Answer to Question #4053 in Mechanics | Relativity for clueless
2*v+1*v=2v2+v1 (assume that friction force is negligible throughout the motion of the two trolleys)
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
Answer t = 45 s
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