Question : A 0.300-kg puck, initially at rest on a horizontal, frictionless surf

Question

Question: A 0.300-kg puck, initially at rest on a horizontal, frictionless surface, is struck by a 0.200-kg puck moving initially along the x axis with a speed of 2.00 m/s. After the collision, the0.200-kg puck has a speed of 1.00 m/s at an angle of = 53.0° to the positive x axis (see the figure below).

(a) Determine the velocity of the 0.300-kg puck after the collision.
( 1.073 ) m/s

(b) Find the fraction of kinetic energy transferred away or transformed to other forms of energy in the collision.

( )

I need the answer of Part (b) ?

Explanation / Answer

(b) Kinetic energy Ki = 1/2 m2 (v2i)2

KEi = 1/2 * 0.20 * (2^2)

KE i = 0.4 J

Kf = 1/2 m2 (v2f)2 +1/2 m1 (v1f)2

Kf = 1/2 * 0.20 * (1^2) +1/2 * 0.3 * (1.07^2

Kf= 0.271 J

K = Ki - Kf = 0.4 J - 0.271 J = 0.129 J

Thus Fraction of kinetic energy lost is

K /Ki= 0.129/0.4= - 0.32

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