Question

Two ice pucks (one orange and one blue) of equal mass are involved in a perfectly elastic glancing collision. The orange puck is initially moving to the right at voi = 6.55 m/s, strikes the initially stationary blue puck, and moves off in a direction that makes an angle of ? = 40.0° with the horizontal axis while the blue puck makes an angle of ? = 50.0° with this axis. Note that for an elastic collision of two equal masses, the separation angle ? + ? = 90.0°. Determine the speed of each puck after the collision in meters per second. v0f = ? vbf = ?

Answer 1

Mass of the pucks m = m

Initial velocity of orange puck, v_oi = 6.55 m/s

Initial velocity of the red puck, v_bi = 0

Angle of orange puck = 40.0^o

Angle of blue puck ? = 50.0°

In vertical direction :

Apply law of conservation of momentum,

0 = mv_of sin + m(- v_bf sin ? )

v_of sin = v_bf sin ?

v_of sin 40 = v_bf sin 50

   v_of = 1.1918 v_bf       ----------( 1)

In horizontal direction:

mv_oi + m(0) = mv_of cos + m(v_bf cos ? )

             v_oi = v_of cos + v_bf cos ?

6.55 = 1.1918*v_bf* cos 40 + v_bf*cos 50

               = 0.9129 v_bf + 0.6428 v_bf

               = 1.56 v_bf

           v_bf = 4.21 m/s

Substitute v_bf value in v_of = 1.1918 V you get ,

    v_of = 1.1918(4.21)

      = 5.02 m/s

i.e.,

v_of = 5.02 m/s
v_bf = 4.21 m/s