A particle (m=5 kg) hits another identical particle. If v^0= 10m/s= initial speed. What are both...

A) by conservation of momentum.. mV0=2mV

V=v0/2=5m/sec

B) the collision is elastic means velocity of separation =velocity of approach

velocity of approach =v0=10m/sec

let the v1 and v2 be the velocities of both particles respectively

then mV0= mv2-mv1

and also v0=v1+v2(as collision is elastic)

so v1=0 v2=10m/sec

C) applying conservation of momentum..

mV0=mV2-m10

v2=20m/sec.

a) conserving momentum

5*10 = 2*5*V( both have same speed)

V = 5m/sec

b) for elastic collision e = 1

  e = U1-U2/V2-V1

and conserving momentum

U1 = V1+V2 ( since masses are same )

by above two equations

V1 = 0 V 2 = 10 m/sec

c) conserving momentum

we get

10=(V2-10)

V2 = 20 m/sec