Chapter 13.4, Problem 13.173P

To determine

(a)

Boulder will land on the road or beyond if $e=0.2$

Answer to Problem 13.173P

The horizontal distance travelled by the rock is $21.31m$. Therefore, the rock will land beyond the road.

Explanation of Solution

Given information:

$\alpha =45°$

Rock A falls from a distance of $20m$.

$\begin{array}{l}h=30m\\ d=20m\end{array}$

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

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

For a uniformly accelerated motion

$x=x_0+v_{0}t+\frac{1}{2}a{t}^2$

In above equation

$x$ - End distance

$x_0$ - Start distance

$v_0$ - Initial velocity

$t$ - Elapsed time

$a$ -Acceleration

Calculation:

Assume $v_0$ as the velocity of rock ‘A’ hits the incline.

$\begin{array}{l}{v}^2=u_0^2+2a\left(x-x_0\right)\\ v_0^2=0+2\left(9.81m/s^2\right)\left(20m\right)\\ v_0=19.8m/s\end{array}$

Apply conservation of linear momentum in direction t.

$m{v}_0\sin 45°=m{\left(v_A{}^1\right)}_t$

Substitute

${\left(v_A{}^1\right)}_t=\left(19.8m/s\right)\sin 45°=14m/s$

Apply co-efficient of restitution equation in direction n.

$0-{\left(v_A{}^1\right)}_n=e\left(-v_0\cos 45°-0\right)$

Substitute

$\begin{array}{l}{\left(v_A{}^1\right)}_n=0.2\left(19.8m/s\right)\cos 45°\\ {\left(v_A{}^1\right)}_n=2.8m/s\end{array}$

Velocity ${\left(v_x\right)}_0$ in x direction

$\begin{array}{l}{\left(v_x\right)}_0={\left(v_A{}^1\right)}_t\cos 45°+{\left(v_A{}^1\right)}_n\sin 45°\\ {\left(v_x\right)}_0=14\cos 45°+2.8\sin 45°=11.879m/s\end{array}$

Velocity in y direction

$\begin{array}{l}{\left(v_y\right)}_0={\left(v_A{}^1\right)}_n\cos 45°-{\left(v_A{}^1\right)}_t\sin 45°\\ {\left(v_y\right)}_0=2.8\cos 45°-14\sin 45°=-7.92m/s\end{array}$

Assume t as the time that takes for the rod to reach the ground.

$\uparrow y=y_0+v_{0}t+\frac{1}{2}a{t}^2$

Substitute

$0=30-7.92t-\frac{1}{2}\left(9.81\right)t^2$

Solve

$t=1.794s$

Find the horizontal distance $x_A$ reached by rock A.

$\leftarrow x_A=x_0+{\left(v_x\right)}_{0}t$

Substitute

$\begin{array}{l}{x}_A=0+\left(11.879m/s\right)\left(1.794s\right)\\ x_A=21.31m\end{array}$

Conclusion:

The horizontal distance travelled by the rock is $21.31m$. Therefore, the rock will land beyond the road.

To determine

(b)

Boulder will land on the road or beyond if $e=0.1$

Answer to Problem 13.173P

The horizontal distance travelled by the rock is $18.796m$. Therefore the rock will land on the road.

Explanation of Solution

Given information:

$\alpha =45°$

Rock A falls from a distance of $20m$.

$\begin{array}{l}h=30m\\ d=20m\end{array}$

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

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

For a uniformly accelerated motion

$x=x_0+v_{0}t+\frac{1}{2}a{t}^2$

In above equation

$x$ - End distance

$x_0$ - Start distance

$v_0$ - Initial velocity

$t$ - Elapsed time

$a$ -Acceleration

Calculation:

Assume $v_0$ as the velocity of rock ‘A’ hits the incline.

$\begin{array}{l}{v}^2=u_0^2+2a\left(x-x_0\right)\\ v_0^2=0+2\left(9.81m/s^2\right)\left(20m\right)\\ v_0=19.8m/s\end{array}$

Apply conservation of linear momentum in direction t.

$m{v}_0\sin 45°=m{\left(v_A{}^1\right)}_t$

Substitute

${\left(v_A{}^1\right)}_t=\left(19.8m/s\right)\sin 45°=14m/s$

Apply co-efficient of restitution equation in direction n.

$0-{\left(v_A{}^1\right)}_n=e\left(-v_0\cos 45°-0\right)$

Substitute

$\begin{array}{l}{\left(v_A{}^1\right)}_n=0.1\left(19.8m/s\right)\cos 45°\\ {\left(v_A{}^1\right)}_n=1.4m/s\end{array}$

Velocity ${\left(v_x\right)}_0$ in x direction

$\begin{array}{l}{\left(v_x\right)}_0={\left(v_A{}^1\right)}_t\cos 45°+{\left(v_A{}^1\right)}_n\sin 45°\\ {\left(v_x\right)}_0=14\cos 45°+1.4\sin 45°=10.89m/s\end{array}$

Velocity in y direction

$\begin{array}{l}{\left(v_y\right)}_0={\left(v_A{}^1\right)}_n\cos 45°-{\left(v_A{}^1\right)}_t\sin 45°\\ {\left(v_y\right)}_0=1.4\cos 45°-14\sin 45°=-8.91m/s\end{array}$

Assume t as the time that takes for the rod to reach the ground.

$\uparrow y=y_0+v_{0}t+\frac{1}{2}a{t}^2$

Substitute

$0=30-8.91t-\frac{1}{2}\left(9.81\right)t^2$

Solve

$t=1.726s$

Find the horizontal distance $x_A$ reached by rock A.

$\leftarrow x_A=x_0+{\left(v_x\right)}_{0}t$

Substitute

$\begin{array}{l}{x}_A=0+\left(10.89m/s\right)\left(1.726s\right)\\ x_A=18.796m\end{array}$

Conclusion:

The horizontal distance travelled by the rock is $18.796m$. Therefore the rock will land on the road.