Starting with an initial speed of 5.00 m/s at a height of 0.315 m, a 1.75-kg bal

Question

Starting with an initial speed of 5.00 m/s at a height of 0.315 m, a 1.75-kg ball swings downward and strikes a 4.60-kg ball that is at rest, as the drawing shows.

(a) Using the principle of conservation of mechanical energy, find the speed of the 1.75-kg ball just before impact.

m/s

(b) Assuming that the collision is elastic, find the velocities (magnitude and direction) of both balls just after the collision.

b1-   m/s (1.75 kg-ball)

b2-   m/s (4.60 kg-ball)

(c) How high does each ball swing after the collision, ignoring air resistance?

c1- m/s (1.75 kg-ball)

c2- m/s (4.60 kg-ball)

Explanation / Answer

Part A)

Apply PE +KE = KE

mgh .5mv2 = .5mv2 (mass cancels)

(.5)(5)2 + (9.8)(.315) = (.5)(v)2

v = 5.58 m/s

Part B)

For elsatic collisions, momntum and KE are conserved

For the KE portion...

.5mv2 = .5mv'2 + .5Mv2

(.5)(1.75)(5.58)2 = (.5)(1.75)(v')2 + (.5)(4.6)(v)2

27.24 = .875v'2 + 2.3v2

For the momentum portion...

mv = mv' + Mv

(1.75)(5.58) = 1.75v' + 4.6v

9.77 = 1.75v' + 4.6v

Solve for v'...

v' = 5.58 - 2.63v

Sub that in for the KE equation...

27.24 = .875(5.58 - 2.63v)2 + 2.3v2

27.24 = 27.245 - 25.7v + 6.05v2 + 2.3v2

Simplify

27.5 = 8.35v

v = 3.29

v' = 5.58 - (2.63)(3.29) = -3.08 m/s

That means that the 1.75 kg ball bounces off and moves backward at 3.08 m/s (your answer key may want that as a negative) and the 4.6 kg ball moves forward at 3.29 m/s

Part C)

For the heights...

mgh = .5mv2 for each ball

(9.8)(h) = .5(3.29)2

h = .552 m for the 4.6 kg ball

(9.8)(h) = .5(3.08)2

h = .484 m for the 1.75 kg ball

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