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?

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

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14