Question

A ball of mass 0.440 kg moving east (+x direction) with a speed of \({\bf{3}}{\bf{.80}}\;{{\bf{m}} \mathord{\left/{\vphantom {{\bf{m}} {\bf{s}}}} \right.\\} {\bf{s}}}\) collides head-on with a 0.220-kg ball at rest. If the collision is perfectly elastic, what will be the speed and direction of each ball after the collision?

Short answer

The final velocities of the first and second balls are \(1.27\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right.\\} {\rm{s}}}\) and \(5.07\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.\\} {\rm{s}}}\), towards the east, respectively.

Step-by-step solution

Define elastic collision

In an elastic collision, the momentum and the total kinetic energy of the system remain conserved.

Given information

The mass of the first ball is\({m_1} = 0.440\;{\rm{kg}}\).

The initial velocity of the first ball is\({v_{1,{\rm{i}}}} = 3.80\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.\\} {\rm{s}}}\).

The mass of the second ball is\({m_2} = 0.220\;{\rm{kg}}\).

The initial velocity of the second ball is \({v_{2,{\rm{i}}}} = 0\).

Apply the law of conservation of linear momentum

Since the collision is perfectly elastic, the coefficient of restitution is\(e = 1\).

Consider the two balls as a system. Also, take the direction towards the east positive and the west negative.

Apply the law of conservation of momentum.

\(\begin{array}{c}{m_1}{v_{1,{\rm{i}}}}+{m_2}{v_{2,{\rm{i}}}} = {m_1}{v_{1,{\rm{f}}}} +{m_2}{v_{2,{\rm{f}}}}\\\left({0.440\;{\rm{kg}}}\right)\left({3.80\;{{\rm{m}}\mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.\\} {\rm{s}}}} \right) + \left( {0.220\;{\rm{kg}}}\right)\left( 0 \right) = \left( {0.440\;{\rm{kg}}} \right){v_{1,{\rm{f}}}} +\left({0.220\;{\rm{kg}}}\right){v_{2,{\rm{f}}}}\\\left({0.440\;{\rm{kg}}}\right){v_{1,{\rm{f}}}}+\left({0.220\;{\rm{kg}}}\right){v_{2,{\rm{f}}}}=1.672\;{{{\rm{kg}}\cdot {\rm{m}}} \mathord{\left/{\vphantom {{{\rm{kg}} \cdot {\rm{m}}} {\rm{s}}}} \right.\\} {\rm{s}}}\end{array}\) … (i)

Calculate the final velocity of the first and second balls

The condition for perfectly elastic collision is as follows:

\(\begin{array}{c}e=\frac{{{v_{2,{\rm{f}}}}{v_{1,f}}}}{{{v_{1,i}}{v_{2,{\rm{i}}}}}}\\1=\frac{{{v_{2,{\rm{f}}}}-{v_{1,f}}}}{{{v_{1,i}}- {v_{2,{\rm{i}}}}}}\\{v_{2,{\rm{f}}}} - {v_{1,f}} = {v_{1,i}} - {v_{2,{\rm{i}}}}\end{array}\)

Substitute the values in the above equation.

\(\begin{array}{c}{v_{2,{\rm{f}}}} - {v_{1,f}} = \left( {3.80\;{{\rm{m}} \mathord{\left/ {\vphantom{{\rm{m}}{\rm{s}}}}\right.\\}{\rm{s}}}}\right)0\\{v_{2,{\rm{f}}}}=\left({3.80\;{{\rm{m}}\mathord{\left/{\vphantom{{\rm{m}}{\rm{s}}}}\right.\\}{\rm{s}}}}\right)+{v_{1,f}}\end{array}\) ……… (ii)

Substitute the value of equation (ii) in equation (i).

\(\begin{array}{c}\left({0.440\;{\rm{kg}}}\right){v_{1,{\rm{f}}}}+\left({0.220\;{\rm{kg}}} \right)\left[ {\left( {3.80\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.\\} {\rm{s}}}} \right) + {v_{1,f}}} \right] = 1.672\;{{{\rm{kg}} \cdot {\rm{m}}} \mathord{\left/{\vphantom {{{\rm{kg}} \cdot {\rm{m}}} {\rm{s}}}} \right.\\} {\rm{s}}}\\\left( {0.440\;{\rm{kg}}} \right){v_{1,{\rm{f}}}} + \left( {0.220\;{\rm{kg}}} \right){v_{1,f}} = \left( {1.672\;{{{\rm{kg}} \cdot {\rm{m}}} \mathord{\left/{\vphantom {{{\rm{kg}} \cdot {\rm{m}}} {\rm{s}}}} \right.\\} {\rm{s}}}} \right) - \left( {0.836\;{{{\rm{kg}} \cdot {\rm{m}}} \mathord{\left/{\vphantom {{{\rm{kg}} \cdot {\rm{m}}}{\rm{s}}}}\right.\\}{\rm{s}}}}\right)\\{v_{1,{\rm{f}}}}=1.27\;{{\rm{m}}\mathord{\left/{\vphantom{{\rm{m}}{\rm{s}}}}\right.\\}{\rm{s}}}\end{array}\)

Substitute this value in equation (ii) to get the final velocity of the second ball.

\(\begin{array}{c}{v_{2,{\rm{f}}}} = \left( {3.80\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.\\}{\rm{s}}}}\right) + {v_{1,f}}\\{v_{2,{\rm{f}}}} = \left( {3.80\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.\\} {\rm{s}}}} \right) + \left( {1.27\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.\\} {\rm{s}}}} \right)\\{v_{2,{\rm{f}}}} = 5.07\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.\\}{\rm{s}}}\end{array}\)

Thus, the final velocity of the first ball is \(1.27\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.\\} {\rm{s}}}\) towards the east, and that of the second ball is \(5.07\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.\\} {\rm{s}}}\), also towards the east.