Linear Momentum Question?

Question

A man (weighing 915 N) stands on a long railroad flatcar (weighing 3270 N) as it rolls at 17.5 m/s in the positive direction of an x axis, with negligible friction. Then the man runs along the flatcar in the negative x direction at 55.00 m/s relative to the flatcar. What is the resulting increase in the speed of the flatcar?

Help, I only have one submission left and still cannot get this.

Answer 1

Initially:

Mass of man, m1 = 915 N

Velocity of man (relative to flatcar), u1 = 0 (since he is standing stationary)

Mass of flatcar, m2 = 3270 N

Velocity of flatcar, u2 = 17.5 m/s

Finally;

Mass of man, m1 = 915 N

Velocity of man (relative to flatcar), v1 = -55 m/s (minus sign indicates that his direction of velocity is opposite to that of the  velocity of the flatcar, which is positive)

Mass of flatcar, m2 = 3270 N

Velocity of flatcar, v2 = ?

Use the Principle of Conservation of Momentum :

m1u1 + m2u2 = m1v1 + m2v2

This gives v2 =  31 m/s

Note that v2 is positive which shows that it is moving in the same direction as before but with a greater speed.

Hope this helps.

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Answer 2

Before he runs (M + m)v is the combined momentum for the car W = Mg = 3270 N, the man w = mg = 915 N, and the velocity v = 17.5 mps wrt ground.  g = 9.81 m/sec^2; note these W and w are weights, not masses.  The goods news, as you shall see is that using W and w instead of M and m would also work.

While he runs Mu + mV is the combined momentum where u = ? the car's velocity wrt ground you are looking for, V = ? mps wrt ground for the running man.  The man's velocity V' = 55 mps relative to the car.  Thus, the V wrt ground is u - V' as the man is running in a direction opposite to the car's direction (plus).  This means his velocity wrt ground will be the difference between the car's velocity and his relative velocity wrt car.

The before and after are equal; so (M + m)v = Mu + mV = Mu + m(u - V') = Mu - mV' + mu = (M + m)u - mV'  Solve for u.  [(M + m)v + mV']/(M + m) = u  Everything on the LHS is known, you can do the math.  Note, as promised, if we muliply the LHS by (g/g) = 1.0, each of the M and m would be converted to W and w respectively.  So we can solve for u using the weights given rather than find the masses.  That is [(W + w)v + wV']/(W + w) = u  is also true.

The physics is of course the conservation of momentum.  But the trick is that momentum is measured with respect to the ground, not wrt the car.  So all the velocities have to be written wrt ground, which means that 55 mps relative to the car has to be converted to velocity wrt ground.

Answer 3

First of all, always try to do momentum problems with frame of reference as ground So make all the velocities wrt ground this is "THE" rule of solving these momentum problems

Secondly apply momentum impulse law !!! :P

m= weight of man M= weight of railroad

let the final velocity of M is V w.r.t ground (+ve x axis)

velocity of m w.r.t ground is  -55+V ( -55 coz -ve x axis)

now

initial momentum= (M+m)17.5

final momentum = M(V) + m( -55+V)

so final V is

((M+m) 17.5 + 55m)/(M+m)

put da values now,

Answer 4

P1 = P2

V1*(W+w) = W*u + (u-55)*w →

u = [V1(W+w)+55w]/(W+w) = [17.5(4185)+55*915]/(4185)

u = +29.53 m/s

So, the INCREASE in the car speed (your actual question) is

∆V = 29.53-17.5 = 12.03 m/s

.........Just thought of an easier way........

Assume initial V = 0;  Then w*(55-∆V) = W*∆V → ∆V = 12.025 m/s