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Question

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)


A

(i) en2gh0, (ii)e2nh0, (iii) h0[1+e21e2], (iv) (1+e1e)2h0g

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B

(i) e2n2gh0, (ii)e4nh0, (iii) h0[1e21+e2], (iv) (1+e1e)2h0g

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C

(i)en2gh0, (ii)e4nh0, (iii)infinite (iv) infinite

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D

None of these

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Solution

The correct option is A

(i) en2gh0, (ii)e2nh0, (iii) h0[1+e21e2], (iv) (1+e1e)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 = e2gh0 [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(11e2)] [As 1 + e2 + e4 + .......= 11e2]

H = h0[1+e21e2]

Total time taken by the ball to stop bouncing

T = t0 + 2t1 + 2t2 + 2t3 + ....... = 2h0g + 22h1g + 22h2g + .......

= 2h0g[1 + 2e + 2e2 + .......] [As h1 = e2h0; h2 = e4h0]

= 2h0g[1 + 2e(1 + e + e2 + e3 + ........)] = 2h0g[1 + 2e(11e)] = 2h0g(11e)

T = (1+e1e)2h0g


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