(a) Derive a formula for the fraction of kinetic energy lost,\(\Delta KE/KE\), in terms of m and M for the ballistic pendulum collision of Example 7–9. (b) Evaluate for\(m = 18.0\;{\rm{g}}\)and mass\(M = 380\;{\rm{g}}\).

The energy possessed by a body by its motion is called kinetic energy.Mathematically it is given by

\(KE=\frac{1}{2}m{v^2}\). … (i)

Here, m is the mass, and v is the velocity of the body.

The linear momentum of a body is defined as the product of its mass and velocity. It is given by \(p = mv\).

The value of mass is\(m = 18\;{\rm{g}}\).

The value of mass is\(M = 380\;{\rm{g}}\).

The initial kinetic energy of mass m moving with an initial velocity v is

\(K{E_{\rm{i}}} = \frac{1}{2}m{v^2}\). … (ii)

The final kinetic energy after the collision of the combined mass of the pendulum,\(m + M\), is

\(K{E_{\rm{f}}} = \frac{1}{2}\left( {m + M} \right)v_{\rm{f}}^2\). … (iii)

From the principle of conservation of linear momentum, if no external force is applied, the total linear momentum remains constant. It can be written as follows:

\(\begin{array}{c}m{v_{\rm{i}}} = \left( {m + M} \right){v_{\rm{f}}}\\{v_{\rm{f}}} = \left( {\frac{m}{{m + M}}} \right){v_{\rm{i}}}\end{array}\) … (iv)

Substitute equation (iv) in (iii).

\(\begin{array}{c}K{E_{\rm{f}}} = \frac{1}{2}\left( {m + M} \right){\left( {\left( {\frac{m}{{m + M}}} \right){v_{\rm{i}}}} \right)^2}\\ = \frac{1}{2}\left( {\frac{{{m^2}}}{{m + M}}} \right)v_{\rm{i}}^2\end{array}\) … (v)

\(\begin{array}{c}\frac{{\Delta KE}}{{KE}} = \frac{{K{E_{\rm{i}}} - K{E_{\rm{f}}}}}{{K{E_{\rm{i}}}}}\\ = \frac{{\frac{1}{2}mv_{\rm{i}}^2 - \frac{1}{2}\left( {\frac{{{m^2}}}{{m + M}}} \right)v_{\rm{i}}^2}}{{\frac{1}{2}mv_{\rm{i}}^2}}\\ = 1 - \frac{m}{{\left( {m + M} \right)}}\\ = \frac{M}{{m + M}}\end{array}\)

Thus, the fraction of kinetic energy lost during collision is\(\frac{{\Delta KE}}{{KE}} = \frac{M}{{m + M}}\).

For the given values of m and M, the expression for a fraction of energy lost is given as follows:

\(\begin{array}{c}\frac{{\Delta KE}}{{KE}} = \frac{{380\;{\rm{g}}}}{{18\;{\rm{g}} + 380\;{\rm{g}}}}\\ = 0.955\end{array}\)

Thus, for the given numeric values of the two masses, the fraction of energy lost is 0.955.