Chapter 6, Problem 44P

A space probe, initially at rest, undergoes an internal mechanical malfunction and breaks into three pieces. One piece of mass m_l = 48.0 kg travels in the positive x-direction at 12.0 m/s, and a second piece of mass m₂ = 62.0 kg travels in the xy-plane at an angle of 105° at 15.0 m/s. The third piece has mass m₃ = 112 kg. (a) Sketch a diagram of the situation, labeling the different masses and their velocities, (b) Write the general expression for conservation of momentum in the x- and y-directions in terms of m₁, m₂, m₃, v₁, v₂ and v₃ and the sines and cosines of the angles, taking θ to be the unknown angle, (c) Calculate the final x-components of the momenta of m₁ and m₂. (d) Calculate the final y-components of the momenta of m₁ and m₂. (e) Substitute the known momentum components into the general equations of momentum for the x- and y-directions, along with the known mass m₃. (f) Solve the two momentum equations for v₃ cos θ and v₃ sin θ, respectively, and use the identity cos² θ + sin² θ = 1 to obtain v₃. (g) Divide the equation for v₃ sin θ by that for v₃ cos θ to obtain tan θ, then obtain the angle by taking the inverse tangent of both sides, (h) In general, would three such pieces necessarily have to move in the same plane? Why?

To determine

Sketch a diagram of the situation labelling the different masses and their velocities.

Answer to Problem 44P

Explanation of Solution

The diagram of the breakage is,

The numerical values of the masses and velocities are,

The numerical values of the masses and velocities are,

$\begin{array}{l}{m}_1=48.0\text{kg}\\ v_1=12.0\text{m/s}\\ m_2=62.0\text{kg}\\ v_2=15.0\text{m/s}\\ m_3=112\text{kg}\end{array}$

Conclusion:

Thus, the diagram of the breakage is,

To determine

The general expression for the conservation of momentum.

Answer to Problem 44P

The general expression for the conservation of momentum in x-direction is $\underset{\_}{m_1{v}_1\cos 0°+m_2{v}_2\cos 105°+m_3{v}_3\cos \theta =0}$ and the general expression for the conservation of momentum in y-direction is $\underset{\_}{m_1{v}_1\sin 0°+m_2{v}_2\sin 105°+m_3{v}_3\sin \theta =0}$.

Explanation of Solution

The general expression for the conservation of momentum in x-direction is,

$\begin{array}{l}\Sigma p_{xf}=\Sigma p_{xi}\Rightarrow \\ m_1{v}_1\cos 0°+m_2{v}_2\cos 105°+m_3{v}_3\cos \theta =0\end{array}$

The general expression for the conservation of momentum in y-direction is,

$\begin{array}{l}\Sigma p_{yf}=\Sigma p_{yi}\Rightarrow \\ m_1{v}_1\sin 0°+m_2{v}_2\sin 105°+m_3{v}_3\sin \theta =0\end{array}$

Conclusion:

Thus, the general expression for the conservation of momentum in x-direction is $\underset{\_}{m_1{v}_1\cos 0°+m_2{v}_2\cos 105°+m_3{v}_3\cos \theta =0}$ and the general expression for the conservation of momentum in y-direction is $\underset{\_}{m_1{v}_1\sin 0°+m_2{v}_2\sin 105°+m_3{v}_3\sin \theta =0}$.

To determine

The final x-components of the momenta of the masses.

Answer to Problem 44P

The final x-component of the momenta of the mass $m_1$ is $\underset{\_}{576\text{kg}\cdot \text{m/s}}$ and the final x-component of the momenta of the mass $m_2$ is $\underset{\_}{-241\text{kg}\cdot \text{m/s}}$.

Explanation of Solution

The final x-component of the momenta of the mass $m_1$ is,

$p_{1x}=m_1{v}_1\cos 0°$

Substitute $48.0\text{kg}$ for $m_1$ and $12.0\text{m/s}$ for $v_1$.

$\begin{array}{c}{p}_{1x}=\left(48.0\text{kg}\right)\left(12.0\text{m/s}\right)\left(1\right)\\ =576\text{kg}\cdot \text{m/s}\end{array}$

The final x-component of the momenta of the mass $m_1$ is,

$p_{2x}=m_2{v}_2\cos 105°$

Substitute $62.0\text{kg}$ for $m_2$, $-0.259$ for $\cos 105°$ and $15.0\text{m/s}$ for $v_2$.

$\begin{array}{c}{p}_{2x}=\left(62.0\text{kg}\right)\left(15.0\text{m/s}\right)\left(-259\right)\\ =-241\text{kg}\cdot \text{m/s}\end{array}$

Conclusion:

Thus, the final x-component of the momenta of the mass $m_1$ is $\underset{\_}{576\text{kg}\cdot \text{m/s}}$ and the final x-component of the momenta of the mass $m_2$ is $\underset{\_}{-241\text{kg}\cdot \text{m/s}}$.

To determine

The final y-components of the momenta of the masses.

