Chapter 11, Problem 26P

To determine

To Find: The speed of the bullet.

Answer to Problem 26P

The speed of the bullet before impact is $596.5\text{ m/s}$ .

Explanation of Solution

Given info:

Mass of bullet $M=25\text{g}=0.025\text{kg}$ .

Mass of block $m=0.6\text{kg}$ .

Spring constant of block $k=\text{}7.7\text{}\times {10}^3\text{N/m}$ .

Amplitude $A=0.215\text{m}$ .

Formula used:

Conservation of energy:

  $E=\frac{1}{2}\times k\times A^2=\frac{1}{2}\times (m+M)\times v_{\text{max}}{}^2$

Here,

  $k$ is spring constant,

  $A$ is amplitude of spring,

  $v_{\max }$ is maximum velocity.

Conservation of momentum;

  $\begin{array}{l}{P}_{\text{before}}=P_{\text{after}}\\ P=m\times v\end{array}$

Here,

  $m$ is the mass,

  $v$ is the velocity.

Calculation:

When collision take place, the energy is conserved. $v_{\text{max}}$ is the combined velocity of the system. The value of combined velocity is determined from the following formula.

  $\begin{array}{l}{v}_{\max }{}^2=\frac{k\times A^2}{m+M}=\frac{7.7\times {10}^3\times {(0.215)}^2}{0.6+0.025}=569.49\text{ m/s}\\ v_{\max }=23.86\text{ m/s}\end{array}$

This is the combined speed of the system after the collision. So, linear momentum is conserved in collision.

Final speed of the bullet before impact is calculated by second formula.

  $\begin{array}{l}{P}_{\text{before}}=P_{\text{after}}\\ m\times v_0=(m+M)\times v_{\max }\\ v_0=\frac{(m+M)\times v_{\max }}{v_0}\\ v_0=\frac{(0.025+0.6)\times 23.86}{0.025}\\ v_0=596.5\text{ m/s}\end{array}$

Conclusion:

Thus, final speed of bullet is $596.5\text{ m/s}$ .