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1. A hockey puck has a mass of 0.115-kg and strikes the pole of the net at (37 m/s)x. It bounces off in the opposite direction at (-25 m/s)x +(5 m/s)y. (A) What is the impulse on the puck? (B) If the collision takes .005s what is the force acted on the puck?

Respuesta :

onyebuchinnaji

(a) The impulse on the puck is 1.323 kgm/s.

(b) The magnitude of the force that acted on the puck is 264.5 N.

Impulse of the puck

The impulse on the puck is equal to the change in momentum of the puck.

The change in th momentum of the puck is calculated as follows;

ΔP = m(u - v)

where;

  • v is the final velocity = (-25x + 5y) m/s
  • u is the initial velocity = 37 m/s

Magnitude of the final velocity

|v| = √[(-25)² + (5)²] = 25.5 m/s

ΔP = 0.115(37 - 25.5)

ΔP = 1.323 kgm/s

Thus, the impulse on the puck is 1.323 kgm/s.

Force on the puck

The magnitude of the force that acted on the puck is calculated as follows;

F = ΔP/t

F = 1.323/0.005

F =264.5 N

Learn more about impulse here: https://brainly.com/question/25700778

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