Question

A 7150-kg railroad car travels alone on a level frictionless track with a constant speed of a 3350-kg load, initially at rest, is dropped onto the car. what will be the car's new speed? the positive direction is the direction of the original motion of the car.

Answer 1

Answer:

v = 10.21 m/s

Explanation:

It is given that,

Mass of the railroad car, m₁ = 7150 kg

Mass of the load, m₂ = 3350 kg

It can be assumed as the speed of the car, u₁ = 15 m/s

Initially, it is at rest, u₂ = 0

Let v is the speed of the car. It can be calculated using the conservation of momentum as :

$m_1u_1+m_2u_2=(m_1+m_2)v$

$v=\dfrac{m_1u_1}{m_1+m_2}$

$v=\dfrac{7150\times 15}{7150+3350}$

v = 10.21 m/s

So, the new speed of the car is 10.21 m/s. Hence, this is the required solution.