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"A 60 g tennis ball is dropped from a 2 m height, strikes a horizontal sidewalk and rebounds to a height of 1 m. Find the average force with which the ball hits the ground if the collision lasts 0.02s."

Respuesta :

NusrathC

Answer:

F = Changing momentum / time

Explanation:

change in momentum = final momentum - initial momentum

but before finding the force we have to find the initial and final vertical velocity of the ball in both cases (1st bounce and 2nd bounce)

so first let's find the vertical velocities of the ball in the first bounce

  • here the ball dropped - it means (initial velocity is zero 0ms-1)
  • but there is a final vertical velocity...

let's find the final vertical velocity

v² = u² + 2as (here a = g = 10ms^-2)

v = √0²+2(10)2

v = 6.32ms^-1

After that let's find the vertical velocities of the ball in the second bounce

so here our final velocity is zero (because vertical velocity in maximum heigh )is zeo

then let's find our initial vertical velocity

v² = u² + 2as

0 = u² + 2(-10)1

(here -10 cause gravitational acceleration act positive only in the downward direction )

u = √20

u = 4.47ms^-1

ok, now we found all the velocities then let's find the force

F = Changing momentum / time

changing momentum = impulse

= mv-m(-u)

= 0.06 (6.32+4.47)

=0.6474 Ns

so we found the chang in momentum

the let's find the force

F = Changing momentum / time

F = 0.6474 / 0.02

F = 32.37N

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