Thanks. You have two air cars (X and Y) each with a mass of 2.0 kg approaching e

Question

Thanks.

You have two air cars (X and Y) each with a mass of 2.0 kg approaching each other on a very long frictionless air track that has a mass of 10.0 kg. The air track itself is resting on a very long frictionless sheet of ice. Initially, car X is moving at 0.70 m/s to the right, car Y is moving at 0.50 m/s to the left, and the air track is motionless. Exactly one second after the cars elastically collide, the air pump fails and both cars abruptly stop on the air track. How far apart are the cars 0.50 second after they collide and in what directions are they moving? What are the velocities of the cars and the track 2.0 seconds after the collision? How would all of your numbers change if the cars had equal and opposite velocities before the collision? (Ignore the electro-mechanics needed to operate the track. All velocities are measured by someone standing still next to the system.)

Explanation / Answer

since there is no friction initially and the collision is elastic ,the blocks exchange there velcoties

so X goes to left with 0.5 m/sec

and Y goes to right with 0.7 m/sec

so 1 second after collision distance between them is (0.5 + 0.7) x 1 = 1.2 m (relative speed x time )

when the air pump fails and the cars stop abruptly ,they stop with respect to air track .Since there is no external force on the system momentum should be conserved .

momentum just after collsion and after air pump fails

2(0.5) - 2(0.7) +10 (0) = (2+2+10) V     they will go toether with common velocity

V = -0.0285 m/sec to the left

if they had equal and opposite speed then intial momentum is zero

then V = 0    all of them are at rest after the air pump fails

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