A bullet of mass 10 g moving horizontally at a speed of 50√ m/s strikes a block of mass 490 g kept on a frictionless track as shown in figure (9-E17). The bullet remains inside the block and the system proceeds towards the semicircular track of radius 0.2 m. Where will the block strike the horizontal part after leaving the semicircular track?
Mass of the block is 490 gm,
Mass of bullet is 10 gm.
Since the bullet embedded inside the block, it is a plastic collision.
Initial velocity of bullet is
Velocity of the block is.
Let final velocity of both is vA.
Therefore, by law of conservation of momentum
When the block losses contact at ‘D’ the component mg will act on it.
Angular momentum acting at point D
___________ (1)
Putting work energy principle
Therefore, angle of projection will be
Then, time taken to reach the ground will be
Thus, distance travelled in horizontal direction
So, total distance will be
Hence, at the junction of the straight and the curved parts, the block will strike the horizontal part after leaving the semicircular track.