A 3.9kg block moving at 2.0 m/s toward the west on a frictionless surface has an

Question

A 3.9kg block moving at 2.0 m/s toward the west on a frictionless surface has an elastic head-on collision with a second 0.80kg block traveling east at 3.0 m/s.

a)Determine the final velocity of first block. Assume due east direction is positive.

b)Determine the final velocity of second block. Assume due east direction is positive.

c)Determine the kinetic energy of first block before the collision

d)Determine the kinetic energy of second block before the collision.

e)Determine the kinetic energy of first block after the collision.

Note: The block with the least initial kinetic energy actually gains energy and the one with the most loses an equal amount. This is analogous to what happens when cool air comes into contact with warm air. The cool air warms (its molecules speed up) and the warm air cools (its molecules slow down).

f)Determine the kinetic energy of second block after the collision.

Explanation / Answer

apply the law of conservtion of moemntum as

m1u1 + m2u2 = m1v1+ m2v2

so

final velocity v1 =   (m1-m2) u1/(m1+m2)

v1 = (3.9-0.8)* 2/(3.9 +0.8)

v1 = 1.319 m/s
---------------------------------

b. V2 = 2m1u1/(m1+m2)

v2 = 2* 3.9 * 2/(3.9 +0.8)

v2 =3.319 m/s

---------------------

c KE initial = 0.5 mv^2

KE i = 0.5 * 3.9 * 2*2 = 7.8 J

--------------------

d. KEi of 2   = 0.5 * 0.8* 3*3   = 3.6 J
-------------------------------

e KE of 1 after collision is

KE = 0.5 * 3.9 * 1.319*1.319

KE   = 3.39 25 J

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