A ball is dropped on to a horizontal plate from a height h = 9 m above it. If the coefficient of restitution is e = 1/2, the total distance travelled before the ball comes to rest is 

Options

A. 10 m
B. 15 m
C. 20 m
D. 25 m

Answer

B

Explanation

The impact between the ball and the plate is direct impact. If u is the velocity of the ball before impact , its veloicty after impact is eu, where e is the coefficient of restitution.
But `u = sqrt(2gh)` since the ball falls from a height h.
So the velocity of the ball at the first rebound
`v_1 = eu = esqrt(2gh)`
The height `h_1` to which it rises after first rebound
`h_1 = (v_1^2)/(2g) = (e^2. 2gh)/(2g) = e^2 h`
The velocity `v_2` after second rebound, `v_2 = ev_1 = e^2u`.
The height `h_2` it rises after second rebound
`h_2 = (v_2^2)/(2g) = (e^4. u^2)/(2g) = (e^4.2gh)/(2g) = e^4h`
In general , the height `h_n` to which the ball rises after `n^(th)` rebound is given by `h_n = r^(2n)h.`
Total distance travelled.
`= h + 2h_1 = 2h_2 + 2h_3 + ....+ 2h_n +.....to oo`
`= h + 2(e^2 h + e^4 h + e^6 h + .......) = h = 2he^2 (1 + e^2 + e^4 + .....)`
`= h + (2he^2)/(1 + e^2) = h (1 + (2e^2)/(1 - e^2)) = (h (1 + e^2))/((1 - e^2)) = (9(1+ (1/2)^2))/((1- (1/2)^2)) = 15 m`