Two wooden pucks approach each other on an ice rink as shown in the figure. Puck

Question

Two wooden pucks approach each other on an ice rink as shown in the figure. Puck #2 has an initial speed of 5.02 m/s and a mass that is some fraction

f =

that of puck #1. Puck #1 is made of a hard wood and puck #2 is made of a very soft wood. As a result, when they collide, puck #1 makes a dent in puck #2 and 18.0% of the initial kinetic energy of the two pucks is lost. Before the collision, the two pucks approach each other in such a manner their momentums are of equal magnitude and opposite directions. Determine the speed of the two pucks after the collision.

v1f

v2f

Explanation / Answer

initial momentum = m2 v2 - m1 v1

= (2 m / 3 x 5.02) - (m v1)

v1 = 3.35 m/s

Applying momentum conservation during collision,

m1 v1f sin40 - m2 v2f sin40 = 0

m v1f = 2m/3 v2f

v2f = 1.5 v1f

Applying energy conservation,

  
(1 - 0.18) (m x 3.35^2 /2 ) + (2m/3 x 5.02^2 / 2) = (m v1f^2 /2 + (2m/3) v2f^2 / 2)

(0.82) ( 28.02) = v1f^2 + (1.5 v1f)^2

v1f = 2.94 m/s ......Ans

v2f = 4.40 m/s ......Ans

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