A 0.100-kg stone rests on a frictionless, horizontal surface. A bullet of mass 6.00 g, traveling horizontally at 350 m/s, strikes the stone and rebounds horizontally at right angles to its original direction with a speed of 250 m/s. (a) Compute the magnitude and direction of the velocity of the stone after it is struck. (b) Is the collision perfectly elastic?

The initial velocity of a bullet in a horizontal direction is ${\mathrm{u}}_{\mathrm{bx}}=350\hspace{0.33em}\mathrm{m}/\mathrm{s}$

The rebound velocity of a bullet in the vertical direction is ${\mathrm{v}}_{\mathrm{by}}=250\hspace{0.33em}\mathrm{m}/\mathrm{s}$

The mass of stone is: $\mathrm{M}=0.100\hspace{0.33em}\mathrm{kg}$

The mass of the bullet is: $\mathrm{m}=6\hspace{0.33em}\mathrm{g}=0.006\hspace{0.33em}\mathrm{kg}$

The magnitude and direction of the velocity of the stone after the collision is found by using momentum conservation along the horizontal and vertical directions.

Apply the momentum conservation along the horizontal direction.

${\mathrm{mu}}_{\mathrm{bx}}=\mathrm{Mv}\mathrm{cos\theta }$

Here, $\mathrm{\theta }$is the direction of the stone and v is the velocity after collision.

Substitute all the values in the above equation.

$$\begin{gathered} \left(0.006\mathrm{kg}\right)\left(350\mathrm{m}/\mathrm{s}\right)=\left(0.100\mathrm{kg}\right)\mathrm{v}\mathrm{cos\theta } \\ \mathrm{v}\mathrm{cos\theta }=\left(21\mathrm{m}/\mathrm{s}\right)........\left(1\right) \end{gathered}$$

Apply the momentum conservation along the vertical direction.

localid="1664450494024" ${\mathrm{mu}}_{\mathrm{bx}}-\mathrm{Mv}\mathrm{sin\theta }=0$

Substitute all the values in the above equation.

$$\begin{gathered} \left(0.006\mathrm{kg}\right)\left(250\mathrm{m}/\mathrm{s}\right)-\left(0.100\mathrm{kg}\right)\mathrm{v}\mathrm{sin\theta }=0 \\ \mathrm{v}\mathrm{sin\theta }=\left(15\mathrm{m}/\mathrm{s}\right)........\left(2\right) \end{gathered}$$

Divide equation (2) by equation(1) to find the direction of stone.

$$\begin{gathered} \frac{\mathrm{vsin}\mathrm{\theta }}{\mathrm{vcos}\mathrm{\theta }}=\frac{15\mathrm{m}/\mathrm{s}}{21\mathrm{m}/\mathrm{s}} \\ \mathrm{tan\theta }=0.71 \\ \mathrm{\theta }=35.4^{°} \end{gathered}$$

Substitute $\mathrm{\theta }=35.4^{°}$in the equation (2) to find velocity of stone after impact.

$$\begin{gathered} v\sin \left(35.4^{°}\right)=15\mathrm{m}/\mathrm{s} \\ \mathrm{v}=26\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the velocity of stone after impact is $\mathrm{v}=26\mathrm{m}/\mathrm{s}$at an angle $35.4^{°}$.

The total kinetic energy before the collision is given as:

$$\begin{gathered} {\mathrm{K}}_1=\frac{1}{2}{{\mathrm{mu}}^2}_{\mathrm{bx}} \\ {\mathrm{K}}_1=\frac{1}{2}(0.006\hspace{0.33em}\mathrm{kg}){(350\hspace{0.33em}\mathrm{m}/\mathrm{s})}^2{\mathrm{K}}_1=367.5\hspace{0.33em}\mathrm{J} \end{gathered}$$

The total kinetic energy after collision is given as:

$$\begin{gathered} {\mathrm{K}}_2=\frac{1}{2}{{\mathrm{mu}}^2}_{\mathrm{by}}+\frac{1}{2}{\mathrm{Mv}}^2 \\ {\mathrm{K}}_2=\frac{1}{2}(0.006\hspace{0.33em}\mathrm{kg}){(250\hspace{0.33em}\mathrm{m}/\mathrm{s})}^2\frac{1}{2}\left(0.1\mathrm{kg}\right){\left(26\mathrm{m}/\mathrm{s}\right)}^2{\mathrm{K}}_2=221.3\mathrm{J} \end{gathered}$$

Since the kinetic energies before and after collision are not equal, so collision of bullet with stone is not perfectly elastic.

Therefore, the collision is not perfectly elastic.