A 0.150 kg stone rests on a frictionless, horizontal surface. A bullet of mass 9.50 g, traveling horizontally at 380 m/s, strikes the stone and rebounds horizontally at right angles to its original direction with a speed of 250 m/s. Required:
a. Compute the magnitude and direction of the velocity of the stone after it is struck.
b. Is the collision perfectly elastic?

(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.

answer: a ball

explanation: because it will bounce