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

12

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_("cannon.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_("carria g e") = 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`.