A 513 g ball strikes a wall at 12.1 m/s and
rebounds at 13.1 m/s. The ball is in contact
with the wall for 0.045 s.
What is the magnitude of the average force
acting on the ball during the collision?
Answer in units of N.

Respuesta :

The average force on the ball is 287.3 N.

Explanation:

The impulse exerted on an object, which is equal to the product between the force exerted and the duration of the collision, is equal to the change in momentum of the object.

If we apply this to the ball, we can write:

[tex]F \Delta t = m(v-u)[/tex]

where

F is the force exerted on the ball

[tex]\Delta t = 0.045 s[/tex] is the duration of the collision

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

u = 12.1 m/s is the initial velocity of the ball

v = -13.1 m/s is the final velocity (negative since the ball rebounds in the opposite direction)

And solving for F, we find:

[tex]F=\frac{m(v-u)}{\Delta t}=\frac{(0.513)(-13.1-12.1)}{0.045}=-287.3 N[/tex]

So, the magnitude of the average force is 287.3 N.

Learn more about impulse and change in momentum:

brainly.com/question/9484203

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