The masses are each initially 1.8m above the ground, and the massless frictionless pulley is 4.8m above the ground. (a) calculate the velocity of the lighter mass at the instance the heavier mass hits the ground. (b) find the maximum height reached by the lighter object (from the ground level) after the system is released. Use the energy method to find the answers.
a) depends on the difference in the two masses.
energy: PE converted to KE:
(M-m)g*4.8
that PE is converted to ke of the two bodies. KE=1/2 M V^2+1/2 m V^2 set that equal to the PE, solve for V
b) knowing the 1/2 m V^2 at release, then the max height above release time can be found by max PE=mg*4.8+1/2 m V^2
then from max PE, =mgh, solve for h.
To solve this problem, we can use the principle of conservation of mechanical energy. The initial potential energy of the system is equal to the final kinetic energy and vice versa.
(a) To calculate the velocity of the lighter mass at the instance the heavier mass hits the ground:
1. Start by determining the initial potential energy of the system. Both masses are initially 1.8m above the ground level, and the potential energy is given by the formula: PE = mgh, where m is the mass, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height.
The potential energy of the lighter mass (m_1) is m_1 * 9.8 * 1.8 J (joules).
The potential energy of the heavier mass (m_2) is m_2 * 9.8 * 1.8 J.
2. As the system is released, the potential energy is converted into kinetic energy. At the instant the heavier mass (m_2) hits the ground, the lighter mass (m_1) will have reached its maximum velocity (v_max). The kinetic energy is given by the formula: KE = 1/2 * m * v^2, where m is the mass and v is the velocity.
The kinetic energy of the lighter mass at the instant the heavier mass hits the ground is 1/2 * m_1 * v_max^2 J.
3. Since energy is conserved, the initial potential energy of the system is equal to the final kinetic energy:
m_1 * 9.8 * 1.8 = 1/2 * m_1 * v_max^2
4. Now, solve the equation for v_max:
9.8 * 1.8 = 1/2 * v_max^2
Simplify the equation:
17.64 = v_max^2
Take the square root of both sides:
v_max = √(17.64) m/s
v_max ≈ 4.2 m/s
Therefore, the velocity of the lighter mass at the instance the heavier mass hits the ground is approximately 4.2 m/s.
(b) To find the maximum height reached by the lighter object (from the ground level) after the system is released:
1. The maximum height reached by the lighter object occurs when it has come to a momentary stop before falling back down. At this point, all the potential energy has been converted into kinetic energy.
Let's denote the maximum height as h_max.
2. The initial potential energy of the system is equal to the final kinetic energy.
m_1 * 9.8 * 1.8 = 1/2 * m_1 * v_max^2 + m_1 * 9.8 * h_max
3. Substitute the known values into the equation:
1.8 * 9.8 = 1/2 * 1.8 * (4.2)^2 + 1.8 * 9.8 * h_max
4. Solve for h_max:
17.64 = 1.8 * (8.82 + 9.8 * h_max)
17.64 = 15.876 + 17.64 * h_max
Rearrange the equation:
17.64 * h_max = 17.64 - 15.876
17.64 * h_max ≈ 1.764
Divide by 17.64:
h_max ≈ 0.1 m
Therefore, the maximum height reached by the lighter object (from the ground level) after the system is released is approximately 0.1 m.
To solve this problem using the energy method, we need to analyze the conservation of energy throughout the motion of the system.
(a) To calculate the velocity of the lighter mass at the instance the heavier mass hits the ground, we can apply the principle of conservation of mechanical energy.
Initially, the system has potential energy due to the masses being above the ground. This potential energy will be converted into kinetic energy as the masses fall.
The potential energy of the system is given by the sum of the potential energies of the two masses:
PE = m₁gh₁ + m₂gh₂
Where:
m₁ = mass of the lighter object
m₂ = mass of the heavier object
g = acceleration due to gravity
h₁ = initial height of the lighter object
h₂ = initial height of the heavier object
At the instant when the heavier mass hits the ground, the lighter mass is still moving. So, its potential energy is zero, and all the initial potential energy is converted into kinetic energy.
Using the principle of conservation of mechanical energy, we can equate the initial potential energy to the kinetic energy at the instant the heavier mass hits the ground:
PE = KE
m₁gh₁ + m₂gh₂ = 1/2(m₁ + m₂)v²
Where:
v = velocity of the lighter mass when the heavier mass hits the ground
Since the masses are connected by a massless, frictionless pulley, their accelerations are equal in magnitude but opposite in direction. Therefore, we can equate their accelerations using Newton's second law:
m₁a = m₂(-a)
Solving for a, we find:
a = m₁g / (m₁ + m₂)
Now, we can express the velocity in terms of acceleration and time using the kinematic equation:
v = -at
Where t is the time taken for the heavier mass to hit the ground. This time can be calculated using the equation:
h₂ = 1/2gt²
Solving for t, we find:
t = √(2h₂ / g)
Substituting the value of t into the equation for v, we can find the velocity.
(b) To find the maximum height reached by the lighter object after the system is released, we need to find the height at which all the initial potential energy is converted entirely into potential energy of the lighter mass.
Using the law of conservation of energy again, we can equate the initial potential energy of the system to the potential energy of the lighter mass at its maximum height:
PE = m₁gh₁ + m₂gh₂ = m₁gh
Where h is the maximum height reached by the lighter object.
Solving for h, we find:
h = (m₁gh₁ + m₂gh₂) / (m₁g)
Substituting the given values, we can find the maximum height reached by the lighter object.