Chapter 7, Problem 72QAP

To determine

(a)

Which would make the block swing higher, a 0.204 Ruger bullet of mass 2.14 g and muzzle speed 1290 m/s or a 7 mm Remington Magnum bullet of mass 9.71 g and muzzle speed 948 m/s? Assume the bullets enter the block right after leaving the muzzle of the rifle?

Answer to Problem 72QAP

Remington bullet will make the block swing higher.

Explanation of Solution

Given info:

  $\begin{array}{l}{m}_{Ruger}=2.14g\\ v_{Ruger}=1290m/s\\ m_{\text{Remington}}=9.71g\\ v_{\text{Remington}}=948m/s\end{array}$

Formula used:

Calculation:

By conservation of momentum,

  $\begin{array}{l}{m}_{bullet}{v}_{ix}=(m_{bullet}+m_{block})v_{fx}\\ v_{fx}=\frac{m_{bullet}{v}_{ix}}{(m_{bullet}+m_{block})}\end{array}$

By conservation of energy,

  $\begin{array}{l}\frac{1}{2}(m_{bullet}+m_{block})v^2{}_{fx}=(m_{bullet}+m_{block})gh\\ h=\frac{v^2{}_{fx}}{2g}=\frac{1}{2g}{\left(\frac{m_{bullet}}{m_{bullet}+m_{block}}\right)}^2{v}^2{}_{ix}\end{array}$

We know that $m_{bullet}<<<m_{block}$

So, we have,

  $h=\frac{v^2{}_{ix}}{2g}{\left(\frac{m_{bullet}}{m_{block}}\right)}^2$

Comparing the height of 2 bullets,

  $\frac{h_{Ruger}}{h_{\text{Remington}}}=\frac{{\left[\frac{v^2{}_{ix}}{2g}{\left(\frac{m_{bullet}}{m_{block}}\right)}^2\right]}_{Ruger}}{{\left[\frac{v^2{}_{ix}}{2g}{\left(\frac{m_{bullet}}{m_{block}}\right)}^2\right]}_{\text{Remington}}}=\frac{{\left[v^2{}_{ix}{m}_{bullet}{}^2\right]}_{Ruger}}{{\left[v^2{}_{ix}{m}_{bullet}{}^2\right]}_{\text{Remington}}}=\frac{{1290}^2\times {2.14}^2}{{948}^2\times {9.71}^2}=0.090$

Here the ratio is less than 1, so the Remington bullet will make the block swing higher.

Conclusion:

Remington bullet will make the block swing higher.

To determine

(b)

Using your answer in part (a), the mass of the block so that when hit by the bullet it will swing through a 60.0° angle? The block hangs from wire of length 1.25 m and negligible mass.

Answer to Problem 72QAP

Mass of block = 2.62 kg

Explanation of Solution

Given info:

Formula used:

Calculation:

From trigonometry the final height of block reached is,

  $h=L-L\cos 60$

By conservation of energy,

  $\begin{array}{l}\frac{1}{2}(m_{bullet}+m_{block})v^2{}_{fx}=(m_{bullet}+m_{block})gh\\ v_{fx}=\sqrt{2gh}=\sqrt{2g(L-L\cos 60)}\end{array}$

By conservation of momentum,

  $\begin{array}{l}{m}_{bullet}{v}_{ix}=(m_{bullet}+m_{block})v_{fx}\\ m_{block}=\frac{m_{bullet}{v}_{ix}-m_{bullet}{v}_{fx}}{v_{fx}}=\frac{m_{bullet}{v}_{ix}-m_{bullet}\sqrt{2g(L-L\cos 60)}}{\sqrt{2g(L-L\cos 60)}}\\ m_{block}=\frac{m_{bullet}{v}_{ix}-m_{bullet}\sqrt{gL}}{\sqrt{gL}}=\frac{9.71\times 948-9.71\times \sqrt{9.81\times 1.25}}{\sqrt{9.81\times 1.25}}=2620g=2.62kg\end{array}$

Conclusion:

Mass of block = 2.62 kg

To determine

(c)

What is the speed of an 8.41-g bullet that causes the block to swing upward through a 30.0° angle?

Answer to Problem 72QAP

Speed of an 8.41-g bullet = 566 m/s

Explanation of Solution

Given info:

Formula used:

Calculation:

From trigonometry the final height of block reached is,

  $h=L-L\cos 30$

By conservation of energy,

  $\begin{array}{l}\frac{1}{2}(m_{bullet}+m_{block})v^2{}_{fx}=(m_{bullet}+m_{block})gh\\ v_{fx}=\sqrt{2gh}=\sqrt{2g(L-L\cos 30)}\end{array}$

By conservation of momentum,

  $\begin{array}{l}{m}_{bullet}{v}_{ix}=(m_{bullet}+m_{block})v_{fx}\\ v_{ix}=\frac{(m_{bullet}+m_{block})v_{fx}}{m_{bullet}}=\frac{(m_{bullet}+m_{block})\sqrt{2g(L-L\cos 30)}}{m_{bullet}}\\ v_{ix}=\frac{(8.41+2620)\sqrt{2\times 9.81(1.25-1.25\cos 30)}}{8.41}=566m/s\end{array}$

Conclusion:

Speed of an 8.41-g bullet = 566 m/s