A ball is dropped from a height h0 on a horizontal floor and keeps rebounding after hitting the floor. Find: (i) speed after n rebounds
(ii) Total distance travelled by the ball before stopping
(iii) Height it rises after nth rebound
(iv) Total time before stopping
(coefficient of restitution is e)
(i) en√2gh0, (ii)e2nh0, (iii) h0[1+e21−e2], (iv) (1+e1−e)√2h0g
If a ball is dropped from a height h0 on a horizontal floor, then it strikes with the floor with a speed.
v0 = √2gh0 [From v2 = u2 + 2gh]
and it rebounds from the floor with a speed
v1 = e v0 = e√2gh0 [As e = velocity after collisionvelocity before collision]
first height of rebound: h1 = v212g = e2h0
∴ hn = e2nh0
Height of the ball after nth rebound: Obviously, the velocity of ball after nth rebound will be
Vn = en V0
therefore the height after nth rebound will be hn = v2n2g = e2n h0
∴ hn = e2n h0
Total distance travelled by the ball before it stops bouncing
H = h0 + 2h1 + 2h2 + 2h3 + ...... = h0 + 2e2h0 + 2e4h0 + 2e6h0 + .......
H = h0[1 + 2e2 + 2e4 + 2e6 + .......]
=h0[1 + 2e2(11−e2)] [As 1 + e2 + e4 + .......= 11−e2]
∴ H = h0[1+e21−e2]
Total time taken by the ball to stop bouncing
T = t0 + 2t1 + 2t2 + 2t3 + ....... = √2h0g + 2√2h1g + 2√2h2g + .......
= √2h0g[1 + 2e + 2e2 + .......] [As h1 = e2h0; h2 = e4h0]
= √2h0g[1 + 2e(1 + e + e2 + e3 + ........)] = √2h0g[1 + 2e(11−e)] = √2h0g(11−e)
∴ T = (1+e1−e)√2h0g