Question

A man (weighing 763 N) stands on a long railroad flatcar (weighing 3513 N) as it rolls at 19.8 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 4.68 m/s relative to the flatcar. What is the resulting increase in the speed of the flatcar?

Answer 1

Answer:

0.8m/s

Explanation:

Weight of mas,F=763 N

Mass of man=$\frac{F}{g}=\frac{763}{9.8}=77.86 kg$

By using $g=9.8m/s^2$

Weight of flatcar=F'=3513 N

Mass of flatcar=$\frac{3513}{9.8}=358.5 Kg$

Total mass of the system=Mass of man+mass of flatcar=77.86+358.5=436.36 kg

Velocity of system=19.8m/s

Let v be the velocity of flatcar with respect to ground

Velocity of man relative to the flatcar=$-4.68m/s$

Final velocity of man with respect to ground=v-4.68

By using law of conservation of momentum

Initial momentum=Momentum of car+momentum of flatcar

$436.36(19.8)=77.86(v-4.68)+358.5v$

$8639.928=77.86v-364.3848+358.5v$

$8639.928+364.3848=436.36 v$

$9004.3128=436.36v$

$v=\frac{9004.3128}{436.36}$

$v=20.6 m/s$

Initial speed of flatcar=Speed of system

Increase in speed=Final speed-initial speed=20.6-19.8=0.8m/s