Three balls `A, B` and `C` of masses `2 kg, 4 kg` and `8 kg`, respectively, move along the same straight line and in the same direction, with velocities `4m//s,1m//s, 3/4m//s`. If `A` collides with `B` and subse-quently `B` collides with `C`, find the velocity of ball `A` and ball `B` after collision, taking the coefficient of restitution as unity.

Options

A. `V_(A)=3,V_(B)=9//4`
B. `V_(A)=0, V_(B)=3`
C. `V_(A)=3, V_(B)=0`
D. `V_(A)=0, V_(B)=0`

Answer

D

Explanation

First consider the collision of balls `A` and `B`. let the velocities of these two balls after their collision be `v` and `v'`.
Momentum after impact `=` Momentum before impact

`:. 2v+4v'=2xx4+4xx1`
`:. 2v+4v'=12`
`implies v+2v'=6`..........i
relative velocity after impact `=-exx` relative velocity before impact
`v-v'=-(4-1)`
`v-v'=(-3)`............ii
substracting eqn ii from eqn i we get
`3v'=9`
`v'=3m//s` Substituting in eqn ii we get
`v_3=-3impliesv=0`

Hence, after the collision ball `A` is brought to rest, while ball `B` will move with a velocity of `3m//s`. now consider the collision of balls `B` and `C`.Let the velocities of these balls after collision be `V` and `V'` respectively.
Total momentum after impact = Total momentum before impact
`4V+8V'=4xx3+8xx3/4=18`
`V+2V'=9/2`............iii
Relative velocity after impact `=-exx` relative velocity before impact
`V-V'=(-1)(3-3/4)`
`V-V'=-9/4`.........iv
Subtracting eq iv from eqn iii we get
`3V'=27/4impliesV'=9/4m//s`
`V=0`
`V_(A)=0,V_(B)=0`