3. Starting with an initial speed of 6.50 m/s at a height of 0.40 m, a 1.80-kg b

Question

3. Starting with an initial speed of 6.50 m/s at a height of 0.40 m, a 1.80-kg ball swings downward and strikes a 4.20-kg ball that is at rest.
(i) What is the speed of the 1.80-kg ball just before the collision?

(ii) Assume the collision is elastic. What are the velocities (magnitude and direction) of both objects after the collision?

(iii) How high does each object swing after the collision?

Starting with an initial speed of 6.50 m/s at a height of 0.40 m, a 1.80-kg ball swings downward and strikes a 4.20-kg ball that is at rest. What is the speed of the 1.80-kg ball just before the collision? Assume the collision is elastic. What are the velocities (magnitude and direction) of both objects after the collision? How high does each object swing after the collision?

Explanation / Answer

A) use conservation of energy:

m1 * g * h = 1/2 * m1 * v^2

v = (2gh)^.5 = 2.8

B) V1 = (m1-m2)/(m1+m2)*Vo = (1.8-4.2)/(1.8+4.2)*6.50 = -2.6

V2 = (2m1)/(m1+m2)*Vo = (2*1.8)/(1.8+4.2) * (6.50) = 3.9

C) use conservation of energy to get heights:

mgh = 1/2mv^2

h=(v^2)/(2g)

h1 = (2.6^2)/(2*9.8)=.34

h2 = (3.9^2)/(2*9.8)=.78

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