A car P is moving with a uniform speed `5sqrt(3) m//s` towards a carriage of mass 9 kg at rest kept on the rails at a point B as shown in the figure. The height AC is 120 m. Cannon balls of 1 kg are fired from the car with an initial velocity 100 m/s with respect to the car at an angle `30^@` with the horizontal.

The first cannon ball hits the stationary carriage after a time `t_0` and sticks to it. Determine `t_0`. (in sec)

Answer

`12s, 15.75 ms^(-1)`

Explanation

Consider the vertical motion of th ecannon ball
`:. S = ut + (1)/(2) at^(2) :. -120 = 50t_(0) - 5t_(0)^(2)`
`rArr 5t_(0)^(2) - 5t_(0) - 120 = 0 rArr t_(0)^(2) - 10 t_(0) - 24 = 0`
`:. t_(0) = -((-10) +- [sqrt(100) - 4(1)(-24))]/(2) = 12` or `-2`
The horizontal velocity of the cannon ball remains the same
`:. v_(x) = 100 cos30^(@) + 5sqrt3 = 55 sqrt3 m//s`

`:.` Apply 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
`mv_(x) + M xx 0 = (m + M) V_(x) :. V_(x) = (mv_(x))/(m + M)`
where `v_(x)` is the velocity of the (cannon ball - trollwy) system
`V_(x) = (1 xx 55 sqrt3)/(1 + 9) = 5.5sqrt3 m//s`
The second ball was projected after 12 second
Horizontal distance covered by the car
`P = 12 xx 5sqrt3 = 60sqrt3m`
Since the second ball also struck the trollwy
`rArr` In time 12 seconds the trolley covers a distance of `60sqrt3`
For trolley in `12 sec`
From `s = ((u + v)/(2)) 60 sqrt3 = ((5.5sqrt3 + v)/(2)) (12) rArr v = 7.8 m//s`
To find the final velocity of the carriage after the second impact we again apply conservation of linear momentum in th ehorizontal direction
`mv_(x) + (M + m) 7.8 = (M + 2m) v_(f)`
`:. 1 xx 55 sqrt3 + (9 + 1) 7.8 = (9 + 2)v_(f)`
`rArr v_(f) = 15.75 m//s`