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

A, B, C

Explanation

When the stone reaches the point Q, the component of
velocity in the +Z direction `(v cos theta)` becomes zero due to
the gravitational force in the -Z direction.

The stone has two velocities at Q
`*v_x in the +X direction (4m//s)`
`* v sin theta in the +Y direction (6 sin 30^@=3m//s)`
The resultant velocity of the stone
`v'=sqrt((v_x)^2+(v sin theta)^2)=sqrt(4^2+3^2)=5 m//s`
(i) Applying conservation of linear momentum at Q for
collision with a mass of equal magnitude
`mxx5=2mxxv`
Since the collision is completely inelastic the two
masses will stick together. v is the velocity of the two masses
just after collision.
`:. v=2.5 m//s`
(ii) When the string is undergoing circular motion, at any
arbitrary position `T-2mgcos alpha=(2mv^2)/l`
Given that, `T=0 when alpha=90^@ :. 0-0=(2mv^2)/l rArr v=0`
`rArr` Velocity is zero when `alpha=90^@,` i.e. in the horizontal
position.
Applying energy conservation
from Q to M, we get
`1/2 2mv^2=2mgl`
`rArr l=(v^2)/(2g)=(2.5)^2/(2xx9.8)=0.318m`