Problem 6.56 A 3.5 kg block moving at 2.0 m/s toward the west on a frictionless
ID: 1419921 • Letter: P
Question
Problem 6.56 A 3.5 kg block moving at 2.0 m/s toward the west on a frictionless surface has an elastic head-on collision with a second 0.60 kg block traveling east at 3.0 m/s. Part A Determine the final velocity of first block. Assume due east direction is positive. Part B Determine the final velocity of second block. Assume due east direction is positive. Part C Determine the kinetic energy of first block before the collision. Part D Determine the kinetic energy of second block before the collision. art 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). Part F Determine the kinetic energy of second block after the collision.
Explanation / Answer
(a)
final velcoity of the first object in an elastic collision is
v1 = (m1-m2)* Vo1 /(m1+m2) + (2m2 V02/(m1+m2)
v1 = (3.5 -0.6) * (-2)/(3.5 +0.6) + (2 * 0.6)/(3.5 + 0.6)
v1 = 1.122 m/s
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(b)
final velcoity of the second object in an elastic collision is
v1 = (2m2)* Vo1 /(m1+m2) + (m2-m1) V02/(m1+m2)
v1 = (2 *0.6) * (-2)/(3.5 +0.6) + (2 * 0.6)/(3.5 + 0.6)
v1 = -0.292 m/s
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c KE initial = 0.5 mv^2
KE i = 0.5 * 3.5* 2*2 =
KEi = 7 J
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d. KEi of 2 = 0.5 * 0.6* 3*3 = 2.7J
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e
KE of 1 after collision is
KE = 0.5 * 3.5 * 1.122*1.122
KE = 2.203 J
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f. KE of second block after coillison
KE = 0.5 * 0.6 * 0.292* 0.292
KE = 0.0255 Joules