Chapter 13.4, Problem 13.182P

Block A is released from rest and slides down the frictionless surface of B until it hits a bumper on the right end of B. Block A has a mass of 10 kg and object B has a mass of 30 kg and B can roll freely on the ground. Determine the velocities of A and B immediately after impact when (a) $e=0$ . (b) $e=0.7$ .

To determine

(a)

The velocities of A and B just after the impact if $e=0$.

Answer to Problem 13.182P

velocities of A and B just after the impact if : $e=0$ is $\begin{array}{l}{\left(v_A{}^1\right)}_x=0\\ {\left(v_B{}^1\right)}_x=0\end{array}$.

Explanation of Solution

Given information:

Mass of A is $10kg.$

Mass of B is $30kg.$

Conclusion:

The total linear momentum of two particles is conserved. Therefore:

$m_A{v}_A+m_B{v}_B=m_A{v}_A{}^1+m_B{v}_B{}^1$

The co-efficient of restitution is defined as:

$v_B{}^1-v_A{}^1=e\left(v_A-v_B\right)$

The principle of conservation of energy is defined as.

When a particle moves under the action of conservation of forces. the sum of kinetic energy and potential energy of that particle remains constant.

$T_1+V_1=T_2+V_2$

Calculation:

Let ${\left(v_A\right)}_x,{\left(v_A\right)}_y,{\left(v_B\right)}_x,{\left(v_B\right)}_y$ velocities before the impact and ${\left(v_A{}^1\right)}_x,{\left(v_A{}^1\right)}_y,{\left(v_B{}^1\right)}_x,{\left(v_B{}^1\right)}_y$ velocities after the impact.

Apply conservation of linear momentum while block is sliding down.

$0+0=m_A{(v_A)}_x+m_B{(v_B)}_x$

Therefore.

${(v_B)}_x=-\frac{m_A}{m_B}{(v_A)}_x$

Apply conservation of linear momentum at impact.

$0+0=m_A{(v_A{}^1)}_x+m_B{(v_B{}^1)}_x$

Therefore.

${(v_B{}^1)}_x=-\frac{m_A}{m_B}{(v_A{}^1)}_x$

Assume. $\frac{m_A}{m_B}=\beta$

According to conservation of energy.

The initial potential energy of A is equal to:

$V_0=m_{A}gh$

For B its zero.

Initial kinetic energy is zero.

And just before the impact.

$T_1=\frac{1}{2}{m}_A{v}_A{}^2+\frac{1}{2}{m}_B{v}_B{}^2$

Therefore, according to:

$T_0+V_0=T_1+V_1$

Substitute.

$\begin{array}{l}{m}_{A}gh=\frac{1}{2}{m}_A{v}_A{}^2+\frac{1}{2}{m}_B{v}_B{}^2\\ m_{A}gh=\frac{1}{2}\left(m_A+m_B{\beta }^2\right)v_A{}^2\\ m_{A}gh=\frac{1}{2}{m}_A\left(1+\beta \right)v_A{}^2\end{array}$

Therefore.

$\begin{array}{l}{v}_A{}^2={\left(v_A\right)}_x{}^2=\frac{2gh}{\left(1+\beta \right)}\\ v_A=\sqrt{\frac{2gh}{\left(1+\beta \right)}}\end{array}$

Therefore. the velocities just before the impact.

$\begin{array}{l}\to v_A=\sqrt{\frac{2gh}{\left(1+\beta \right)}}\\ \leftarrow v_B=\beta \sqrt{\frac{2gh}{\left(1+\beta \right)}}\end{array}$

Apply co-efficient of restitution equation.

$v_B{}^1-v_A{}^1=e\left(v_A-v_B\right)$

Substitute.

$\begin{array}{l}{\left(v_B{}^1\right)}_x-{\left(v_A{}^1\right)}_x=e\left[{\left(v_A\right)}_x-{\left(v_B\right)}_x\right]\\ -\beta {\left(v_A{}^1\right)}_x-{\left(v_A{}^1\right)}_x=e\left[{\left(v_A\right)}_x-\beta {\left(v_A\right)}_x\right]\end{array}$

Therefore:

${\left(v_A{}^1\right)}_x=-e{\left(v_A\right)}_x$

But we know that:

$e=0$

As a result of it:

$\begin{array}{l}{\left(v_A{}^1\right)}_x=0\\ {\left(v_B{}^1\right)}_x=0\end{array}$

Conclusion:

The velocities of A and B just after the impact. if $e=0$.

$\begin{array}{l}{\left(v_A{}^1\right)}_x=0\\ {\left(v_B{}^1\right)}_x=0\end{array}$

To determine

(b)

The velocities of A and B just after the impact if $e=0.7$.

Answer to Problem 13.182P

$\begin{array}{l}\leftarrow {\left(v_A{}^1\right)}_x=1.201m/s\\ \to {\left(v_B{}^1\right)}_x=0.4002m/s\end{array}$

Explanation of Solution

Given information:

Mass of A is $10kg$

Mass of B is $30kg$

Conclusion:

According to sub part ‘a’ we have found.

$\begin{array}{l}\to v_A=\sqrt{\frac{2gh}{\left(1+\beta \right)}}\\ \leftarrow v_B=\beta \sqrt{\frac{2gh}{\left(1+\beta \right)}}\end{array}$

Where.

$\frac{m_A}{m_B}=\beta$

Calculation:

Find the exact value of $v_A$

For that.

$\beta =\frac{m_A}{m_B}=\frac{10}{30}=0.3333$

Then.

$\to v_A=\sqrt{\frac{2gh}{\left(1+\beta \right)}}=\sqrt{\frac{2\left(9.81m/s^2\right)\left(0.2m\right)}{\left(1+0.3333\right)}}=1.7155m/s$

But according to co-efficient of restitution equation.

${\left(v_A{}^1\right)}_x=-e{\left(v_A\right)}_x$

Therefore.

${\left(v_A{}^1\right)}_x=-e{\left(v_A\right)}_x=-0.7\left(1.7155m/s\right)=1.201m/s$

Then.

${\left(v_B{}^1\right)}_x=-\beta {\left(v_A{}^1\right)}_x=-0.3333\left(-1.201m/s\right)=0.4002m/s$

Conclusion:

The velocities of A and B just after the impact. if $e=0.7$

$\begin{array}{l}\leftarrow {\left(v_A{}^1\right)}_x=1.201m/s\\ \to {\left(v_B{}^1\right)}_x=0.4002m/s\end{array}$