A cart is moving along +x direction with a velocityof `4 m//s`. A person on the cart throws a stone with a velocity of `6 m//s` relative to himself. In the frame of reference of the cart the stone is thrown in y-z plane making an angle of `30^@` with vertical z-axis. At the highest point of its trajectory, the stone hits an object of equal mass hung vertically from the branch of a tree by means of a string of length L. A completely inelastic collision occurs, in which the stone gets embedded in the object. Determine :
(i) The speed of the combined mass immediately after the collision with respect to an observer on the ground,
(ii) The length L of the string such that the tension in the string becomes zero when the string becomes horizontal during the subsequent motion of the combined mass.

Answer

2.5

Explanation

When the stone reaches the point A, the component of velocity in the +z direction `(u cos theta)` becomes zero due to the gravitational force in the –z direction.
The stone has two velocities at A
`u_(x)` in the +x direction (4m/s)
`u_(y) = u sin theta` in the +y direction
`(6 sin 30^@ = 3m//s)`
The resultant velocity of the stone
`v = sqrt((u_x)^2 + (u sin theta)^2) = sqrt(4^2 + 3^2) = 5m//s`
Applying conservation of linear momentum at A just before and just after collision. Let velocity of combined mass just after collision is V.
`m xx 5 =(m + m) xx V`
`:. V = 2.5 m//s`
Now combined mass undergoes circular motion,
`implies` Velocity of combined mass is zero when it is in horizontal orientation. Applying conservation of mechanical energy from A to B, we get.
`1/2(2m)v^2 = 2mgl implies 1 = (v^2)/(2g) = ((2.5)^2)/(2 xx 10) = 5/16 m`.