Question

Two hockey pucks are moving to the right with puck 1 behind puck 2. Puck 1 is moving twice as fast as puck 2, but its mass is half that of puck 2. If puck 2's Vi is 8 m/s, then what is the velocity of each puck after they collide?

Answer 1

m₁ = mass of puck 1 = m

m₂ = mass of puck 2 = 2m

V_1i = initial velocity of puck 1 = 16 m/s

V_2i = initial velocity of puck 2 = 8 m/s

V_1f = final velocity of puck 1

V_2f = final velocity of puck 2

Using conservation of momentum

m₁ v_1i + m₂ V_2i = m₁ V_1f + m₂ V_2f

m (16) + (2m) (8) = m V_1f + (2m) V_2f

32 = V_1f + 2 V_2f

V_1f = 32 - 2 V_2f                               eq-1

Using conservation of kinetic energy ::

(0.5) m₁ v²_1i + (0.5)m₂ V²_2i = (0.5)m₁ V²_1f + (0.5)m₂ V²_2f

m (16)² + (2m) (8)² = m V²_1f + (2m) V²_2f

384 = (32 - 2 V_2f)² + 2 V²_2f                        using eq-1

V_2f = 8 or 13.33

V_1f = 32 - 2 V_2f     

so V_1f = 16 or 5.34 m/s