A 2050 kg railroad flatcar, which can move with negligible friction, is motionle

Question

A 2050 kg railroad flatcar, which can move with negligible friction, is motionless next to a platform. A 259 kg sumo wrestler runs at 5.3 m/s along the platform (paale to the track) and then jumps onto the flatcar (a) What is the speed of the flatcar if he then stands on it? m/s (b) What is the speed of the flatcar if he then runs at 5.3 m/s relative to it in his original direction? m/s (c) What is the speed of the flatcar if he then turns and runs at 5.3 m/s relative to the flatcar opposite his original direction? m/s

Explanation / Answer

Let the direction of motion of Sumo be towards right and we take this direction as positive.

Let m be the mass of sumo and u be the velocity of sumo

Let M be the mass of railroad flatcar and v be the velocity of the railroadcar

m = 259 Kg ; ui = 5.3 m/s (vlocity before landing on flatcar)

M = 2050 Kg ; vi = o m/s (since the railroad car is at rest)

If we consider the sumo and the railroad flatcar as the system ,then there is no external force acting on the system.Hence momentum will be conserved.

Total momentum of the system before jumping = Total momentum of the system after jumping

(a) If after jumping the sumo stands on the railroad car they both move with the same velocity i.e the velocity of the combined system(sumo + railraod car )will be same

mui + Mvi = (m+M)V

259X5.3 + 2050X0 = (259+2050)V

solving it we get

V = 0.59 m/s.

The speed of the flatcar if he stands on it is 0.59 m/s.

(b) After jumping the sumo moves with a velocity 5.3 m/s relative to the flatcar in his original direction

Vsr = 5.3 m/s (velocity of sumo w.r.t railroad car)

Vsr = Vs - Vr = 5.3

Vs = 5.3 + Vr

i.e uf = 5.3 + vf

where uf is the velocity of sumo w.r.t ground after jumping

vf is the velocity of the railroad car after jumping

Total momentum of the system before jumping = Total momentum of the system after jumping

mui + Mvi = muf + Mvf

Putting values we have

259 X 5.3 + 2050 X 0 = 259 X (5.3 +vf) + 2050 X vf

1372.7 = 1372.7 +259 vf + 2050 vf

Which gives vf = 0 m/s.

The speed of the railroad flatcar when he runs at 5.3 m/s relative to it in his original direction is 0 m/s. (i.e railroad flatcar is at rest )

(c)

After jumping the sumo moves with a velocity 5.3 m/s relative to the flatcar in opposite direction

Vsr = -5.3 m/s (velocity of sumo w.r.t railroad car) (velocty of sumo w.r.t car is towards left and hence negative)

Vsr = Vs - Vr = -5.3

Vs = -5.3 + Vr

i.e uf = - 5.3 + vf

where uf is the velocity of sumo w.r.t ground after jumping

vf is the velocity of the railroad car after jumping

Total momentum of the system before jumping = Total momentum of the system after jumping

mui + Mvi = muf + Mvf

Putting values we have

259 X 5.3 + 2050 X 0 = 259 X (-5.3 +vf) + 2050 X vf

1372.7 = -1372.7 +259 vf + 2050 vf

2745.4 = 2309vf

vf = 1.19 m/s

The speed of the railroad flatcar when he runs at 5.3 m/s relative to it in opposite direction is 1.19 m/s. (i.e positive 1.19 m/s means it will move towards right i.e in the direction of original speed of sumo)

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