Step 1

We are given the mass of disk A is ${m_A} = 4\;{\rm{kg}}$, the mass of disk B is ${m_B} = 2\;{\rm{kg}}$, and the coefficient of restitution is $e = 0.8$.

We are asked to determine the speeds of disks just after impact.

 
Step 2

The free body diagram of the system is shown as:

The line of impact joining the centers of disks A and B is taken as y-axis in the above figure.

We have the initial velocity of disk A is ${v_A} = 15\;{\rm{m/s}}$.

We have the initial velocity of disk B is ${v_B} = 8\;{\rm{m/s}}$.

 
Step 3

The expressions to calculate velocities of disks along x and y directions are,

\[\begin{array}{c} {\left( {{v_B}} \right)_{1y}} = {v_B}\\ {\left( {{v_A}} \right)_{1y}} = \frac{3}{5}{v_A}\\ {\left( {{v_A}} \right)_{1x}} = \frac{4}{5}{v_A}\\ {\left( {{v_A}} \right)_{1y}} = 0 \end{array}\]
 
Step 4

Substitute the values in the above expression.

\[\begin{array}{c} {\left( {{v_B}} \right)_{1y}} = 8\;{\rm{m/s}}\\ {\left( {{v_A}} \right)_{1y}} = \frac{3}{5}\left( {15\;{\rm{m/s}}} \right) = 9\;{\rm{m/s}}\\ {\left( {{v_A}} \right)_{1x}} = \frac{4}{5}\left( {15\;{\rm{m/s}}} \right) = 12\;{\rm{m/s}}\\ {\left( {{v_A}} \right)_{1y}} = 0 \end{array}\]
 
Step 5

The formula to calculate relation between the final velocities of disks by coefficient of restitution is,

\[e = \frac{{{{\left( {{v_B}} \right)}_{2y}} - {{\left( {{v_A}} \right)}_{2y}}}}{{{{\left( {{v_A}} \right)}_{1y}} - {{\left( {{v_B}} \right)}_{1y}}}}\]
 
Step 6

Substitute the values in the above expression.

\[\begin{array}{c} 0.8 = \frac{{{{\left( {{v_B}} \right)}_{2y}} - {{\left( {{v_A}} \right)}_{2y}}}}{{ - 9\;{\rm{m/s}} - 8\;{\rm{m/s}}}}\\ {\left( {{v_B}} \right)_{2y}} - {\left( {{v_A}} \right)_{2y}} = - 13.6\;{\rm{m/s}} \end{array}\] .... (1)
 
Step 7

The formula to calculate relation between the final velocities of disks by momentum conservation is,

\[{m_A}{\left( {{v_A}} \right)_{1y}} + {m_B}{\left( {{v_B}} \right)_{1y}} = {m_A}{\left( {{v_A}} \right)_{2y}} + {m_B}{\left( {{v_B}} \right)_{2y}}\]
 
Step 8

Substitute the values in the above expression.

\[\begin{array}{c} \left( {4\;{\rm{kg}}} \right)\left( { - 9\;{\rm{m/s}}} \right) + \left( {2\;{\rm{kg}}} \right)\left( {8\;{\rm{m/s}}} \right) = \left( {4\;{\rm{kg}}} \right){\left( {{v_A}} \right)_{2y}} + \left( {2\;{\rm{kg}}} \right){\left( {{v_B}} \right)_{2y}}\\ 2{\left( {{v_A}} \right)_{2y}} + {\left( {{v_B}} \right)_{2y}} = - 10\;{\rm{m/s}} \end{array}\] .... (2)
 
Step 9

On subtracting equations (1) and (2), we get:

\[\begin{array}{c} 2{\left( {{v_A}} \right)_{2y}} + {\left( {{v_B}} \right)_{2y}} - {\left( {{v_B}} \right)_{2y}} + {\left( {{v_A}} \right)_{2y}} = - 10\;{\rm{m/s}} + 13.6\;{\rm{m/s}}\\ 3{\left( {{v_A}} \right)_{2y}} = 3.6\;{\rm{m/s}}\\ {\left( {{v_A}} \right)_{2y}} = 1.2\;{\rm{m/s}} \end{array}\]
 
Step 10

Substitute the value of ${\left( {{v_A}} \right)_{2y}}$ in equations (1), we get:

\[\begin{array}{c} {\left( {{v_B}} \right)_{2y}} - 1.2\;{\rm{m/s}} = - 13.6\;{\rm{m/s}}\\ {\left( {{v_B}} \right)_{2y}} = - 12.4\;{\rm{m/s}} \end{array}\]
 
Step 11

Since, no impact occurs along the x-axis, the component of velocity of each disk remains constant before and after the impact.

Therefore, ${\left( {{v_A}} \right)_{2x}} = {\left( {{v_A}} \right)_{1x}} = 12\;{\rm{m/s}}$ and ${\left( {{v_B}} \right)_{2x}} = {\left( {{v_B}} \right)_{1x}} = 0$.

 
Step 12

The formula to calculate the magnitude of velocity of disk A just after the impact is,

\[{\left( {{v_A}} \right)_2} = \sqrt {{{\left[ {{{\left( {{v_A}} \right)}_{2x}}} \right]}^2} + {{\left[ {{{\left( {{v_A}} \right)}_{2y}}} \right]}^2}} \]
 
Step 13

Substitute the values in the above expression.

\[\begin{array}{l} {\left( {{v_A}} \right)_2} = \sqrt {{{\left( {12\;{\rm{m/s}}} \right)}^2} + {{\left( {1.2\;{\rm{m/s}}} \right)}^2}} \\ {\left( {{v_A}} \right)_2} = 12.06\;{\rm{m/s}} \end{array}\]
 
Step 14

The formula to calculate the magnitude of velocity of disk B just after the impact is,

\[{\left( {{v_B}} \right)_2} = \sqrt {{{\left[ {{{\left( {{v_B}} \right)}_{2x}}} \right]}^2} + {{\left[ {{{\left( {{v_B}} \right)}_{2y}}} \right]}^2}} \]
 
Step 15

Substitute the values in the above expression.

\[\begin{array}{l} {\left( {{v_B}} \right)_2} = \sqrt {{{\left( 0 \right)}^2} + {{\left( {12.4\;{\rm{m/s}}} \right)}^2}} \\ {\left( {{v_B}} \right)_2} = 12.4\;{\rm{m/s}} \end{array}\]