Question

A rifle bullet with mass 8.00 g strikes and embeds itself in a block with a mass 0.992 kg that rests on a frictionless, horizontal surface and is attached to a coil spring (Fig. P8.79). The impact compresses the spring 15.0 cm. Calibration of the spring shows that a force of 0.750 N is required to compress the spring 0.250 cm. (a) Find the magnitude of the block’s velocity just after impact. (b) What was the initial speed of the bullet?

Short answer

(a) The block’s velocity after impact is $\mathbf{2}\mathbf{.}\mathbf{60}\mathbf{\hspace{0.33em}}\mathbf{m}\mathbf{/}\mathbf{s}$.

(b) The initial velocity of a rifle bullet is $\mathbf{325}\mathbf{\hspace{0.33em}}\mathbf{m}\mathbf{/}\mathbf{s}$.

Step-by-step solution

Determination of magnitude of block’s velocity after impact.(a)Given Data:

The force for compression of spring is$\mathrm{F}=0.750\hspace{0.33em}\mathrm{N}$

The compression for calibration of spring is $\mathrm{X}=0.250\hspace{0.33em}\mathrm{cm}=0.00250\hspace{0.33em}\mathrm{m}$

The compression of spring is $\mathrm{x}=15\hspace{0.33em}\mathrm{cm}=0.15\hspace{0.33em}\mathrm{m}$

The mass of the block is: $\mathrm{M}=0.992\hspace{0.33em}\mathrm{kg}$

The mass of a rifle bullet is: $\mathrm{m}=8\hspace{0.33em}\mathrm{g}=0.008\hspace{0.33em}\mathrm{kg}$

concept

The magnitude of the velocity of the block after impact is calculated by the energy conservation principle. The momentum conservation is used to find the initial velocity of a rifle bullet.

The force constant of the spring is given as:

$\mathrm{k}=\frac{\mathrm{F}}{\mathrm{X}}$

Here, k is the force constant of spring in calibration.

Substitute all the values in the above equation.

$$\begin{gathered} \mathrm{k}=\frac{0.750\hspace{0.33em}\mathrm{N}}{0.00250\hspace{0.33em}\mathrm{m}} \\ \mathrm{k}=300\hspace{0.33em}\mathrm{N}/\mathrm{m} \end{gathered}$$

Apply the energy conservation to find the block’s velocity after impact.

$\frac{1}{2}(\mathrm{m}+\mathrm{M}){\mathrm{v}}^2=\frac{1}{2}{\mathrm{kx}}^2$

Substitute all the values in the above equation.

$$\begin{gathered} \frac{1}{2}\left(0.008\mathrm{kg}+0.992\mathrm{kg}\right){\mathrm{v}}^2=\frac{1}{2}\left(300\mathrm{N}/\mathrm{m}\right){\left(0.15\mathrm{m}\right)}^2 \\ \mathrm{v}=2.60\mathrm{m}/\mathrm{s} \end{gathered}$$.

Therefore, the block’s velocity after impact is $2.60\mathrm{m}/\mathrm{s}$

Determination of initial velocity of rifle bullet

Apply the conservation of momentum for block and rifle bullets.

$\mathrm{mu}=(\mathrm{m}+\mathrm{M})\mathrm{v}$

Here, u is the initial velocity of a rifle bullet.

Substitute all the values in the above equation.

$$\begin{gathered} \left(0.008\mathrm{kg}\right)\mathrm{u}=\left(0.008\mathrm{kg}+0.992\mathrm{kg}\right)\left(2.60\mathrm{m}/\mathrm{s}\right) \\ \mathrm{u}=325\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the initial velocity of rifle bullet is $\mathrm{u}=325\mathrm{m}/\mathrm{s}$.