## A 1.0-kg cart and a 0.50-kg cart sit at different positions on a low-friction track. You push on the 1.0-kg cart with a constant 4.0-N force

Question

A 1.0-kg cart and a 0.50-kg cart sit at different positions on a low-friction track. You push on the 1.0-kg cart with a constant 4.0-N force for 0.20 m. You then remove your hand, and the cart slides 0.35 m and strikes the 0.50-kg cart. What is the work done by you on the two-cart system? How far does the system’s center of mass move while you are pushing the 1.0-kg cart? By what amount does your force change the kinetic energy of the system’s center of mass?

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1 month 2021-08-04T09:21:19+00:00 1 Answers 3 views 0

a) b) c) Explanation:

From the question we are told that:

Mass of cart 1 Mass of cart 1 Force on  cart 1 Push Distance of cart 1 Slide Distance of cart 1 a)

Generally the equation for work-done is mathematically given by b)

The systems center of mass moved a net totally of (while being pushed)

Mass 1 =0.20m

Mass 2=0

Therefore   c)

Since work-done is equal to K.E energy of cart 1

Therefore    Therefore Kinetic energy before collision is   Generally from the equation for conservation  of momentum the Velocity of cart 2 is mathematically given by   Therefore the final K.E is mathematically given by   Generally the Change in K.E is mathematically given by   Therefore the will force change the kinetic energy of the system’s center of mass by 