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At a baseball game, a ball of m = 0.15 Kg moving at a speed of v = 30m/s is caught by a fan.

a. Show that the impulse supplied to bring the ball to rest is 4.5 N x s. Remember to use the correct units (0.25pts)

b. Show that if the ball is stopped in 0.02 s, then the average force of the ball on the catcher’s hand is 225 N. Remember to use the correct units. (0.25pts)

Respuesta :

A. The impulse supplied to bring the ball to rest 4.5 Ns. Details below

B The average force of the ball is 225 N. Details below

What is impulse?

This is defined as the change in momentum of an object.

Impulse = change in momentum

Impulse = final momentum – Initial momentum

Impule = m(v - u)

Where

  • m is the mass
  • u is the initial velocity
  • v is the final velocity

Also,

Impulse = force × time

Therefore

Force × time = final moment – Initial momentum

A. How to determine the impulse

The impulse can be obtainedas illustrated below

  • Mass (m) = 0.15 Kg
  • Initial velocity = 30 m/s
  • Final velocity = 0 m/s
  • Impulse =?

Impulse = m(v + u) since the came to rest

Impulse = 0.15 × (0 + 30)

Impulse = 0.15 × 30

Impulse = 4.5 Ns

Thus, the impulse is 4.5 Ns

B. How to determine the average force

The average force can be obtained as follow:

  • Impulse = 4.5 Ns
  • Time = 0.02
  • Average force = ?

Impulse = force × time

4.5 = average force × 0.02

Divide both sides by 0.02

Average force = 4.5 / 0.02

Average force = 225 N

Thus, the average force is 225 N

Learn more about impulse:

https://brainly.com/question/14079304

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