Question

Agent Arlene devised the following method of measuring the muzzle velocity of a rifle (Fig. 11–52). She fires a bullet into a 4.148-kg wooden block resting on a smooth surface, and attached to a spring of spring constant k= 162.7 N/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 9.460 cm. What is the speed v of the bullet?

Short answer

The speed v of the bullet is \(311.49\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}\).

Step-by-step solution

Determination of the speed of the bullet

In order to find the relation between velocities before and after the collision, use the concept of conservation of momentum in which the total momentum before collision will be equal to the total momentum after the collision.

To find the speed of the bullet, use the concept of conservation of energy in which the kinetic energy will be equal to the spring potential energy.

Given information

Given data:

The mass of the wooden block is \(M = 4.148\;{\rm{kg}}\).

The spring constant is \(k = 162.7\;{{\rm{N}} \mathord{\left/{\vphantom {{\rm{N}} {\rm{m}}}} \right.} {\rm{m}}}\).

The mass of the bullet is \(m = 7.870\;{\rm{g}}\).

The maximum distance that the block compresses the spring is \(x = 9.460\;{\rm{cm}}\).

Evaluation of equation of initial and final momentum

The initial momentum of the bullet can be calculated as:

\(\begin{aligned}{c}{p_i} &= mv\\{p_i} &= \left( {7.870\;{\rm{g}} \times \frac{{{{10}^{ - 3}}\;{\rm{kg}}}}{{1\;{\rm{g}}}}} \right)v\\{p_i} &= \left( {7.87 \times {{10}^{ - 3}}\;{\rm{kg}}} \right)v\end{aligned}\) … (i)

After the collision, the final velocity of the block-bullet \(\left( {v'} \right)\) behaves as a combined system. Hence, the final momentum of the bullet can be calculated as:

\(\begin{aligned}{c}{p_f} &= \left( {m + M} \right)v'\\{p_f} &= \left( {\left( {7.870\;{\rm{g}} \times \frac{{{{10}^{ - 3}}\;{\rm{kg}}}}{{1\;{\rm{g}}}}} \right) + \left( {4.148\;{\rm{kg}}} \right)} \right)v'\\{p_f} &= \left( {4.155\;{\rm{kg}}} \right)v'\end{aligned}\) … (ii)

Evaluation of the speed of bullet

To find the relation between velocities before and after the collision, use the concept of conservation of momentum by equating equations (i) and (ii), you get:

\(\begin{aligned}{c}\left( {7.87 \times {{10}^{ - 3}}\;{\rm{kg}}} \right)v &= \left( {4.155\;{\rm{kg}}} \right)v'\\v' &= 1.89 \times {10^{ - 3}}v\end{aligned}\)

The kinetic energy of the combined system can be calculated as:

\(\begin{aligned}{c}K &= \frac{1}{2}\left( {m + M} \right){{v'}^2}\\K &= \frac{1}{2}\left( {\left( {7.870\;{\rm{g}} \times \frac{{{{10}^{ - 3}}\;{\rm{kg}}}}{{1\;{\rm{g}}}}} \right) + \left( {4.148\;{\rm{kg}}} \right)} \right) \times {\left( {1.89 \times {{10}^{ - 3}}v} \right)^2}\\K &= 7.42 \times {10^{ - 6}}{v^2}\end{aligned}\) … (iii)

The spring energy due to compression can be calculated as:

\(\begin{aligned}{c}U &= \frac{1}{2}k{x^2}\\U &= \frac{1}{2} \times \left( {162.7\;{{\rm{N}} \mathord{\left/{\vphantom {{\rm{N}} {\rm{m}}}} \right.} {\rm{m}}}} \right) \times {\left( {9.460\;{\rm{cm}} \times \frac{{{{10}^{ - 2}}\;{\rm{m}}}}{{1\;{\rm{cm}}}}} \right)^2}\\U &= 0.72\;{\rm{J}}\end{aligned}\) … (iv)

Now, to find the speed v of the bullet, use the conservation of energy by equating equations (iii) and (iv), you get:

\(\begin{aligned}{c}7.42 \times {10^{ - 6}}{v^2} &= 0.72\;{\rm{J}}\\{v^2} &= 97.03 \times {10^3}\;{{{{\rm{m}}^2}} \mathord{\left/{\vphantom {{{{\rm{m}}^2}} {{{\rm{s}}^2}}}} \right.} {{{\rm{s}}^2}}}\\v = 311.49\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}\end{aligned}\)

Hence, the speed v of the bullet is \(311.49\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}\).