Question

(1) A table tennis ball is dropped

onto the floor from a height of
4m and it rebounds to a height of
3m. If the time of contact with the floor is 0.01s, what is the magnitude and direction of the acceleration during the contact.

Answer 1

Here, h1=4.00m,h92)=3.00m,Delta t =0.01 s.Letv1 be the velocity of the ball (actind downwards) just before striking the floor an dv20 be the velocity of the ball (acting upwaeds) ust after striking the floor. Then, change in velocity of the ball in time Δtv2−(−v1)v2+v1

:. acceleration, a=v2+v1Δt ..(i)

When body falls from height h1,

then u=o,v1,a=gandS=h1

As, v21=u2+2aS,

:. v_(1) ^(2) =0 + g h_(10 or v12–√gh1

Taking motion of the ball after striking the floor, then u=v2,v=0,a=−g,S=h2

As, v2=u22as,∴v21=0+2gh1orv_(1) =sqrt 32 g h_(1)Tak∈gmotionoftheballa>erstrik∈gthe⌊,⌋thenu=v_(2), v=0, a=- g, S=h_(2)As,v2=u2+2aS,wehave0 = v_(2)^(2) +2 (-g) h_92) otr v2=2–√gh2

Putting values in (i) we get,

a=2–√gh2+2–√gh1Δt)

a=2–√×9.8×3+2–√×9.8×40.01

= 1652m/s2.

Explanation: