91Ó°ÊÓ

One part of the bullet's kinetic energy is lost during the collision; therefore, the remaining energy is lost due to work done against the friction force and the stored potential energy in spring due to compression.

The mass of the block is \(M = 0.999\;{\rm{kg}}\).

The mass of the bullet is \(m = 0.0010\;{\rm{kg}}\).

The spring constant is \(k = 140\;{\rm{N/m}}\).

The compression of the spring is \(x = 0.050\;{\rm{m}}\).

Let v be the initial velocity of the bullet.

Let \(v'\) be the speed of the bullet and wooden block's combined mass after the collision.

Part (a)

Now, the total momentum before the collision is:

\(\begin{array}{c}{P_{\rm{b}}} = mv + \left( {M \times 0} \right)\\ = mv\end{array}\)

The total momentum of the masses after the collision is \(\left( {m + M} \right)v'\).

Now, using the momentum conservation, you get:

\(\begin{array}{c}\left( {M + m} \right)v' = mv\\v' = \frac{m}{{M + m}}v\end{array}\) … (i)

The total kinetic energy of the bullet and wooden block system just after the collision is:

\(\begin{array}{c}{E_{\rm{B}}} = \frac{1}{2}\left( {M + m} \right){\left( {v'} \right)^2} + 0\\ = \frac{1}{2}\left( {M + m} \right){\left( {v'} \right)^2}\end{array}\)

Now, the box moves x distance before coming to rest.

Therefore, the normal force on the combined system is \(N = \left( {M + m} \right)g\).

The friction force on the combined system is:

\(\begin{array}{c}f = \mu N\\ = \mu \left( {M + m} \right)g\end{array}\)

Therefore, the work done against the friction force before coming to rest is:

\(\begin{array}{c}W = fx\\ = \mu x\left( {M + m} \right)g\end{array}\)

When the total combined mass comes to rest, the potential energy stored in the spring is:

\(E = \frac{1}{2}k{x^2}\)

Now, using energy conservation, you get:

\(\begin{array}{c}{E_{\rm{B}}} = E + W\\\frac{1}{2}\left( {M + m} \right){\left( {v'} \right)^2} = \frac{1}{2}k{x^2} + \mu x\left( {M + m} \right)g\\{\left( {v'} \right)^2} = \frac{{k{x^2}}}{{M + m}} + 2\mu xg\\v' = \sqrt {\frac{{k{x^2}}}{{M + m}} + 2\mu xg} \end{array}\) … (ii)

Here,

\(\begin{array}{c}\left( {M + m} \right) = \left( {0.999\;{\rm{kg}}} \right) + \left( {0.0010\;{\rm{kg}}} \right)\\ = 1.000\;{\rm{kg}}\end{array}\)

Now, comparing the values of \(v'\), you get:

\(\begin{array}{c}\frac{m}{{M + m}}v = \sqrt {\frac{{k{x^2}}}{{M + m}} + 2\mu xg} \\v = \frac{{M + m}}{m}\sqrt {\frac{{k{x^2}}}{{M + m}} + 2\mu xg} \\v = \frac{{1.000\;{\rm{kg}}}}{{0.0010\;{\rm{kg}}}}\sqrt {\frac{{\left( {140\;{\rm{N/m}}} \right) \times {{\left( {0.050\;{\rm{m}}} \right)}^2}}}{{1.000\;{\rm{kg}}}} + \left\{ {2 \times 0.50 \times \left( {0.050\;{\rm{m}}} \right) \times \left( {9.80\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)} \right\}} \\v \approx 920\;{\rm{m/s}}\end{array}\)

Hence, the initial velocity of the bullet is \(920\;{\rm{m/s}}\).

Part (b)

The initial kinetic energy before the collision is:

\(\begin{array}{c}{K_{\rm{i}}} = \frac{1}{2}m{v^2} + \left( {\frac{1}{2}m \times {0^2}} \right)\\ = \frac{1}{2}m{v^2}\end{array}\)

The final kinetic energy after the collision is:

\(\begin{array}{c}{K_{\rm{f}}} = \frac{1}{2}\left( {M + m} \right){\left( {v'} \right)^2}\\ = \frac{1}{2}\left( {M + m} \right){\left( {\frac{m}{{M + m}}v} \right)^2}\\ = \frac{1}{2}\frac{{{m^2}}}{{M + m}}{v^2}\\ = \frac{1}{2}m{v^2}\left( {\frac{m}{{M + m}}} \right)\end{array}\)

Now, the dissipated kinetic energy during the collision is:

\(\begin{array}{c}{K_{\rm{i}}} - {K_{\rm{f}}} = \frac{1}{2}m{v^2} - \frac{1}{2}m{v^2}\left( {\frac{m}{{M + m}}} \right)\\ = \frac{1}{2}m{v^2}\left( {1 - \frac{m}{{M + m}}} \right)\\ = \frac{1}{2}m{v^2}\left( {\frac{{M + m - m}}{{M + m}}} \right)\\ = \frac{1}{2}m{v^2}\frac{M}{{M + m}}\end{array}\)

After further calculation, you get:

\(\begin{array}{c}\frac{{{K_{\rm{i}}} - {K_{\rm{f}}}}}{{{K_{\rm{i}}}}} = \frac{{\frac{1}{2}m{v^2}\frac{M}{{M + m}}}}{{\frac{1}{2}m{v^2}}}\\ = \frac{M}{{M + m}}\\ = \frac{{0.999\;{\rm{kg}}}}{{1.000\;{\rm{kg}}}}\\ = 0.999\end{array}\)

Hence, the dissipated energy is 0.999 times the initial kinetic energy of the bullet.