A hockey puck of mass 0.16 kg, sliding on a nearly frictionless surface of ice with a velocity of 2,0 m/s[E], strikes a second puck at rest with a mass of 0.17kg. The first puck has a velocity of 1.5 m/s [N 31 E] after the collision. Determine the velocity of the second puck after the collision.
Ash
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Momentum is conserved in elastic collision
Let velocity in East and North direction be positive, while West and South direction be negative
Conservation of momentum in East-West direction
m₁ux₁ + m₂ux₂ = m₁vx₁ + m₂vx ₂
(0.16 kg)(2.0 m/s) + (0.17 kg)(0) = (0.16 kg)(1.5sin31 m/s) + (0.17 kg)(vx ₂)
0.32 = 0.24sin31 + 0.17 vx ₂
vx ₂ = (0.32 - 0.24sin31)/0.17
vx ₂ = 1.16 m/s
Conservation of momentum in North-South direction
m₁uy₁ + m₂uy₂ = m₁vy₁ + m₂vy ₂
(0.16 kg)(0) + (0.17 kg)(0) = (0.16 kg)(1.5cos31 m/s) + (0.17 kg)(vy ₂)
0 = 0.24cos31 + 0.17 vy ₂
vy ₂ = (- 0.24cos31)/0.17
vy ₂ = -1.2 m/s
Resultant velocity = √[(1.16)² + (-1.2)²] = 1.67 m/s
angle θ = arctan(-1.2/1.16) = -46°
This represents E 46 S
The second puck moves, after collision, at 1.67 m/s at [E 46 S]