Two small spheres of mass 421 g and 675 g are suspended from the ceiling at the
ID: 2145685 • Letter: T
Question
Two small spheres of mass 421 g and 675 g aresuspended from the ceiling at the same point
by massless strings of equal length 10.1 m.
The lighter sphere is pulled aside through an
angle of 57 from the vertical and released.
a. At what speed will the lighter mass m1 hit
the heavier mass m2? The acceleration due to
gravity is 9.8 m/s2 .
Answer in units of m/s
b. After the lighter sphere is released and col-
lides with the heavier sphere at the bottom of
its swing, the two spheres immediately bind
together.What is the speed of the combined system
just after the collision?
Answer in units of m/s
c. What is the maximum angle of deflection of
the two bound objects?
Answer in units of
Explanation / Answer
given: m1 = .421 kg m2 = .675 kg length of string = L = 10.1 m angle = x = 57 find the height the lighter sphere is pulled back: (use TRIG) y = 10.1 - 10.1cos57 y = 4.6 m find the PE stored in the lighter sphere: PE = mgy PE = .421(9.8)(4.6) PE = 18.98 J This PE is converted to KE by the time it is just about to strike the second sphere PE = KE = 18.98 J Use the KE to find the velocity in which the lighter sphere will contact the 2nd sphere with: KE = 0.5mv^2 18.98 = 0.5(.421)(v^2) v = 9.5 m/s (ANSWER TO PART A) Use conservation of momentum to determine the speed of the combined system: m1v1 + m2v2 = (m1+m2)(v3) .421(9.5) + .675(0) = (.421+.675)(v3) v3 = 3.65 m/s (ANSWER TO PART B) Use this speed to find the KE of the system: KE = 0.5(m1+m2)(v3)^2 KE = 0.5(.421 + .675)(3.65)^2 KE = 7.29 J This KE will be converted to PE by the time it reaches its maximum deflection PE = KE = 7.29 Use this PE to find the maximum height PE = mgy 7.29 = (.421 + .675)(9.8)(y) y = .679 m Use this and TRIG to find the maximum deflection angle: cos(x) = (10.1 - .679)/(10.1) x = 21.1 degrees (ANSWER TO C) BOL
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