Question

28. Three identical train cars, coupled together are rolling east at 2m/s. A fourth car travelling éast at 4 m/s catches up with the three and couples to make a four car train. The four car train rolls down the track without losing speed untill they collide with and couple to a fifth car that is stationary on the train tracks. if the fifth car is identical to other four, what is the speed of five- car train.​

Answer 1

Given :

There are in total 5 identical train cars which collide with each other after some interval of time

• Let the Train Cars be : A,B,C,D and E
• So, Velocity of A,B and C = 2m/s
• Velocity of D = 4m/s
• Velocity of E = 0m/s

Let's Assume: Let's Assume that the mass of all the carts is 'x', as all are identical Coaches

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To Find :

The common Velocity of the 5 Car train (A+B+C+D+E) if they collide with each other

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Law To be used :

We will use the law of conservation of momentum, which states that the initial momenta of objects = Final momenta of objects

${\implies\displaystyle \rm m_1 u_1 +m_2 u_2 = m_1 v_1+m_2 v_2}$

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Solution :

Given, Mass of Train Cars = x (Assumed)

Velocity of A,B,C = 2m/s

So, they are travelling on the track, so if they collide with D, their Initial momenta will be the same as final momenta.

As Given, their final velocity are equal . So,

${\implies \displaystyle \rm a_{ip} + b_{ip} +c_{ip} + d_{ip} = a_{vp} + b_{vp}+c_{vp} + d_{vp}}$

Where, IP = Initial momentum

VP = final momentum

${\implies \displaystyle \rm (2 \times x) + (2 \times x) + (2 \times x) + (4 \times x) =(x \times v) + (x \times v) + (x \times v) + (x \times v) }$

${\implies \displaystyle \rm 2x + 2x + 2x + 4x = v(x + x + x + x)}$

$\implies \displaystyle \rm 10x = 4xv$

$\implies \displaystyle \rm v = \frac{10x}{4x}$

$\implies \displaystyle \rm v = 2.5mps$

Hence, The final velocity of (A,B,C,D) or Four Car train = 2.5 m/s

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Now, After sometime travelling with same velocity, they collide with a stationary train car of same mass and it also start moving with same velocity .

${\implies \displaystyle \rm a_{ip} + b_{ip} +c_{ip} + d_{ip} + \displaystyle \rm e_{ip} = a_{vp} + b_{vp}+c_{vp} + d_{vp} + \displaystyle \rm e_{vp} }$

${\implies \displaystyle \rm (10x) + (0 \times x) = (v \times x) + (v \times x) + (v \times x) + (v \times x) + (v \times x)}$

${\implies \displaystyle \rm (10x) + 0 = v(x + x + x + x + x)}$

${\implies \displaystyle \rm 10x = 5xv}$

$\implies \displaystyle \rm v = \frac{10x}{5x}$

$\blue{ \underline{ \boxed{ \therefore \displaystyle \rm v = 2mps}}}$

Hence, The speed of 5 Car train (A+B+C+D+E) = 2m/s

Final Answer :

If the train moves with common Velocity, then:

• the velocity of 5 Car train = 2m/s