oh and the rest of the question...
B) After the lighter sphere is released and collides 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◦
Two small spheres of mass 451 g and 695 g are suspended from the ceiling at the same point by massless strings of equal length 10.9 m . The lighter sphere is pulled aside through an angle of 65 degrees from the vertical and let go. At what speed will the lighter mass m1 hit the heavier mass m2? The acceleration of gravity is 9.8 m/s 2 . Answer in units of m/s
5 answers
Well, the question is pretty long but since your name sounds like a fishing vessel I will answer.
Use conservation of energy for the first part
10.9 cos 65 = 4.61
10.9 - 4.61 = 6.29 = distance ball is pulled up from bottom of swing
(1/2) m v^2 = m g h
v^2 = sqrt (2gh) = sqrt (2*9.8*6.29)
v = 11.1 m/s = answer part A
Use conservation of energy for the first part
10.9 cos 65 = 4.61
10.9 - 4.61 = 6.29 = distance ball is pulled up from bottom of swing
(1/2) m v^2 = m g h
v^2 = sqrt (2gh) = sqrt (2*9.8*6.29)
v = 11.1 m/s = answer part A
Part B, conservation of momentum
before collision
m v = .451 * 11.1 = 5 kg m/s
so after the collision it remains 5
5 = (.451+.695) v
v = 4.37 m/s answer part B
before collision
m v = .451 * 11.1 = 5 kg m/s
so after the collision it remains 5
5 = (.451+.695) v
v = 4.37 m/s answer part B
kinetic energy again (some lost because collision was not elastic)
(1/2) m v^2 = m g h
v^2 = 4.37^2 = 2*9.8*h
h = .974 meter
10.9 - .974 = 9.93
10.9 cos T = 9.93
T = 24.3 deg
(1/2) m v^2 = m g h
v^2 = 4.37^2 = 2*9.8*h
h = .974 meter
10.9 - .974 = 9.93
10.9 cos T = 9.93
T = 24.3 deg
Thank you so much!!!! That is extremely helpful! Our names sound like a fishing vessel? Leah and I are twins so we post one and figure out how to do it and then do the others a well :D
1 answer