Chapter 7, Problem 75P

(a)

To determine

The speed of the shooter after collision.

Answer to Problem 75P

The speed of the shooter after collision is $2.4\text{m}/\text{s}$.

Explanation of Solution

The initial speed of the shooter is $3.2\text{m}/\text{s}$. The mass of the shooter is 3m and mass of the marble is m.

Write the expression for the conservation of kinetic energy.

$\frac{1}{2}\left(3m\right)v_i{}^2=\frac{1}{2}\left(3m\right)\left(v_x{}^2+v_y{}^2\right)+\frac{1}{2}m{V}^2$

Here, $v_i$ is the initial speed of the shooter, $v_x$ is the x-component of the final speed of the shooter, $v_y$ is the y-component of the final speed of the shooter and V is the final speed of the marble.

Rewrite the above equation in terms of $v_i$.

$3v_i{}^2=\left(3\right)\left(v_x{}^2+v_y{}^2\right)+V^2$ (I)

Write the expression for the conservation of momentum in x-direction.

$\left(3m\right)v_x-mV\sin 40°=0$

Rewrite the above equation in terms of V.

$\begin{array}{c}V\sin 40°=3v_x\\ V=\frac{3v_x}{\sin 40°}\end{array}$ (II)

Write the expression for the conservation of momentum in y-direction.

$3m{v}_i=3m{v}_y+mV\cos 40°$

Rewrite the above equation in terms of $v_i$.

$3v_i=3v_y+V\cos 40°$ (III)

Substitute (II) in (III) to calculate $v_y$.

$\begin{array}{c}3v_i=3v_y+\left(\frac{3v_x}{\sin 40°}\right)\cos 40°\\ v_i=v_y+v_x\cot 40°\\ v_y=v_i-v_x\cot 40°\end{array}$ (IV)

Substitute (II) and (IV), $3.2\text{m}/\text{s}$ for $v_i$ in (I) to calculate $v_x$.

$\begin{array}{c}\left(3\right)\left(v_x{}^2+{\left(v_i-v_x\cot 40°\right)}^2\right)+{\left(\frac{3v_x}{\sin 40°}\right)}^2=3v_i{}^2\\ v_x{}^2+v_i{}^2+v_x{}^2{\cot }^{2}40°-2v_i{v}_x\cot 40°+\frac{3v_x{}^2}{{\sin }^{2}40°}=v_i{}^2\\ v_x\left(1+{\cot }^{2}40°+\frac{3}{{\sin }^{2}40°}\right)=2\left(3.2\text{m}/\text{s}\right)\cot 40°\\ v_x\left(9.70\right)=7.63\\ v_x=0.77\text{m}/\text{s}\end{array}$

Substitute the above expression in (IV) and $3.2\text{m}/\text{s}$ for $v_i$ to calculate $v_y$.

$\begin{array}{c}{v}_y=3.2\text{m}/\text{s}-\left(0.77\text{m}/\text{s}\right)\cot 40°\\ =2.28\text{m}/\text{s}\end{array}$

Write the expression to calculate the speed of shooter.

$v=\sqrt{v_x{}^2+v_y{}^2}$

Here, v is the speed of the shooter.

Substitute $2.28\text{m}/\text{s}$ for $v_y$ and $0.77\text{m}/\text{s}$ for $v_x$ in the above equation to calculate v.

$\begin{array}{c}v=\sqrt{{\left(2.28\text{m}/\text{s}\right)}^2+{\left(0.77\text{m}/\text{s}\right)}^2}\\ =2.4\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the speed of the shooter after collision is $2.4\text{m}/\text{s}$.

(b)

To determine

The speed of the marble after collision.

Answer to Problem 75P

The speed of the marble after collision is $3.6\text{m}/\text{s}$.

Explanation of Solution

Substitute $0.77\text{m}/\text{s}$ for $v_x$ in (II) to calculate V.

$\begin{array}{c}V=\frac{3\left(0.77\text{m}/\text{s}\right)}{\sin 40°}\\ =3.6\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the speed of the marble after collision is $3.6\text{m}/\text{s}$.

(c)

To determine

The magnitude of angle $\theta$.

Answer to Problem 75P

The magnitude of angle $\theta$ is $18.7°$.

Explanation of Solution

Write the expression to calculate the angle $\theta$.

$\theta ={\tan }^{-1}\left(\frac{v_y}{v_y}\right)$

Substitute $2.28\text{m}/\text{s}$ for $v_y$ and $0.77\text{m}/\text{s}$ for $v_x$ in the above equation to calculate $\theta$.

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{0.77\text{m}/\text{s}}{2.28\text{m}/\text{s}}\right)\\ =18.7°\end{array}$

Conclusion:

Therefore, the magnitude of angle $\theta$ is $18.7°$.