0.140 kg stone rests on a frictionless, horizontal surface. A bullet of mass 5.0

Question

0.140 kg stone rests on a frictionless, horizontal surface. A bullet of mass 5.00g , traveling horizontally at 360m/s , strikes the stone and rebounds horizontally at right angles to its original direction with a speed of 220m/s .
Compute the magnitude of the velocity of the stone after it is struck.
Compute the direction of the velocity of the stone after it is struck. = http://session.masteringphysics.com/render?tex=%5E%5Ccirc%5C%3Bfrom%5C%3Bthe%5C%3Binitial%5C%3Bdirection%5C%3Bof%5C%3Bthe%5C%3Bbullet

Is the collision perfectly elastic? And why?

Explanation / Answer

Conservation of momentum along horizontal

5*360 = 140*Vx  Vx = 180/14 = 90/7

Along vertical

0 = 5*220 + 140*Vy   Vy = -55/7

The velocity of stone = (Vx^2 + Vy^2)^0.5 = 105.475m/s

the angle of stone = tan-1(Vy/Vx) = tan-1(-11/18)

An elastic collision is an encounter between two bodies in which the total kinetic energy of the two bodies after the encounter is equal to their total kinetic energy before the encounter. Elastic collisions occur only if there is no net conversion of kinetic energy into other forms.During the collision of small objects, kinetic energy is first converted to potential energy associated with arepulsive force between the particles (when the particles move against this force, i.e. the angle between the force and the relative velocity is obtuse), then this potential energy is converted back to kinetic energy (when the particles move with this force, i.e. the angle between the force and the relative velocity is acute).

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