queenkortney172
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A 513 g ball strikes a wall at 14.7 m/s and rebounds at 11.3 m/s. The ball is in contact with the wall for 0.038 s. What is the magnitude of the average force acting on the ball during the collision? Answer in units of N.

Respuesta :

consider the velocity of the ball towards the wall as negative and away from the wall as positive.

m = mass of the ball = 513 g = 0.513 kg

v₀ =  initial velocity of the ball towards the wall before collision = - 14.7 m/s

v = final velocity of the ball away from the wall after collision = 11.3 m/s

t = time of contact with the wall = 0.038 sec

F = average force acting on the ball

using impulse-change in momentum equation , average force is given as

F = m (v - v₀)/t

inserting the values

F = (0.513) (11.3 - (- 14.7))/0.038

F = 351 N


asuwafohjames

The Force acting on the ball during collision is -45.9 N.

To calculate the magnitude of the average force, we use the formula below.

Formula:

  • F = m(v-u)/t............... Equation 1

Where:

  • F = Average force acting on the ball
  • m = mass of the ball
  • v = Final velocity
  • u = Initial velocity
  • t = time.

From the question,

Given:

  • m = 513 g = 0.513 kg
  • v = 11.3 m/s
  • u = 14.7 m/s
  • t = 0.038 s

Substitute these given values into equation 1

  • F = 0.513(11.3-14.7)/0.038
  • F = -45.9 N

Hence, The Force acting on the ball during collision is -45.9 N.
Learn more about force here: https://brainly.com/question/12970081

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