PLEZZz I NEEED HELPPP WITH PHYSICSS PROBLEM

A 0.30 kg puck, initially at rest on a frictionless horizontal surface, is struck by a 0.20 kg puck that is initially moving along the x axis with a velocity of 2.0 m/s. After the collision, the 0.20 kg puck has a speed of 0.9 m/s at an angle of θ = 53° to the positive x axis.
(a) Determine the velocity of the 0.30 kg puck after the collision.
____________m/s at ____________ ° from +x axis

(b) Find the fraction of kinetic energy lost in the collision.
|ΔKE| / KEi = ______________ %

a) call 0.3 kg m1 , and 0.2 kg m2 , then :

m1 u1 + m2 u2 = m1v1 + m2v2

taking the x component u get :

0 + 0.2 (2) = 0.3 (v1 cosθ2) + 0.2 (0.9) cos53

hence 0.4 = 0.3 (v1 cosθ2) + 0.108

so v1 cosθ1 = 0.97 --------- (1)

taking the y component u get :

0 = 0.3 (v1 sinθ2) + 0.2(0.9) sin53

so v1 sinθ1 = - 0.48 -------------- (2)

dividing (2) by (1) u get :

tanθ1 = - 0.495

hence θ1 = arctan( - 0.495) = -26.3 degrees with the positive x axis. <<<<<<<<<<<<<

plug in -26.3 for θ1 in (1) u get :

v1 cos( - 26.3) = 0.97

=> v1 = 0.97 / 0.896 = 1.08 m/s (approx) <<<<<<<<<<<<,

so the answer is 1.08 m/s at -26.3° from +x axis <<<<<<<<<<<<<<<<<<

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b) KEi = 1/2 (0.3) (0) + 1/2(0.2) (4) = 0.4 J

....KEf = 1/2 (0.3) (1.08)^2 + 1/2 (0.2)(0.9)^2 = 0.256 J

so |ΔKE| = 0.144 J

and |ΔKE| / KEi = 0.144 / 0.4 = 0.36 = 36 % (approx) <<<<<<<<<<<

Hope my calculations were correct .
good luck :)