Answer:
(a)Magnitude=28.81 m/s
Direction=33.3 degree below the horizontal
(b) No, it is not perfectly elastic collision
Explanation:
We are given that
Mass of stone, M=0.150 kg
Mass of bullet, m=9.50 g=
Initial speed of bullet, u=380 m/s
Initial speed of stone, U=0
Final speed of bullet, v=250m/s
a. We have to find the magnitude and direction of the velocity of the stone after it is struck.
Using conservation of momentum

Substitute the values





Magnitude of velocity of stone
=
|V|=28.81 m/s
Hence, the magnitude and direction of the velocity of the stone after it is struck, |V|=28.81 m/s
Direction

=

=33.3 degree below the horizontal
(b)
Initial kinetic energy


Final kinetic energy

=

Initial kinetic energy is not equal to final kinetic energy. Hence, the collision is not perfectly elastic collision.