Question

A 11600 kg railroad car is coasting on a level, frictionless track at a speed of 16.0 m/s when a 5420 kg load is dropped onto it. If the load is initially at rest, find the new speed of the car and the % change of the kinetic energy.

Answer 1

Answer:

The speed of the car when load is dropped in it is 10.90 m/s.

Explanation:

Given that,

Mass of the railroad car, m₁ =  11600kg

Mass of the load, m₂ =  5420kg

It can be assumed as the speed of the car, u₁ = 16 m/s

Initially, it is at rest, u₂ = 0

The additional load is dropped onto the car.

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

$m_1u_1+m_2u_2=(m_1+m_2)vv=\dfrac{m_1u_1}{m_1+m_2}v=\dfrac{11600\times 16}{11600+5420}v = 10.90 m/s$

So, the speed of the car when load is dropped in it is 10.90 m/s.

Answer 2

The speed of the car when the load is dropped in it is 10.90 m/s.

Calculation of the speed of the car as follows:

Since

Mass of the railroad car, m₁ =  11600kg

Mass of the load, m₂ =  5420kg

It can be assumed as the speed of the car, u₁ = 16 m/s

So, it is at rest, u₂ = 0

Here we assume the speed of the car be v

So, here we used the conservation of energy

So

m1v1 + m2v2 = (m1v1)/ (m1 + m2)

= 11600*16 / 11600 + 5420

= 10.90 m/s

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