# Answer to Question #4053 in Mechanics | Relativity for clueless

Question #4053
The distance between 2 trolleys is D. The mass of 1 trolley is 2 unit, the other is 1 unit. at t= 0 seconds, the distance of the 1 unit trolley is also D distance from the wall. The two trolleys are both moving at the same speed V towards each other initially and they collide head on with each other at t = 5s. Immediately after the collision, both trolleys bounce off in opposite direction and the 1-unit mass trolley reaches its original position at t = 9s. Both trolleys continue their motion and the 1-unit mass trolley eventually hits the wall and bounces back at the same speed at which it hits the wall. At what time will the 1-unit mass trolley hit the 2-unit mass trolley again? (You may assume that friction force is negligible throughout the motion of the two trolleys)
1
2011-08-23T14:04:39-0400
Firstly, let&#039;s find the velocities of trolleys after collision
2*v+1*v=2v2+v1 (assume that friction force is negligible throughout the motion of the two trolleys)
3*v=2v2+v1

We can find v1 = (D/2)/(9-5) = D/8
Also =(D/2)/5=D/10 (before collision trolleys cover distance D/2 each since they have equal velocities)
Hence v2 = (3v-v1)/2 = (3D/10-D/8)/2 = 7D/80
Now we can write the expressions for time of motion of each trolley and then equate these times.
Let the distance covered by 2 unit trolley after collision be x.
So, the total time of motion of 2 unit trolley before the second hitting is
t2 = 5+x/(v2) = 5 + x/(7D/80) = 5+80x/7D
total time of the motion of 1 unit trolley before the second hitting
t1 = 5 + 4 + 4 + (D/2 + x)/v1 = 13 + (D/2 + x)/(D/8) = 13 + 4 + 8x/D = 17 + 8x/D

17 + 8x/D = 5 + 80x/7D

x/D = 7/2

Hence t2 = t1 = t = 5+80x/7D = 5+80/7*7/2 = 45 s

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