Disks A and B have a mass of 16 kg and 10 kg respectively If

Disks A and B have a mass of 16 kg and 10 kg, respectively. If they are sliding on a smooth horizontal plane with the velocities shown, determine their speeds just after impact. The coefficient of restitution between them is e = 0.6

Solution

Angle of line of impact = atan (4/3) = 53.13 deg

Tangential velocity of A, Vta = 10* sin 53.13 = 8 m/s

Normal velocity of A, Vna = 10* cos 53.13 = 6 m/s

Tangential velocity of B, Vtb = -8* sin 53.13 = -6.4 m/s

Normal velocity of B, Vnb = -8* cos 53.13 = -4.8 m/s

By momentum conservation in tangential direction,

ma*Vta = ma*Vta\' and mb*Vtb = mb*Vtb\'

Thus, Vta\' = 8 m/s, Vtb\' = -6.4 m/s

By momentum conservation in normal direction,

ma*Vna + mb*Vnb = ma*Vna\' + mb+Vnb\'

16*6 + 10*(-4.8) = 16*Vna\' + 10*Vnb\'

16*Vna\' + 10*Vnb\' = 48.........eqn1

Coeff of restitution = (Vnb\' - Vna\') / (Vna - Vnb)

0.6 = (Vnb\' - Vna\') / (6 - (-4.8))

(Vnb\' - Vna\') = 6.48.............eqn2

Solving eqns 1 and 2 we get,

Vna\' = -0.646 m/s and Vnb\' = 5.834 m/s

Thus, Va\' = sqrt (Vna\'² + Vta\'²)

Va\' = sqrt (0.646² + 8²)

Va\' = 8.026 m/s

Vb\' = sqrt (Vnb\'² + Vtb\'²)

Vb\' = sqrt (5.834² + 6.4²)

Vb\' = 8.66 m/s