Question

A man weighing 70 kg runs alongside railroad tracks with a velocity of

Answer 1

Answer:

7.4 km/h

Step-by-step explanation:

Since this involves momentum conservation, apply the principle of linear momentum.

The principle of linear momentum states that:

$m_{1}u_{1}$ + $m_{2} u_{2}$ = ($m_{1}$ + $m_{2}$)V

where: $m_{1}$ is the mass of the man, $u_{1}$ is the velocity of the man,  $m_{2}$ is the mass of the car, $u_{2}$ is the velocity of the car, and V is the common velocity of the man and car.

But, since the car was initially at rest, $u_{2}$ = 0.

70 x 18 + 100 x 0 = (70 + 100)V

1260 = 170V

V = $\frac{1260}{170}$

  = 7.4118

V = 7.4 km/h

The car and man would start to move with a velocity of 7.4 km/h.