Physics bullet Vs Stone

0.110kg stone rests on a frictionless, horizontal surface. A bullet of mass 9.50g , traveling horizontally at 310m/s , strikes the stone and rebounds horizontally at right angles to its original direction with a speed of 220 m/s

(A) Compute the magnitude of the velocity of the stone after it is struck.

(B) Compute the direction of the velocity of the stone after it is struck.

(C) Is the collision perfectly elastic?

Hello

momentum is conserved.
m1,v1 = mass, initial velocity of bullet
m2, v2 = mass, initial velocity of stone
v1' = final velocity of bullet
v2' = final velocity of stone
a = angle of stone with + x axis., assuming that the bullet is shot in direction of + x-axis, and that the bullet is directed into - y axis.
the x and y components of momentum are calculated separately.

x direction:
0.0095*310 + 0.11*0 = 0.0095*0*cos0 + 0.11*v2'*cos(a)
0.0095*310 = 0.11*v2' *cos(a)

y direction:
0 + 0 = 0.0095*220*sin (-90) + 0.11*v2' * sin(a)
0 = - 0.0095*220 + 0.11*v2'*sin(a)
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the solutions of the two equations are
v2' = 32.8295 m/s <--- velocity of stone <--- ans.
a = 35.362° with + x-axis. <--- angle of stone <--- ans.
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the initial kinetic energy = 1/2*m*v^2 = 1/2*0.0095*310^2 = 456.475 Nm
final kinetic energy = 1/2*0.0095*220^2 + 1/2*0.11*32.8295^2 = 289.177 Nm
kinetic energy is not conserved

Regards
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