\right)\left(v_{1f}-1\right)=0\end{array}$

Solve the equation to obtain the values of $v_{1f}$

$\begin{array}{l}{v}_{1f}-0.8=0\text{ or }{v}_{1f}-1=0\\ v_{1f}=0.80\text{ m/s}\text{ or }{v}_{1f}=1\text{m/s}\\ v_{1f}=80\text{ cm/s}\text{ or }{v}_{1f}=100\text{cm/s}\end{array}$

Since the 90 g ball collides with a stationary ball thus it will impart some its energy to the stationary ball due to which its velocity will decrease.

Therefore, $v_{1f}=0.80\text{ m/s}$.

Consider the expression for final velocity of Body B after the collision

$v_{2f}=9-9v_{1f}$

Substitute $0.80\text{ m/s}$ for $v_{f1}$

$\begin{array}{c}{v}_{2f}=9-9\left(0.80\text{ m/s}\right)\\ =9-7.2\\ =1.8\text{m/s}\end{array}$

Conclusion:

Therefore, the speed of $90\text{ g}$ ball after impact is $0.80\text{ m/s}$ and the speed of $90\text{ g}$ ball after impact is $1.8\text{m/s}$.

To determine

The speed of two balls of masses 10 g and 90 g after collision, considering that the $90\text{ g}$ ball strikes the $10\text{ g}$ stationary ball with a speed of $100\text{ cm/s}$, given that the coefficient of restitution for the collision is $0.90$.

Answer to Problem 48SP

Solution: The speed of the $90\text{ g}$ ball after impact is $0.81\text{ m/s}$ and the speed of the $90\text{ g}$ ball after impact is $1.71\text{ m/s}$.

Explanation of Solution

Given data:

The mass of the ball moving with $100\text{ cm/s}$ is $90\text{ g}$.

The mass of the stationary ball is $10\text{ g}$.

Both the balls stick together after the collision.

The coefficient of restitution is $0.90$.

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

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

Explanation:

Consider the ball moving with a velocity of $100\text{ cm/s}$ as Body A and the stationary ball as Body B.

Consider the expression for final momentum of Body A after the collision

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

Substitute $90\text{g}$ for $m_1$

$\begin{array}{c}{p}_{1f}=\left(90\text{g}\right)v_{1f}\\ =\left(90\text{g}\right)v_{1f}\left(\frac{{10}^{-3}\text{ kg}}{1\text{ g}}\right)\\ =0.09v_{1f}\end{array}$

Consider the expression for the momentum of Body B after the collision

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

Substitute $10\text{ g}$ for $m_2$

$\begin{array}{c}{p}_{2f}=\left(10\text{ g}\right)v_{2f}\\ =\left(10\text{ g}\right)v_{2f}\left(\frac{{10}^{-3}\text{ kg}}{1\text{ g}}\right)\\ =0.01v_{2f}\end{array}$

Consider the expression for conservation of momentum for the collision

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

Substitute $0.09v_{1f}$ for $p_{1f}$, $0.01v_{2f}$ for $p_{2f}$, $0\text{ kg}\cdot \text{m/s}$ for $p_{2i}$, and $0.09\text{ kg}\cdot \text{m/s}$ for $p_{1i}$

$\begin{array}{c}0.09\text{ kg}\cdot \text{m/s}+0\text{ kg}\cdot \text{m/s}=0.09v_{1f}+0.01v_{2f}\\ 0.09=0.09v_{1f}+0.01v_{2f}\\ 0.09v_{1f}+0.01v_{2f}=\mathrm{0.09.....................}(1)\end{array}$

Consider the expression for coefficient of restitution for the collision

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

Substitute $0.90$ for $e$, $100\text{ cm/s}$ for $v_{1i}$, and $0\text{ m/s}$ for $v_{2i}$

$\begin{array}{c}0.90=\frac{v_{2f}-v_{1f}}{100\text{ cm/s}-\left(0\text{ m/s}\right)}\\ 0.90=\frac{v_{2f}-v_{1f}}{\left(100\text{ cm/s}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)-\left(0\text{ m/s}\right)}\\ 0.90=\frac{v_{2f}-v_{1f}}{1\text{ m/s}-\left(0\text{ m/s}\right)}\\ 0.90=\frac{v_{2f}-v_{1f}}{1\text{ m/s}}\end{array}$

Further solve,

$\begin{array}{c}{v}_{2f}-v_{1f}=0.90\\ v_{2f}=0.90+v_{1f}\mathrm{..........................}(2)\end{array}$

Rewrite Equation (1) as,

$0.09v_{1f}+0.01v_{2f}=0.09$

Substitute $0.90+v_{1f}$ for $v_{2f}$

$\begin{array}{c}0.09v_{1f}+0.01\left(0.90+v_{1f}\right)=0.09\\ 0.09v_{1f}+0.01\left(0.90\right)+0.01v_{1f}=0.09\\ 0.09v_{1f}+0.009+0.01v_{1f}=0.09\\ 0.1v_{1f}+0.009=0.09\end{array}$

Further solve,

$\begin{array}{c}0.1v_{1f}=0.09-0.009\\ 0.1v_{1f}=0.081\\ v_{1f}=\frac{0.081}{0.1}\\ v_{1f}=0.81\text{ m/s}\end{array}$

Rewrite Equation (2) as,

$v_{2f}=0.90+v_{1f}$

Substitute $0.81\text{ m/s}$ for $v_{1f}$

$\begin{array}{c}{v}_{2f}=0.90+0.81\\ =1.71\text{ m/s}\end{array}$

Conclusion:

Therefore, the speed of the $90\text{ g}$ ball after impact is $0.81\text{ m/s}$ and the speed of the $90\text{ g}$ ball after impact is $1.71\text{ m/s}$.