Question

A railroad car is moving along a straight frictionless track. In each of the following cases, the car initially has a total mass (car and contents) of 200 kg and is travelling with a velocity of 4 m/s. Find the final velocity of the car in each of the three cases.
(a) A 20 kg mass is thrown sideways out of the car with a velocity of 2 m/s relative to the car.
(b) A 20 kg mass is thrown backward out of the car with a velocity of 4 m/s relative to the car.
(c) A 20 kg mass is thrown into the car with a velocity of 6 m/s relative to the ground and opposite in direction to the velocity of the car.

Answer 1

Hint:In this question, we will use conservation of momentum. When there is no external force acting on an object , momentum is conserved. Here, there is no external force acting on the object so momentum will be conserved.

Complete step by step solution:
In the question, it is given that a railroad car is moving a track and there is no friction.
m 1 = 200 kg (Given)
v 1 = 4 m/s (Given)
Now let us consider the given three cases one by one:
(a) In this case, the object is thrown sideways.
According to the law of conservation of momentum, when either two or more bodies are made to act upon one another, the total momentum remains same or constant (i.e. conserved) provided that no external forces are acting. Thus in the present case:
$
{m_1}{v_1} = {m_2}{v_2} \\
200 \times 4 = 200 \times {v_1} \\
{v_1} = \dfrac{{800}}{{200}} = 4m/s \\
$
The railroad car is moving on the straight track and there is no change in momentum in the direction of the motion of the car. So, when a body is thrown sideways, no change will take place in the velocity of the car.
(b) As we know, \[p = mv\]
So, initial momentum would be:
\[p = 200 \times 4\left[ {east} \right]\]
Final momentum will include two terms (1 is the car, and 2nd is the other mass); let us assume east is the positive direction. Thus, final momentum would be:
\[p = {m_1}{v_1} + {m_2}{v_2}\]
According to law of conservation of momentum, initial momentum should be equal to the final momentum, thus:
$
200 \times 4 = 200 \times {v_1} + 20 \times 2 \\
{v_1} = 3.8m/s \\
$
As the body is thrown backward out of the car then the velocity of the car will be3.8m/s
(c) Again, we will apply conservation of momentum. First we will find the initial momentum which includes two terms. Initial momentum would be:
${m_1}{v_{1i}} + {m_2}{v_{2i}} = 150 \times 4.60 - 21 \times 6$
Final momentum would be:
$\left( {{m_1} + {m_2}} \right){v_f} = (150 + 21){v_f}$
By conservation of momentum:
$
150 \times 4.60 - 21 \times 6 = (150 + 21){v_f} \\
{v_f} = 3.3m/s \\
 $

Note: Conservation of momentum is a direct consequence of Newton's third law. Momentum is not conserved in case there is gravity, friction, or net force (net force means just the total amount of force).