A70.0 kg70.0 kg ice hockey goalie, originally at rest, has a 0.170 kg0.170 kg hockey puck slapped at him at a velocity of 41.5 m/s.41.5 m/s. suppose the goalie and the puck have an elastic collision, and the puck is reflected back in the direction from which it came. what would the final velocities of the goalie and the puck be in this case? assume that the collision is completely elastic.

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

i think it is non rem sleep

i believe the answer is electric potential.

explanation: