# 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)

Expert's answer

Firstly, let's find the velocities of trolleys after collision

2*v+1*v=2v

3*v=2v

We can find v

Also =(D/2)/5=D/10 (before collision trolleys cover distance D/2 each since they have equal velocities)

Hence v

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

t

total time of the motion of 1 unit trolley before the second hitting

t

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

x/D = 7/2

Hence t

Answer t = 45 s

2*v+1*v=2v

_{2}+v_{1}(assume that friction force is negligible throughout the motion of the two trolleys)3*v=2v

_{2}+v_{1}_{}We can find v

_{1 }= (D/2)/(9-5) = D/8Also =(D/2)/5=D/10 (before collision trolleys cover distance D/2 each since they have equal velocities)

Hence v

_{2 }= (3v-v_{1})/2 = (3D/10-D/8)/2 = 7D/80Now 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

t

_{2 }= 5+x/(v_{2}) = 5 + x/(7D/80) = 5+80x/7Dtotal time of the motion of 1 unit trolley before the second hitting

t

_{1 }= 5 + 4 + 4 + (D/2 + x)/v_{1 }= 13 + (D/2 + x)/(D/8) = 13 + 4 + 8x/D = 17 + 8x/D17 + 8x/D = 5 + 80x/7D

x/D = 7/2

Hence t

_{2 }= t_{1 }= t = 5+80x/7D = 5+80/7*7/2 = 45 sAnswer t = 45 s

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