Question

A 60.0 kg person running at an initial speed of +3.40 m/s jumps onto a 120 kg cart that is initially at rest. The person slides on the cart's top surface and finally comes to rest relative to the cart. The coefficient of kinetic friction between the person and the cart is 0.445. Friction between the cart and the ground can be neglected.

(e) Determine the displacement of the person relative to the ground while he is sliding on the cart.
m

Answer 1

For this question, do i have to add the displacement of the person and the cart?

Answer 2

Displacement of person relative to ground = displacement of person relative to cart + displacement of cart relative to ground

Answer 3

is the displacement of person relative to cart= 3.4t+0.5at^2, where t is the time taken for the guy to stop sliding?

Answer 4

how did you get that?

Answer 5

a=friction/mass, for t, i used momentum conservation and equation of motion. and so s=ut+0.5at^2. however, this gives me the displacement of the person relative to the earth and not relative to the cart, which is what i don't understand.

Answer 6

by any chance is there a diagram for this problem?

Answer 7

@surry99

Answer 8

I am off to bed soon but here are the hints:

1) draw a FBD of the man when he is on the cart. From that, you determine his acceleration and you will be able to determine how far he slides relative to the cart.
(you know vinitial and vfinal and can solve for distance)
2) Draw a FBD of the cart. you will again be able to determine the acceleration of the cart but it will be less than in 1). From this FBD, you will be able to determine how far the cart moves relative to the ground. (you will know vinitial and vfinal and can solve for distance)
3) Adding 1) and 2) should give you the answer you seek.

Answer 9

For 1, if i use s=ut+0.5at^2, i get a different value to what i got using v^2=u^2+2as, why is it like that?

Answer 10

never mind

Answer 11

did you get everything to work out?

Answer 12

it was confusing because the answers that i got were different from the answers in yahoo answers.

Answer 13

where they just did s=ut+0.5at^2 to calculate the displacement and did not add displacement of the cart

Answer 14

hmm...I can work out my answers and send them to you if you like?

Answer 15

wait. ill post how i did and maybe you can check that

Answer 16

@surry99

Answer 17

Note: the actual question says the person was traveling at 4ms and coefficient of 0.4

Answer 18

ok I will update the parameters and will get back to you later today.
I have found a few variations of this question online.

Answer 19

@surry99

Answer 20

question 2B