Q29P A 0.280-kg croquet ball makes an... [FREE SOLUTION]

Chapter 7: Q29P (page 170)

A 0.280-kg croquet ball makes an elastic head-on collision with a second ball initially at rest. The second ball moves off with half the original speed of the first ball.

(a) What is the mass of the second ball?

(b) What fraction of the original kinetic energy \(\left( {\frac{{{\rm{\Delta KE}}}}{{{\rm{KE}}}}} \right)\) gets transferred to the second ball?

Short Answer

Expert verified

The mass of the second ball is \(0.84\;{\rm{kg}}\), and the fraction of the original kinetic energy transferred to it is 0.75.

Step by step solution

Given Data

The mass of croquet ball is\({m_{\rm{A}}} = 0.280\;{\rm{kg}}\).

For the first ball, let the initial speed be\({v_{\rm{A}}}\)and the final speed be\({v'_{\rm{A}}}\).

The initial speed of the second ball is\({v_{\rm{B}}} = 0{\rm{ m/s}}\).

The final speed of the second ball is \({v'_{\rm{B}}} = \frac{{{v_{\rm{A}}}}}{2}\).

Step 2: Relationship between relative velocities in a perfectly elastic collision

The relation for relative velocities in a perfectly elastic collision is

\(\begin{array}{c}{v_{\rm{A}}} - {v_{\rm{B}}} = - \left( {{{v'}_{\rm{A}}} - {{v'}_{\rm{B}}}} \right)\\{v_{\rm{A}}} - 0 = - \left( {{{v'}_{\rm{A}}} - \frac{{{v_{\rm{A}}}}}{2}} \right)\end{array}\)\({v'_{\rm{A}}} = - \frac{{{v_{\rm{A}}}}}{2}.\)… (i)

Evaluation of the mass of the second ball

The relation obtained from the momentum conservation equation is

\(\begin{array}{c}P = P'\\{m_{\rm{A}}}{v_{\rm{A}}} + {m_{\rm{B}}}{v_{\rm{B}}} = {m_{\rm{A}}}{{v'}_{\rm{A}}} + {m_{\rm{B}}}{{v'}_{\rm{B}}}.\end{array}\)

Here, \({m_{\rm{A}}}\) is the mass of the first ball, and \({m_{\rm{B}}}\) is the mass of the second ball.

Plugging the values in the above equation and using equation (i), you get

\(\begin{array}{c}{m_{\rm{A}}}{v_{\rm{A}}} + 0 = - \frac{1}{2}{m_{\rm{A}}}{v_{\rm{A}}} + \frac{1}{2}{m_{\rm{B}}}{v_{\rm{A}}}\\{m_{\rm{B}}} = 3{m_{\rm{A}}}\\{m_{\rm{B}}} = 3\left( {0.280\;{\rm{kg}}} \right)\\{m_{\rm{B}}} = 0.84\;{\rm{kg}}{\rm{.}}\end{array}\)

Thus, \({m_{\rm{B}}} = 0.84\;{\rm{kg}}\) is the required mass of the second ball.

Calculation for the fraction of kinetic energy

The fraction of the kinetic energy transferred to ball B is the ratio of the final kinetic energy of ball B and the total initial kinetic energy of balls A and B. It is given as

\(\frac{{{\rm{K}}{{{\rm{E'}}}_{\rm{B}}}}}{{{\rm{K}}{{\rm{E}}_{\rm{A}}} + {\rm{K}}{{\rm{E}}_{\rm{B}}}}} = \frac{{\frac{1}{2}{m_{\rm{B}}}{{v'}_{\rm{B}}}^2}}{{\frac{1}{2}{m_{\rm{A}}}{v_{\rm{A}}}^2 + 0}}\).

Plugging the values in the above equation, you get

\(\begin{array}{l}\frac{{{\rm{K}}{{{\rm{E'}}}_{\rm{B}}}}}{{{\rm{K}}{{\rm{E}}_{\rm{T}}}}} = \frac{{\left( {3{m_{\rm{A}}}} \right){{\left( {\frac{1}{2}{v_{\rm{A}}}} \right)}^2}}}{{{m_{\rm{A}}}{v_{\rm{A}}}^2}}\\\frac{{{\rm{K}}{{{\rm{E'}}}_{\rm{B}}}}}{{{\rm{K}}{{\rm{E}}_{\rm{T}}}}} = \frac{3}{4}.\end{array}\)

Thus, \(\frac{{{\rm{K}}{{{\rm{E'}}}_{\rm{B}}}}}{{{\rm{K}}{{\rm{E}}_{\rm{T}}}}} = \frac{3}{4} = 0.75\) is the required fraction.