Question 13.153: A 1-oz bullet is traveling with a velocity of 1400 ft/s when...

A 1-oz bullet is traveling with a velocity of 1400 ft/s when it impacts and becomes embedded in a 5-lb wooden block. The block can move vertically without friction. Determine (a) the velocity of the bullet and block immediately after the impact, (b) the horizontal and vertical components of the impulse exerted by the block on the bullet.

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Weight and mass.     Bullet:   \quad w=1 \mathrm{\ oz}=\frac{1}{16} \mathrm{\ lb} \quad m=0.001941 \mathrm{\ lb} \cdot \mathrm{s}^{2} / \mathrm{\ ft}.

Block:        W=5 \mathrm{\ lb} \quad M=0.15528 \mathrm{\ lb} \cdot \mathrm{s}^{2} / \mathrm{\ ft}.

(a) Use the principle of impulse and momentum applied to the bullet and the block together.

\Sigma m \mathbf{v} _1+\Sigma \operatorname{Imp} p _{1 \rightarrow 2}=m \mathbf{v} _2

Components ↓: \quad m \nu_0 \cos 30^{\circ}+0=(m+M) \nu^{\prime}

\begin{aligned}&\begin{aligned}& \nu^{\prime}=\frac{m \nu_0 \cos 30^{\circ}}{m+M}=\frac{(0.001941)(1400) \cos 30^{\circ}}{0.157221} \\& \nu^{\prime}=14.968  \mathrm{\ ft} / s\end{aligned}\\&&&\mathbf{v}^{\prime}=14.97 \mathrm{\ ft}/ s ↓\blacktriangleleft\end{aligned}

(b) Use the principle of impulse and momentum applied to the bullet alone.

x-components:  \quad \quad -m \nu_0 \sin 30^{\circ}+R_x \Delta t=0

\begin{array}{rlr}R_x \Delta t=m \nu_0 \sin 30^{\circ} & =(0.001941)(1400) \sin 30^{\circ} & \\& =1.3587 \ lb \cdot s & R_x \Delta t=1.359 \ lb \cdot s \blacktriangleleft\end{array}

y-components: \quad \quad -m \nu_0 \cos 30^{\circ}+R_y \Delta t=-m \nu^{\prime}

\begin{aligned}R_y \Delta t & =m\left(\nu_0 \cos 30^{\circ}-\nu^{\prime}\right) \\& =(0.001941)\left(1400 \cos 30^{\circ}-14.968\right) \quad R_y \Delta t=2.32 \ lb \cdot s \blacktriangleleft\end{aligned}

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