Raghav Singh
Last Activity: 6 Years ago
Take right hand side to be positive(+) and left hand side to be negative(-)
Initially, Speed of 4kg block= 2m/s (left to right) and Speed of 6kg block= 4m/s (right to left)
Let speed after collision to be v1 and v2 respectively (left to right both)
Conserving momentum we have
Pinitial = Pfinal
4(2) – 6(4) = 4(v1) + 6(v2)
-16 = 4(v1) + 6(v2)
2(v1) + 3(v2) = -8 ----equation (i)
Now as collison is elastic e=1, i.e
1= velocity of separation after collision/velocity of approach before collision
1= (v2 – v1)/(4+2)
6= v2 – v1 i.e
2(v2) – 2(v1) = 12 ------equation (ii)
adding equations (i) and (ii) we have
5(v2) = 4
v2 = 4/5 (as it is +ve it means it is moving from left to right)
Putting value of v2 in equation (ii) we get v1 = -26/5 (as it is -ve it means it is moving from right to left)