Answer to Problem 44P

The final y-component of the momenta of the mass $m_1$ is $\underset{\_}{0\text{kg}\cdot \text{m/s}}$ and the final y-component of the momenta of the mass $m_2$ is $\underset{\_}{898\text{kg}\cdot \text{m/s}}$.

Explanation of Solution

The final y-component of the momenta of the mass $m_1$ is,

$p_{1y}=m_1{v}_1\sin 0°$

Substitute $48.0\text{kg}$ for $m_1$ and $12.0\text{m/s}$ for $v_1$.

$\begin{array}{c}{p}_{1y}=\left(48.0\text{kg}\right)\left(12.0\text{m/s}\right)\left(0\right)\\ =0\text{kg}\cdot \text{m/s}\end{array}$

The final y-component of the momenta of the mass $m_1$ is,

$p_{2y}=m_2{v}_2\sin 105°$

Substitute $62.0\text{kg}$ for $m_2$, $0.966$ for $\sin 105°$ and $15.0\text{m/s}$ for $v_2$.

$\begin{array}{c}{p}_{2x}=\left(62.0\text{kg}\right)\left(15.0\text{m/s}\right)\left(0.966\right)\\ =898\text{kg}\cdot \text{m/s}\end{array}$

Conclusion:

Thus, the final y-component of the momenta of the mass $m_1$ is $\underset{\_}{0\text{kg}\cdot \text{m/s}}$ and the final y-component of the momenta of the mass $m_2$ is $\underset{\_}{898\text{kg}\cdot \text{m/s}}$.

To determine

The final x and y-components of the momenta of mass $m_3$.

Answer to Problem 44P

The final x and y-components of the momenta of mass $m_3$ are $\underset{\_}{\text{576}\text{kg}\cdot \text{m/s}-241\text{kg}\cdot \text{m/s}+\left(112\text{kg}\right)v_3\cos \theta =0}$ and $\underset{\_}{0+898\text{kg}\cdot \text{m/s}+\left(112\text{kg}\right)v_3\sin \theta =0}$ respectively.

Explanation of Solution

The final x-component of the momenta of mass $m_3$ is $\text{576}\text{kg}\cdot \text{m/s}-241\text{kg}\cdot \text{m/s}+\left(112\text{kg}\right)v_3\cos \theta =0$

The final y-component of the momenta of mass $m_3$ is $0+898\text{kg}\cdot \text{m/s}+\left(112\text{kg}\right)v_3\sin \theta =0$.

Conclusion:

Thus, the final x and y-components of the momenta of mass $m_3$ are $\underset{\_}{\text{576}\text{kg}\cdot \text{m/s}-241\text{kg}\cdot \text{m/s}+\left(112\text{kg}\right)v_3\cos \theta =0}$ and $\underset{\_}{0+898\text{kg}\cdot \text{m/s}+\left(112\text{kg}\right)v_3\sin \theta =0}$ respectively.

To determine

Solve the two momentum equations to find $v_3$.

Answer to Problem 44P

The velocity $v_3$ is $\underset{\_}{8.56\text{m/s}}$.

Explanation of Solution

In the x-direction,

$\begin{array}{c}{v}_3\cos \theta =\frac{-\text{576}\text{kg}\cdot \text{m/s+}241\text{kg}\cdot \text{m/s}}{112\text{kg}}\\ =-2.99\text{m/s}\end{array}$

In the y-direction,

$\begin{array}{c}{v}_3\sin \theta =\frac{-898\text{kg}\cdot \text{m/s}}{112\text{kg}}\\ =-8.02\text{m/s}\end{array}$

Squaring and adding the equations,

$\begin{array}{c}{v}_3^2\left({\cos }^2\theta +{\sin }^2\theta \right)={\left(-2.99\text{m/s}\right)}^2+{\left(-8.02\text{m/s}\right)}^2\\ v_3=\sqrt{73.3{\text{m}}^2{\text{/s}}^2}\\ =8.56\text{m/s}\end{array}$

Conclusion:

Thus, the velocity $v_3$ is $\underset{\_}{8.56\text{m/s}}$.

To determine

Obtain $\tan \theta$ and thus find $\theta$.

Answer to Problem 44P

The angle $\theta$ is $\underset{\_}{250°}$.

Explanation of Solution

The tangent of the angle is,

$\begin{array}{c}\tan \theta =\frac{v_3\sin \theta }{v_3\cos \theta }\\ =\frac{-8.02\text{m/s}}{-2.99\text{m/s}}\\ =2.68\end{array}$

Thus, the angle is,

$\begin{array}{c}\theta ={\tan }^{-1}\left(2.68\right)+180°\\ =250°\end{array}$

The angle must be in third quadrant. So, angle $180°$ is introduced.

Conclusion:

Thus, the angle $\theta$ is $\underset{\_}{250°}$.

To determine

Would the three pieces have to move in the same plane.

Answer to Problem 44P

All three pieces have to move in the same plane.

Explanation of Solution

The momentum of the third fragment must be equal in magnitude and must be in the opposite direction to the resultant of the other two fragments momenta. So, all three pieces have to move in the same plane.

Conclusion:

Thus, all three pieces have to move in the same plane.