The correct option is D m√2gh(1+e1−e)
When the ball is dropped (u=0) from height h, time taken to reach the floor will be,
t0=√2hg
Its speed while striking floor is
v0=√2gh
After 1stcollision with floor, its rebound speed will be:
v1=ev0
Total time (t1) between 1st & 2nd collision is equal to time of flight of ball, after 1st collision.
∴t1=2v1g
Similarly, total time (t2) between 2nd & 3rd collision is equal to time of flight of ball, after 2nd collision.
t2=2v2g, where v2=ev1
(since ball will hit the ground at same velocity with which its starts)
Now, v2=ev1=e(ev0)=e2v0
Similarly, v3=e3v0
Change in momentum of ball after 1st collision,
Δp1=mv1−(−mv0)=mv1+mv0
Change in momentum of ball after 2nd collision,
Δp2=mv2−(−mv1)=mv2+mv1
Δp3=mv3−(−mv2)=mv3+mv2
Similarly change in momentum of ball after nth collision,
Δpn=mvn−(−mvn−1)=mvn+mvn−1
Therefore, change in momentum of ball:
Δp=Δp1+Δp2+Δp3+......+Δpn ...(i)
=mv0+2mv1+2mv2+.....+2mvn−1+mvn
=mv0+2m(ev0)+2m(e2v0)+.....+2m(en−1v0)+m(env0)
where number of collisions with floor n→∞
∴Δp=mv0+2mev0(1+e+e2+e3+...∞) ...(ii)
Applying formula for sum of infinite GP, where a=1, r=e
∑=a1−r=11−e ...(iii)
From Eq.(ii) & (iii):
Δp=mv0[1+2e(11−e)]
Δp=mv0(1+e1−e)
∴Δp=m√2gh(1+e1−e)