An m = 65.0-kg person running at an initial speed of v = 3.50 m/s jumps onto an

Question

An m = 65.0-kg person running at an initial speed of v = 3.50 m/s jumps onto an M = 130-kg cart initially at rest (figure below). 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 0.320. Friction between the cart and ground can be ignored. (Let the positive direction be to the right.)

(a) Find the final velocity of the person and cart relative to the ground. (Indicate the direction with the sign of your answer.)

(b) Find the friction force acting on the person while he is sliding across the top surface of the cart. (Indicate the direction with the sign of your answer.) N

(c) How long does the friction force act on the person? s

(d) Find the change in momentum of the person. (Indicate the direction with the sign of your answer.) N · s

Find the change in momentum of the cart. (Indicate the direction with the sign of your answer.) N · s

(e) Determine the displacement of the person relative to the ground while he is sliding on the cart. (Indicate the direction with the sign of your answer.) m

(f) Determine the displacement of the cart relative to the ground while the person is sliding. (Indicate the direction with the sign of your answer.) m

(g) Find the change in kinetic energy of the person. J

(h) Find the change in kinetic energy of the cart. J

(i) Explain why the answers to (g) and (h) differ. (What kind of collision is this one, and what accounts for the loss of mechanical energy?)

Explanation / Answer

person, m1 = 65.0-kg

initial speed of v1 = 3.50 m/s

cart, m2 = 130-kg

The coefficient of kinetic friction between the person and the cart is 0.320.

a)
by conservation of momentum,
v1m1 = v(m1+m2)
v = v1m1/(m1+m2)
v = 65*3.5/195 = 1.166 m/s

b)

friction force = mg = 65*9.81*0.320 = 204.048 N

d)

loss in momentum of person = m1v1 - m1v = 65(3.5-1.166) = 151.671 kgm/s

gain of momentum of cart = m2v = 130*1.166 = 151.671 kgm/s

c)

impulse = change of momentum = friction force*time
151.671 Ns = 204.048 N*t
t = 151.671/204.048 = 0.74s

g)
Ekin = 1/2 mvi^2 - 1/2 mvf^2
= 1/2*65*3.5^2 - 1/2*65*(1.166)^2 = 353.93 kg.m2/s2

h)
Ekin = 1/2 mv^2 = 1/2*130*(1.166)^2 = 88.46 kg.m2/s2

e)

total loss of kinetic energy = 353.93 kg.m2/s2 - 88.46 kg.m2/s2 = 265.47Kg.m2/s2
This loss can only be due to friction work:
friction work = friction force*distance:
265.47 = 204.048 N* d
d = 265.47/204.048 N = 1.301 m (along the top of the cart, not against ground).

for the cart we have
v = at
1.166 = a*0.74s
a = (1.166)/0.74 = 1.58 m/s^2 and with
s = 1/2 at^2
s = 1/2*1.58*0.74^2
s = 0.432 m
so total displacement of person against ground = 1.301 m + 0.432 m = 1.733 m

f)

displacement of the cart, s=0.432 m

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