6 5.55/11.11 points 1 Previous Answers SerPOPS 8 P My A 0.300 kg puck, initially

Question

6 5.55/11.11 points 1 Previous Answers SerPOPS 8 P My A 0.300 kg puck, initially at rest on a horizontal, frictionless surface, is the 0.200 kg puck has a speed of 1.00 m/s at an angle of -S4.0. to the positive x axis (Fig. 8.11). struck by a 0.200 kg puck moving intially along the x axis with a speed of 2.00 m/s. After the collsion, Active Figure 8.11 (a) Determine the speed of the 0.300 kg puck after the collision 1.07 b) Find the fraction of kinetic energy lost in the colision m/s Your response differs from the correct answer by more than 10%. Double check your calculations. Need Help? os

Explanation / Answer

(a)Applying momentum conservation,

(0.300 x 0) + (0.200 x 2 i) = 0.200 (cos54i + sin54j) + 0.300 v

v = 0.941 i - 0.54 j  

speed = sqrt(0.94^2 + 0.54^2) = 1.08 m/s .......Ans

(B) Ki = 0.20 x 2^2 /2 =0.400 J  

Kf = (0.20 x 1^2 /2 ) + (0.30 x 1.08^2 /2)

= 0.275 J  

klost = Ki - Kf = 0.125 J

fraction = 0.125 / 0.400 = 0.31

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