Question

Two 78.5-kg hockey players skating at 6.47 m/s collide and stick together. If the angle between their initial directions was 115 ∘, what is their speed after the collision?

Answer 1

Let the two players travel along x axis after collision and y axis be perpendicular to the final direction. Then before collision one player will travel at an angle of 57.5(=115/2) degree and other at an angle of -57.5 degree to the x axis. As finally the players move along x axis so final momentum along y axis is zero and before collision the y component of the momenta of the players will be equal and opposite making total momentum before collision also zero along y direction.

Now total momenta along x direction before collision

Pi = [78.5*6.47*cos(57.5)+78.5*6.47*cos(-57.5)]

= (272.89 + 272.89) kg.m/s

= 545.78 kg.m/s

Thus, by conservation of momentum total momentum along x direction after collision will be

Pf = 545.78 kg.m/s

But as the players stick together after collision so total mass of the players after collision is

M = (78.5 + 78.5) kg = 157.0 kg

If V be the speed of the entangled players after collision, then, Pf = MV

Or, 545.78 = 157*V

Or, V = 545.78/157 = 3.48

Hence final speed of the players = 3.48 m/s