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
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
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
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
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◦
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
To find the speed at which the lighter mass m1 will hit the heavier mass m2, we can use the principle of conservation of mechanical energy, assuming that there is no loss of energy due to friction or other factors.
The mechanical energy of a system is given by the sum of kinetic energy (KE) and potential energy (PE). At the initial position, when the lighter mass is pulled aside, all of its energy is in the form of potential energy.
The potential energy of an object is given by the equation:
PE = m * g * h
Where:
m is the mass of the object
g is the acceleration due to gravity
h is the height of the object from a reference point
In this case, the reference point will be at the bottom of the motion. So, the initial potential energy (PE1) of the lighter mass m1 can be calculated as:
PE1 = m1 * g * h1
Where:
m1 = 451 g = 0.451 kg (converting grams to kilograms)
g = 9.8 m/s^2
h1 is the vertical displacement of m1 = L * (1 - cosθ), where L is the length of the string and θ is the angle from the vertical (65 degrees).
L = 10.9 m
θ = 65 degrees = 65 * (π/180) radians (converting degrees to radians)
Now we can calculate h1:
h1 = L * (1 - cosθ)
= 10.9 * (1 - cos(65 * (π/180)))
Next, we need to find the final potential energy (PE2) when m1 reaches the initial vertical position of m2.
At this position, the potential energy of m2 will be equal to m1's initial potential energy. So:
PE2 = m2 * g * h2
Where:
m2 = 695 g = 0.695 kg
g = 9.8 m/s^2
h2 is the vertical displacement of m2 = L
Now we can equate PE1 and PE2:
m1 * g * h1 = m2 * g * h2
Solving for h2:
h2 = (m1 * h1) / m2
Now, with h2, we can find the final kinetic energy (KE2) of m1 at the vertical position:
KE2 = m1 * v^2 / 2
Where v is the speed at which m1 hits m2.
Equating PE2 and KE2:
m2 * g * h2 = m1 * v^2 / 2
Solving for v:
v^2 = 2 * m2 * g * h2 / m1
Finally, we can solve for v:
v = √(2 * m2 * g * h2 / m1)
Substituting the given values and calculating:
m1 = 0.451 kg
m2 = 0.695 kg
g = 9.8 m/s^2
L = 10.9 m
θ = 65 degrees = 65 * (π/180) radians
Calculating h1:
h1 = L * (1 - cosθ)
= 10.9 * (1 - cos(65 * (π/180)))
≈ 4.108 m
Calculating h2:
h2 = (m1 * h1) / m2
= (0.451 * 4.108) / 0.695
≈ 2.670 m
Finally, calculating the speed v:
v = √(2 * m2 * g * h2 / m1)
= √(2 * 0.695 * 9.8 * 2.670 / 0.451)
≈ 7.798 m/s
Therefore, the speed at which the lighter mass m1 will hit the heavier mass m2 is approximately 7.798 m/s.