Question

A hockey puck (1) of mass 130 g is shot west at a speed of 8.40 m/s. It strikes a second puck (2), initially at rest, of mass 124 g. As a result of the collision, the first puck (1) is deflected at an angle of 32° north of west and the second puck (2) moves at an angle of 40° south of west. What is the magnitude of the velocity of puck (1) after the collision?

Answer 1

Consider the east-west direction along x-axis with east being in positive x-direction.

Consider the north-south direction along y-axis with north being in positive y-direction.

m₁ = mass of puck 1 = 130 g = 0.130 kg

v_1i = initial velocity of puck 1 = - 8.40 i + 0 j

v_1f = final velocity of puck 1 = - (v Cos32) i + (v Sin32) j

m₂ = mass of puck 2 = 124 g = 0.124 kg

v_2i = initial velocity of puck 2 = 0 i + 0 j

v_1f = final velocity of puck 1 = - (v' Cos40) i - (v' Sin40) j

Using conservation of kinetic energy

m₁ v²_1i + m₂ v²_2i = m₁ v²_1f + m₂ v²_2f

(0.130) (8.40)² + (0.124) (0)² = (0.130) v² + (0.124) v'²

(0.130) v² + (0.124) v'² = 9.2

v'² = (9.2 - (0.130) v² )/(0.124)   

v' = sqrt((9.2 - (0.130) v² )/(0.124))              eq-1                        

Using conservation of momentum

m₁v_1i + m₂v_2i = m₁v_1f + m₂v_2f

(0.130) (- 8.40 i + 0 j) + (0.124) (0 i + 0 j) = (0.130) (- (v Cos32) i + (v Sin32) j) + (0.124) (- (v' Cos40) i - (v' Sin40) j)

comparing the coefficient of "i" both side

(0.130) (- 8.40) = (0.130) (- (v Cos32)) + (0.124) (- (v' Cos40) )

- 1.092 = - (0.11) v - (0.095) v'

using eq-1

- 1.092 = - (0.11) v - (0.095) (sqrt((9.2 - (0.130) v² )/(0.124)))

v = 2.97 m/s