Momentum, friction, and distance?

Question

Fred (mass 75 kg) is running with the football at a speed of 5.3 m/s when he is met head-on by Brutus (mass 120 kg), who is moving at 4.0 m/s. Brutus grabs Fred in a tight grip, and they fall to the ground. How far do they slide? The coefficient of kinetic friction between football uniforms and Astroturf is 0.30.

I have:

P1 = (75)(5.3) = 397.5
P2 = (120)(-4.0) = -480

Total Momentum System: 5.3m - 4.0m = 0.7m


I am lost as to where to go from there. Do I need the sum of forces in the x-direction?


Thanks

Answer 1

good start.

total inital momentum = 480-397.5

this is an inelasitc collision, and aferwards you have
both masses travelling at velocity v:

(75+120)v = 480-397.5

solve for v. This is the velocity at which they both slide.
The inital energy now is 1/2(75+120)v^2

This energy is converted to frictional energy:

inital kinetic energy = total work of friction = Fr(distance)

Fr = uN = u(120+75)g = .3(195)(9.8)

so we have

.3(195)(9.8)(distance slided) = 1/2(195)v^2

You know v form above. So find distance slided.

Answer 2

Total Momentum =P1-P2=(m1 + m2) * V
then you obtain V which is initial velocity of the men.
acceleration =(m1 + m2) * g * coefficient of kinetic friction /(m1 + m2) then
a=g * coefficient of kinetic friction
now you have acceleration and initial velocity. final speed is 0.
V2^2 - V1^2 = 2 a x
x=V^2/(2a)

Answer 3

You can use conservation of energy.
The resultant kinetic energy of the collision is total mass moving at .7m/s
The loss of energy is the force of friction times the displacement.

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