A 60.0 kg person running at an initial speed of 3.25 m/sjumps onto a 120 kg cart

Question

A 60.0  kg person running at an initial speed of 3.25  m/sjumps 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 0.405. Friction between the cart and the ground can be neglected.

I NEED HELP ON PARTS C, E, F AND G.

Part A

Find the final speed of the person and cart relative to the ground

1.08 m/s

Part B

For how long does the frictional force act on the person?

.546 s

Part C

Determine the magnitude of the displacement of the person relative to the ground while he is slipping on the cart.

Part D

Determine the magnitude of the displacement of the cart relative to the ground while the person is slipping on the cart.

.296 m

Part E

Use the definition of the work done by a constant force (that definition is F r ) to find the work done on the person by the force of kinetic friction (on the person by the cart) while he is slipping on the cart. (The frame of reference for this question is that of a stationary observer on the Earth.)

Part F

Use the definition of the work done by a constant force (that definition is F r ) to find the work done on the cart by the force of kinetic friction (on the cart by the person) while the person is slipping on the cart. (The frame of reference for this question is that of a stationary observer on the Earth.)

Part G

How much kinetic energy was lost in this perfectly inelastic collision?

1.08 m/s

Part B

For how long does the frictional force act on the person?

.546 s

Part C

Determine the magnitude of the displacement of the person relative to the ground while he is slipping on the cart.

Part D

Determine the magnitude of the displacement of the cart relative to the ground while the person is slipping on the cart.

.296 m

Part E

Use the definition of the work done by a constant force (that definition is F r ) to find the work done on the person by the force of kinetic friction (on the person by the cart) while he is slipping on the cart. (The frame of reference for this question is that of a stationary observer on the Earth.)

Part F

Use the definition of the work done by a constant force (that definition is F r ) to find the work done on the cart by the force of kinetic friction (on the cart by the person) while the person is slipping on the cart. (The frame of reference for this question is that of a stationary observer on the Earth.)

Part G

How much kinetic energy was lost in this perfectly inelastic collision?

Explanation / Answer

a)
by conservation of momentum,
v1m1 = v(m1+m2)
v = v1m1/(m1+m2)
v = 60*3.25/180 = 1.083 m/s
b)
friction force = mg = 60*9.81*0.405 = 238.4 N

c)
loss in momentum of person = mvi - mvf = 60(3.25 - 1.083) = 130 kgm/s
gain of momentum of cart = mv = 120*1.083= 130kgm/s
impulse = change of momentum = friction force*time
130 Nm =238.4 N*t
t = .545s

change in kinetic energy of the person

Ekin = 1/2 mvi^2 - 1/2 mvf^2
= 1/2*60*3.25^2 - 1/2*60*(1.083)^2 = 281.68 Nm

change in kinetic energy of the cart.

Ekin = 1/2 mv^2 = 1/2*120*(1.08)^2 = 69.98 Nm

total loss of kinetic energy = 281.68 Nm - 69.98 Nm = 211.7 Nm
This loss can only be due to friction work:
friction work = friction force*distance:
211.7 Nm = 238.4 N* d
d = .888 m (along the top of the cart, not against ground).

for the cart we have
v = at
1.083 = a*0.545s
a =1.98 m/s^2 and with
s = 1/2 at^2
s = 1/2*1.98*0.545^2
s = 0.289 m

so total displacement of person against ground = .888 m + 0.289 m = 1.177m

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