Chapter 7, Problem 47QAP

To determine

The final velocities of the resulting pieces in terms of $v_0$.

Answer to Problem 47QAP

The final velocities of the resulting pieces in terms of $v_0$.

  $\begin{array}{l}{v}_{1x}=1.55v_0\cos {45}^{\circ }=1.10v_0\\ v_{1y}=-1.55v_0\sin {45}^{\circ }=-1.10v_0\\ v_{2x}=1.10v_0\cos {30}^{\circ }=0.953v_0\\ v_{2y}=-1.55v_0\sin {30}^{\circ }=0.550v_0\end{array}$

Explanation of Solution

Given data:

Mass of object $=3M$

Object breaks into masses of $=2M$ and $M$

  $\begin{array}{l}{\theta }_1={45}^{\circ }\\ {\theta }_2={30}^{\circ }\end{array}$

Formula used:

Momentum, $p_x=m{v}_x$

Where m= mass and $v_x$ = velocity in x- direction.

And principle of conservation of linear momentum

Calculation:

  

After the piece break into two pieces, Piece $1$has a mass of M and travels off at an angle of ${45}^{\circ }$ below the x-axiswith a speed of $v_1$.

Piece $2$has a mass of $2M$ and travels off at an angle of ${30}^{\circ }$above thex-axis with a speed of $v_2$.

We can use conservation of momentum in order to calculate thefinal velocities of the two pieces.

Since this is a two-dimensional problem, we will need to splitthe momenta into components and solve the x and y component equations.

y component:

  $\begin{array}{l}{P}_{\text{i}y}=P_{\text{f}y}\\ P_{\text{i}y}=p_{2\text{f}y}+p_{1\text{f}y}\\ 0=\left(2M\right)v_2\sin ({30}^{\circ })-\left(M\right)v_1\sin ({45}^{\circ })\\ v_2=\frac{v_1\sin ({45}^{\circ })}{2\sin ({30}^{\circ })}=\frac{v_1}{\sqrt{2}}\end{array}$

x component:

  $\begin{array}{l}{P}_{\text{i}x}=p_{2\text{f}x}+p_{1\text{f}x}\\ \left(3M\right)v_0=\left(2M\right)v_2\cos ({30}^{\circ })+\left(M\right)v_1\cos ({45}^{\circ })\\ 3v_0=2v_2\cos ({30}^{\circ })+v_1\cos ({45}^{\circ })=2\left(\frac{v_1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\right)+\frac{v_1}{\sqrt{2}}\\ v_1=\frac{3\sqrt{2}}{\sqrt{3}+1}{v}_0=1.55v_0\\ v_2=\frac{v_1}{\sqrt{2}}=\frac{3}{\sqrt{3}+1}{v}_0=1.10v_0\end{array}$

Therefore,

  $\begin{array}{l}{v}_{1x}=1.55v_0\cos {45}^{\circ }=1.10v_0\\ v_{1y}=-1.55v_0\sin {45}^{\circ }=-1.10v_0\\ v_{2x}=1.10v_0\cos {30}^{\circ }=0.953v_0\\ v_{2y}=-1.55v_0\sin {30}^{\circ }=0.550v_0\end{array}$

Conclusion:

Thus, for piece $1$velocities in terms of $v_0$are: -

  $\begin{array}{l}{v}_{1x}=1.55v_0\cos {45}^{\circ }=1.10v_0\\ v_{1y}=-1.55v_0\sin {45}^{\circ }=-1.10v_0\end{array}$

Thus, for piece $2$velocities in terms of $v_0$are: -

  $\begin{array}{l}{v}_{2x}=1.10v_0\cos {30}^{\circ }=0.953v_0\\ v_{2y}=-1.55v_0\sin {30}^{\circ }=0.550v_0\end{array}$