A car P is moving with a uniform speed `5sqrt3 m//s` towards a carriage of mass 9 kg at rest kept on the rails at a point B as shown in figure. The height AC is 120 m. Cannon balls of 1 kg are fired from the car with an initial velocity `100m//s` at an angle `30^@` with the horizontal. The first cannon hall hits the stationary carriage after a time`t_0` and sticks to it. Determine `t_0`.

At `t_0`, the second cannon ball is fired. Assume that the resistive force between the rails and the carriage is constant and ignore the vertical motion of the carriage throughout. If the second ball also hits and sticks to the carriage, what will be the horizontal velocity of the carriage just after the second impact?

Answer

A, B

Explanation

Consider the vetical motion of the cannon ball
`u_y=+100 sin 30^@`
`s_y =-120m`
`a_y=-10m//s^2`
`t=t_0`
`s=ut+1/2at^2 :. -120=50t_0-5t_0^2`
`rArr t_0^2-10t_0-24=0`
`:. t_0=((-10)+-sqrt(100)-4(1)(-24))/2=12 or -2` [Not valid]
`:. t_0=12 sec.`

The horizontal velocity of the cannon ball remains the same
`:. u_x=100 cos 30^@=50 sqrt3ms^-1`
`:.` 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,
`mu_x+Mxx0=(m+M)v_x`
`:. v_x=(m u_x)/(m+M)` where `v_x` is the velocity of the
cannon ball-trolley system.
`:. v_x=(1xx50sqrt3)/(1+9)=5sqrt3m//s`
The second ball also struck the trolley,
Therefore, in time 12 seconds, the trolley covers a distance
of `60sqrt3m.`
For trolley after 12 seconds,
`u=5sqrt3 m//s,v=?, t=12s`

`s=60sqrt3m, s=ut+1/2at^2`
`rArr 60sqrt3=5sqrt3xx12+1/2xxaxx144`
`:. a=0 m//s^2 :. v=u+at=5sqrt3 m//s.`
To find the final velocity of the carriage after the second
impact we again apply conservation of linear momentum in
the horizontal direction
`mu_x+(M+m)v_x=(M+2m)v_f`
`:. 1xx50sqrt3+(9+1)5sqrt3=(9+2)v_f`
`rArr v_f=15.75 m//s`