Two identical blocks `A` and `B` connected by massless string, are placed on a frictionless horizontal plane. A bullet havig the same mass, moving with speed `u` strickes block `B` from behind as shown. If the bullet gets embedded into block `B` then find

a. the velocity of `A,B,C` after collision.
b. impulse on `A` due to tension in the string,
c. impulse on `C` due to normal force of collision,
d. impulse on `B` due to normal force of collision.

Explanation

(a) After collision, the blocks & the bullet will move together with same velocity, say `v`
By conservation of linear momentum `m u=3m u`
`v=(u)/(3)`
(b) Net impulse on `A` is due to tension force,
Impulse on `A=P_(f)-P_(i)`
`intTdt=(m u)/(3)-0`
(c) On the bullet `C`, net impulse is due to `N`
`-intNdt=P_(f)-P_(i)`
`=(m u)/(3)-m u`
`=(-2m)/(3)`
(d) On `B` two impulsive forces act i.e. Normal & Tension.
`vec"J"=int(N-T)dt=intN" "dt-intT" "dt=(m u)/(3)`
`rArr" "intN" "dt=(2m u)/(3)`
The impulse due to normal force on both the colliding bodies is equal. Thus we can directly say impulse on `B` due to normal is same as impulse on `C` due to normal.