Question

A railway carriage of mass 9000 kg moving with a speed of 36 km h-1 collides
with a stationary carriage of the same mass. After the collision, the carriages
get coupled and move together. Calculate their common speed after collision.
What is the nature of this collision? Give reason for your answer.

Answer 1

use conservation Law of linear momentum ,
Pi = Pf

Pi = initial momentum = mV
Pf = final momentum = ( m + m) V1

so, mV = ( m + m )V1
V1 = V/2 = 36/2 = 18 km/h
speed after collision = 18 km/h

this is an example of enelastic collision in this collision , energy not conserved , e
g loss

Answer 2

solution : -

Mass of the cariage (mc) = 9000 kg = 90,00,000 gram

velocity of carriage (vc) = 35 km/hr = 9.7 m/s

Total momentum of carriage (p1) = mc vc = 9000000 x 9.7 = 87300000 kg.m.s-1

and,

Total momentum of carriage at rest (p2) = 0

Hence, total momentum= 87300000 kg.m.s-1+0= 87300000 kg.m.s-1

i.e. total momentum before collision = 87300000 kg.m.s-1......... (1)

Similarly,

Let us consider the commmon velocity of the both the carriage after collision be 'v',

So, total momentum of the system after collision= 9000000 .v +9000000 .v

or, total momentum after collision = 18000000.v............ (2)

Using principle of conservation of momentum, we get,

total momentum before collision = total momentum after collision

or, 18000000.v = 87300000

∴ v = 4.85 m/s ans



ans: This collision is inelastic collision.