Question

Agent Arlene devised the following method of measuring the muzzle velocity of a rifle (the figure below). She fires a bullet into a 4.293-kg wooden block resting on a smooth surface, and attached to a spring of spring constant k = 162.8N/m. The bullet, whose mass is 7.870 g, remains embedded in the wooden block. She measures the maximum distance that the block compresses the spring to be 9.460cm.

What is the speed v of the bullet?

Answer 1

Concepts and reason

The concept required to solve the problem are elastic collision and conservation of total momentum and the total energy of the spring-mass system. 

Initially, calculate the maximum velocity of the spring-mass oscillating system using conservation of momentum and finally calculate the speed of bullet before collision using conservation of energy.

Fundamentals

The collision in which both linear momentum and kinetic energy were conserved is known as elastic collision.

Assume two objects of masses ${m_1}$ moving with initial velocity ${u_1}$ and mass ${m_2}$ moving with initial velocity ${u_2}$ collide with each other and then move with final velocities ${v_1}$ and ${v_2}$ respectively. Then, the expression for conservation of momentum is given as follows:

${m_1}{u_1} + {m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2}$

And conservation of kinetic energy is given by following expression.

$\frac{1}{2}{m_1}{u_1}^2 + \frac{1}{2}{m_2}{u_2}^2 = \frac{1}{2}{m_1}{v_1}^2 + \frac{1}{2}{m_2}{v_2}^2$

The potential energy $P{E_{spring}}$ stored in a spring when it stretched by a distance x is given by following expression.

$P{E_{{\rm{spring}}}} = \frac{1}{2}k{x^2}$

The momentum that combined after collision is stored in the spring when it is fully compressed is,

${P_{{\rm{before}}}} = {P_{{\rm{after}}}}$

Here,  ${P_{{\rm{before}}}}$  is the momentum before compressed and  ${P_{{\rm{after}}}}$  is the momentum after compression.

Substitute, $m{v_0}$ for ${P_{before}}$ and $\left( {m + M} \right){v_{\max }}$ for ${P_{after}}$ .

Here,  ${v_0}$ is the speed of the bullet, ${v_{\max }}$  is the maximum speed of the oscillating system,  $m$ is the mass of the bullet, and  $M$ is the mass of the wooden block.

$m{v_0} = \left( {m + M} \right){v_{\max }}$

Rearrange the equation for ${v_{\max }}$ .

${v_{\max }} = \frac{m}{{\left( {m + M} \right)}}{v_0}$

The potential energy stored in the spring is,

$P{E_{{\rm{spring}}}} = \frac{1}{2}k{x^2}$

Here, k is the spring constant, and x is the displacement in the spring.

This energy must come from the kinetic energy of the block, so elastic potential energy is equal to the kinetic energy of the bullet block system.

Equation for kinetic energy of the block-bullet system is,

$KE = \frac{1}{2}\left( {m + M} \right){v_{\max }}^2$

Here,  ${v_{\max }}$  is the maximum speed of the oscillating system,  $m + M$ is the mass of the oscillating system ( $m$ is the mass of the bullet, and  $M$ is the mass of the wooden block).

From the law of conservation of energy,

$\frac{1}{2}\left( {m + M} \right){v_{\max }}^2 = \frac{1}{2}k{A^2}$

Here,  $A$  is the amplitude,  $m$  is the mass of the body, and  $k$ is the spring constant.

Substitute ${v_{\max }} = \frac{m}{{\left( {m + M} \right)}}{v_0}$ in the above equation.

$\frac{1}{2}\left( {m + M} \right){\left( {\frac{m}{{\left( {m + M} \right)}}{v_0}} \right)^2} = \frac{1}{2}k{A^2}$

Rewrite the equation for ${v_0}$ ,

${v_0} = \frac{A}{m}\sqrt {k\left( {m + M} \right)}$

Convert the units of amplitude from cm to meter as follows:

$\begin{array}{c}\\A = 9.460\,{\rm{cm}}\left( {\frac{{1\;{\rm{m}}}}{{100\,{\rm{cm}}}}} \right)\\\\ = 9.460 \times {10^{ - 2}}\,{\rm{m}}\\\end{array}$

Convert the units of mass of the bullet from g to kg as follows:

$\begin{array}{c}\\{\rm{ }}m = 7.87{\rm{ g}}\left( {\frac{{1\,{\rm{kg}}}}{{1000\;{\rm{g}}}}} \right)\\\\ = 7.87 \times {10^{ - 3}}\,{\rm{kg}}\\\end{array}$

Substitute, $9.460 \times {10^{ - 2}}{\rm{ m}}$ for $A$ , $7.87 \times {10^{ - 3}}{\rm{kg}}$ for $m$ , $162.8{\rm{ N/m}}$ for $k$ and $4.293{\rm{ kg}}$ for $M$ .

$\begin{array}{c}\\{v_0} = \frac{{9.460 \times {{10}^{ - 2}}{\rm{ m}}}}{{7.87 \times {{10}^{ - 3}}{\rm{ kg}}}}\sqrt {162.8{\rm{ N/m}}\left( {7.87 \times {{10}^{ - 3}}{\rm{ kg}} + 4.293{\rm{ kg}}} \right)} \\\\ = 318.06{\rm{ m/s}}\\\end{array}$

Therefore, the speed of the bullet is $318{\rm{ m/s}}{\rm{.}}$

Ans:

The speed of the bullet is $318{\rm{ m/s}}$ .