Question

A railroad car having a mass of 17.5 Mg is coasting at 1.5 m/s on a horizontal track. At the same time another car having a mass of 12 Mg is coasting at 0.75 m/s in the opposite direction. The cars meet and couple together.
Determine the speed of both cars just after the coupling.

Answer 1

Answer:

Then final velocity of the coupled system of cars will be 0.58 m/s

Explanation: Let us consider the car-1 is moving towards right (say + ve direction) and car-2 is moving towards left (say in - ve direction), accordingly velocity are considered +ve and -ve.

$m_{1} = 17.5 \ Mg = 17.5 \times 10^{6} \ g = 17.5 \times 10^{3} \ Kg = 17500 \ Kg$

$v_{1} = + \ 1.5 \ m/s$

$m_{2} = 12 \ Mg = 12 \times 10^{6} \ g = 12 \times 10^{3} \ Kg = 12000 \ Kg$

$v_{2} = - \ 0.75 \ m/s$

Applying the conservation of momentum, and let the final velocity of combined system is V m/s

$m_{1} \times v_{1} + m_{2} \times v_{2} = (m_{1} + m_{2}) \times V$

$17500 \times 1.5 \ + 12000 \times (-0.75) = 29500 \times V$

26250 - 9000 = 29500 $\times$ V

V = $\frac{17250}{29500}$ = 0.58 m/s

Then final velocity of the coupled system of cars will be 0.58 m/s

Answer 2

Answer:

The speed of both cars just after the coupling is 0.584 m/s.

Explanation:

Given that,

Mass of car = 17.5 Mg

Speed of car= 1.5 m/s

Mass of another car = 12 Mg

Speed of another car = 0.75 m/s

We need to calculate the speed of both cars just after the coupling

Using conservation of momentum

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

$v=\dfrac{m_{1}u_{1}+m_{2}u_{2}}{(m_{1}+m_{2})}$

Where, m₁ = mass of one car

m₂ = mass of another car

v₁  = velocity of one car

v₂ = velocity of another car

Put the value into the formula

$v=\dfrac{17.5\times10^{3}\times1.5-12\times10^{3}\times0.75}{17.5\times10^{3}+12\times10^{3}}$

$v=0.584\ m/s$

Hence, The speed of both cars just after the coupling is 0.584 m/s.