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A 0.111 kg hockey puck moving at 55 m/s is caught by a 80. kg goalie at rest. with what speed does the goalie slide on the frictionless ice?​

Respuesta :

MathPhys

Answer:

0.076 m/s

Explanation:

Momentum is conserved:

m v = (m + M) V

(0.111 kg) (55 m/s) = (0.111 kg + 80. kg) V

V = 0.076 m/s

After catching the puck, the goalie slides at 0.076 m/s.

pstnonsonjoku

The speed with which the goalie slide on the frictionless ice is  0.076 m/s.

From the principle of conservation of linear momentum;

Momentum before collision = momentum after collision

Mass of the  hockey puck = 0.111 kg

Speed of the  hockey puck =  55 m/s

Mass of the goalie = 80 kg

Speed of the goalie = 0 m/s

Following the law of conservation of linear momentum;

(0.111 kg × 55 m/s) + (80 kg  × 0 m/s) = (0.111 kg + 80 kg) V

V = (0.111 kg × 55 m/s) + (80 kg  × 0 m/s)/ (0.111 kg + 80 kg)

V = 6.105/80.111

V = 0.076 m/s

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