A car `A` moves with velocity `20 m s^(-1)` and car `B` with velocity `15 m s^(-1)` as shown is. Find the relative velocity of B` w.r.t. `A`, and A w.r.t. `B`.
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Answer

3

Explanation

Consider the vertical motion of the cannon ball
`u_y = + 100 sin 30^@ = 50 m//s , s_y = -120 m`
and `a_y = -10m//s^2`
Using `s_y = u_yt + 1/2 a_y t^2`
`:. -120 = 50 t_0 - 5t_0^2`
`:. 5t_0^2 - 50 t_0 - 120 = 0 implies t_0^2 - 10 t_0 - 24 = 0`
`:. t_0 = (-(-10) +- sqrt(100 - 4(1) (-24)))/(2) = 12 "or" -2`
Hence `t_0 = 12 "sec"`
Horizontal component of cannon ball `(V_("common"))_(x) = (V_("can non.car"))_(x) + (V_("car"))_x`
`:. u_(x) = 100 cos 30^@ + 5sqrt(3) = 55 sqrt(3)m//s`
The horizontal component of the velocity of cannon ball remains the same
`:.` Applying conservation of linear momentum to the cannon ball – trolley system in horizontal direction. If m is the mass of cannon ball and M is the mass of the trolley then.
`m u_(x) + M xx 0 = (m + M)_(v_x)`
`:. v_(x) = (m u_x)/(m + M)` where `v_x` is the velocity of the cannon ball - trolley system.
`implies v_x = (1 xx 55 sqrt(3))/(1 + 9) = 5.5 sqrt(3) m//s`
The second ball was projected after 12 second.

Horizontal distance covered by the car `P = 12 xx 5sqrt(3) = 60sqrt(3)m`.
Since the second ball also struck the trolley.
The cannon lands on the carriage again second time. This is only possible if the carriage moves the same distance as that moved by car in 12 sec.
The distance moved by the carriage in 12 seconds
`s_("carriage") = (5.5sqrt(3))12 - 1/2 a xx 12^2`
Where a is the retardation given by the resistive force. The distance by the car in 12 second `= 5sqrt(3) xx 12m`
`s_("car") = 60sqrt(3) , s_("carriage") = s_("car")`
`(5.5) sqrt(3) xx 12 - 1/2 a xx 12^2 = 60sqrt(3) , a = (sqrt3)/(12) m//s^2`
Velocity of carriage after 12 seconds ,
`v' = 5.5sqrt(3) - a xx 12 = 5.5sqrt(3) - (sqrt3)/(12) xx 12 = 4.5sqrt(3) m//s`
After the second impact, by the conservation of linear momentum
`[(4.5sqrt(3)) 10 + 55sqrt(3)] = 11v^(n) , " "v^(n) = (100 sqrt(3))/(11) m//s`.