Question

A 16000 kg railroad car travels along on a level frictionless track with a constant speed of 23.0 m/s. A 5400 kg additional load is dropped onto the car. What then will be its speed in meters/second?

Answer 1

Answer:

The speed of the car when load is dropped in it is 17.19 m/s.

Explanation:

It is given that,

Mass of the railroad car, m₁ = 16000 kg

Speed of the railroad car, v₁ = 23 m/s

Mass of additional load, m₂ = 5400 kg

The additional load is dropped onto the car. Let v will be its speed. On applying the conservation of momentum as :

$m_1v_1=(m_1+m_2)v$

$v=\dfrac{m_1v_1}{m_1+m_2}$

$v=\dfrac{16000\ kg\times 23\ m/s}{(16000+5400)\ kg}$

v = 17.19 m/s

So, the speed of the car when load is dropped in it is 17.19 m/s. Hence, this is the required solution.