A cart of mass m moves at a speed u on a frictionless surface. At regular intervals of length l, blocks of mass 2 m m drops vertically into the cart. How much time is taken to cover a distance of l 2 9 ?
Answers
I hope there is a small error with the question and the right question would be,
A cart of mass M moves at a speed u on a frictionless surface. At regular intervals of length L, blocks of mass m = M/2 drops vertically into the cart. How much time is taken to cover a distance of 9/2 L?
Answer :
Take two points A and B,
From A to B speed is U,
so time t AB = L / 4 .... (1)
where L is the distance between the points.
From B to C,
By conservation of momentum,
Mu = ( M + M/2 ) u₁ ⇒ u₁ =2/3 u
so, time from B to C is t BC = L / ( 2/3 u ) ..... (2)
From C to D,
By conservation of momentum,
3/2 Mu₁ = 2 Mu₂
3/2 M * 2/3u = 2 Mu₂
u₂ = u/2
Time C to D
t CD = L / (u/2) ............(3)
From D to E,
By conservation of momentum,
2 Mu₂ = 5M/2 u₃
2 M U/2 =5M /2 u₃ ⇒ u₃ = 2/5 u
Time from D to E
t DE = L / (2u/5) ..............(4)
From E to F,
By conservation of momentum,
5M/2 u₃ = 3M u₄
5M/2 * 2u/5 = 3M u₄ ⇒ u₄ =u/3
Time from E to F,
t EF = L/2 / u₄ = L/2(u/3) ..............(5)
Total time,
t = t AB + t BC + t CD + t DE + t EF
= L / u + 3L / 2u + 2L / u + 5L / 2u + 3L / 2u = 17 L / 2u
Time taken to cover a distance of 9/2 L = 17 L / 2u