Two bodies hang on a string 2.1m long. The two bodies have a mass of 3.75kg and 2.4kg respectively. If the 3.75kg mass is raised to a height of 0.545m and released, and if upon impact the mass adhere.
•Mass1 = 3.75 kg
•Mass2 = 2.4 kg
•String length = 2.1 m
•M1 raised = 0.545m
a) Find the velocity of the 3.75kg mass before impact.
b) Find the common velocity of the mass after impact.
c) Find the lost of kinetic energy on impact.
d) Find the height through which the center of gravity of the system will rise.
e) Find the tension on the string of the 3.75kg mass before impact.
Michel, luc, daniel, Henry, or whoever --
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To find the answers to these questions, we can use the principles of conservation of energy and conservation of momentum.
a) Find the velocity of the 3.75kg mass before impact:
We can start by finding the potential energy of the 3.75kg mass when it is raised to a height of 0.545m. The potential energy is given by the formula PE = mgh, where m is the mass, g is the acceleration due to gravity (approximately 9.8m/s^2), and h is the height.
PE = (3.75kg)(9.8m/s^2)(0.545m) = 18.63 J
Next, we can equate this potential energy to the kinetic energy of the mass when it is released. The kinetic energy is given by the formula KE = 0.5mv^2, where m is the mass and v is the velocity. Rearranging the formula, we get:
v = √(2KE/m)
Substituting the potential energy (PE) for the kinetic energy (KE) and the mass (m) of the 3.75kg mass into the equation, we can solve for v:
v = √(2PE/m) = √(2(18.63J)/(3.75kg)) ≈ 3.43 m/s
Therefore, the velocity of the 3.75kg mass before impact is approximately 3.43 m/s.
b) Find the common velocity of the mass after impact:
Since the two masses adhere upon impact, they move together as one combined mass. We can use the principle of conservation of momentum to find their common velocity.
The momentum before the impact is given by the formula p = mv, where p is the momentum, m is the mass, and v is the velocity. The momentum of the 3.75kg mass before impact is (3.75kg)(3.43m/s) = 12.86 kg·m/s.
After impact, the total momentum is still conserved. Let the common velocity of the combined mass be v', and the total mass be (3.75kg + 2.4kg). Thus, the momentum after impact is (5.15kg)(v').
Since momentum is conserved, we have:
12.86 kg·m/s = (5.15kg)(v')
Solving for v', we get:
v' = 12.86 kg·m/s / 5.15kg ≈ 2.5 m/s
Therefore, the common velocity of the mass after impact is approximately 2.5 m/s.
c) Find the lost of kinetic energy on impact:
The kinetic energy before impact is given by the formula KE = 0.5mv^2. For the 3.75kg mass, using the information from part b, KE = 0.5(3.75kg)(3.43m/s)^2 ≈ 17.93 J.
After impact, the kinetic energy is given by KE' = 0.5(5.15kg)(2.5m/s)^2 ≈ 16.5 J.
The lost kinetic energy on impact is the difference between these two values:
Lost kinetic energy = KE - KE' ≈ 17.93 J - 16.5 J ≈ 1.43 J
Therefore, the lost kinetic energy on impact is approximately 1.43 J.
d) Find the height through which the center of gravity of the system will rise:
When the masses adhere, their combined center of gravity will move upward. We can find the height the center of gravity rises by using the principle of conservation of energy.
The potential energy of the combined system is equal to the sum of the potential energy of the masses individually.
PE_system = (mass1 + mass2)gh
Substituting the values, we get:
PE_system = (3.75kg + 2.4kg)(9.8m/s^2)(h)
Since the potential energy of the system is equal to the lost kinetic energy on impact (in this case), we can equate the two values:
Lost kinetic energy = (mass1 + mass2)gh
Solving for h, we get:
h = (Lost kinetic energy) / ((mass1 + mass2)g)
Plugging in the values, we have:
h = 1.43 J / ((3.75kg + 2.4kg)(9.8m/s^2)) ≈ 0.023 m
Therefore, the height through which the center of gravity of the system will rise is approximately 0.023 m.
e) Find the tension on the string of the 3.75kg mass before impact:
To find the tension on the string, we need to consider the forces acting on the 3.75kg mass. At the point where the string is attached, the mass is moving in a circular path. Therefore, the net force acting on it must be directed towards the center of the circle, providing the necessary centripetal force.
The tension in the string provides this centripetal force. We can use the formula for centripetal force, Fc = mv^2/r, where Fc is the centripetal force, m is the mass, v is the velocity, and r is the radius of the circle (which is equal to the length of the string).
Rearranging the formula, we have:
Tension = mv^2/r
Substituting the values, we get:
Tension = (3.75kg)(3.43m/s)^2 / 2.1m ≈ 20.81 N
Therefore, the tension on the string of the 3.75kg mass before impact is approximately 20.81 N.