Part A Find the magnitude of the velocity of the red puck after the collision. E

Question

Part A Find the magnitude of the velocity of the red puck after the collision. Express your answer using two significant figures. Part A Find the magnitude of the velocity of the red puck after the collision. Express your answer using two significant figures. Part A Find the magnitude of the velocity of the red puck after the collision. Express your answer using two significant figures. Part B Find the direction of the velocity of the red puck after the collision. Part B Find the direction of the velocity of the red puck after the collision. Part B Find the direction of the velocity of the red puck after the collision. Part C Express your answer using two significant figures.





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Explanation / Answer

Using the law of conservation of momentum

M1V1 + M2V2 = M1V3 + M2V4

where

M1 = mass of the blue puck = 0.038 kg.
V1 = initial velocity of the blue puck = 0.250 m/sec.
M2 = mass of the red puck
V2 = initial velocity of the red puck = 0 (at rest)
V3 = final velocity of the blue pack = 0.080 m/sec.
V4 = final velocity of the red puck

Substituting values,

0.038(0.250) + 0 = 0.038(.08) + (M2)(V4)

(M2)(V4) = 0.00646 --- call this Equation 1

Since the kinetic energy is conserved,

(1/2)(.038)(.25)^2 = (1/2)(0.038)(.08)^2 + (1/2)(M2)(V4)^2

(M2)(V4)^2 = 0.00228

The above can be rewritten as

(M2*V4)(V4) = 0.0021318

Since (M2*V4) = 0.00646(from Equation 1), then

0.00646(V4) = 0.0021318

and solving for "V4"

V4 = 0.33 m/sec

Since the velocity calculated above is positive, then the direction of the red puck is the same as the direction of the blue pack after collision.


From Equation 1,

M2 = 0.00646/0.33

M2 = 0.0196 kg.

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