Answer:
0.2012 m/s
- 41.3 m/s
Explanation:
M = mass of ice hockey goalie = 70 kg
V = initial velocity of the hockey goalie = 0 m/s
V' = final velocity of hockey goalie after collision = ?
m = mass of hockey puck = 0.170 kg
v = initial velocity of hockey puck = 41.5 m/s
v' = final velocity of hockey puck = ?
Using conservation of momentum
M V + m v = M V' + m v'
(70) (0) + (0.170) (41.5) = (70) V' + (0.170) v'
7.055 = (70) V' + (0.170) v'
V' = 0.1008 - 0.00243 v' eq-1
Using conservation of kinetic energy
(0.5) M V² + (0.5) m v² = (0.5) M V'² + (0.5) m v'²
M V² + m v² = M V'² + m v'²
(70) (0)² + (0.170) (41.5)² = (70) V'² + (0.170) v'²
292.8 = (70) V'² + (0.170) v'²
Using eq-1
292.8 = (70) (0.1008 + 0.00243 v')² + (0.170) v'²
v' = - 41.3 m/s
Using eq-1
V' = 0.1008 - 0.00243 v'
V' = 0.1008 - 0.00243 (- 41.3)
V' = 0.2012 m/s
