Two 0.75m -long pendulums hang side by side. The masses of the pendulum bobs are 65g and 85g . The lighter bob is pulled aside until its string is horizontal and is then released from rest. It swings down and collides elastically with the other bob at the bottom of its arc.To what height does the lighter bob rebound?
momentum is conserved, energy is conserved in an elastic collision.
first pendulum:
m g h= at the bottom 1/2 m v^2
v= sqrt (2g*.75) that is velocity of m1 at the bottom. solve for v
Now, the collision
momentum:
.065*v=.065v'+.085v2
energy:
1/2 .065v'^2=1/2 .065v'^2+ 1/2 .085 v2^2
so in those , you have two unknowns v' and v2
two equations, two unknowns. Solve for v' in the first equation, put it in the second equation, and solve for v2. Have a blank pad with lots of room to do the agebra.
Now, knowing v2, the height it goes:
1/2 .085 v2^2= .085 h * g
solve for h.
To find the height to which the lighter bob rebounds after the collision, we can use the principle of conservation of energy.
Step 1: Calculate the potential energy of the lighter bob at its initial position.
Given that the string is horizontal, the height of the lighter bob from its initial position is zero (h = 0). The potential energy (PE) at its initial position is given by:
PE1 = m1 * g * h
where m1 is the mass of the lighter bob and g is the acceleration due to gravity.
PE1 = (0.065 kg) * (9.8 m/s^2) * 0 = 0 J
Step 2: Calculate the kinetic energy of the lighter bob just before the collision.
The lighter bob is released from rest, so its initial velocity (u1) is zero. Its kinetic energy (KE) just before the collision is given by:
KE1 = (1/2) * m1 * u1^2
Since u1 = 0, the kinetic energy just before the collision is also zero.
KE1 = (1/2) * (0.065 kg) * 0^2 = 0 J
Step 3: Calculate the potential energy of the lighter bob at the maximum height it reaches after the collision.
After the collision, the lighter bob rebounds and reaches a maximum height (h2). The potential energy (PE) at its maximum height is given by:
PE2 = m1 * g * h2
Step 4: Calculate the kinetic energy of the lighter bob just after the collision.
The kinetic energy (KE) just after the collision is given by:
KE2 = (1/2) * m1 * v1^2
where v1 is the velocity of the lighter bob just after the collision.
Since the collision is elastic, both momentum and kinetic energy are conserved. Therefore, the velocity of the lighter bob just after the collision (v1) is the same as the velocity of the heavier bob just before the collision.
Step 5: Calculate the velocity of the heavier bob just before the collision.
To find the velocity of the heavier bob just before the collision, we need to use the principle of conservation of momentum.
According to the principle of conservation of momentum:
m1 * u1 + m2 * u2 = m1 * v1 + m2 * v2
where u1 and u2 are the initial velocities of the lighter and heavier bobs, respectively, and v1 and v2 are their velocities just after the collision.
Since the lighter bob is released from rest (u1 = 0) and the heavier bob is initially at rest (u2 = 0), the equation becomes:
0 + 0 = m1 * v1 + m2 * v2
v1 = -((m2 / m1) * v2)
where the negative sign indicates that the lighter bob moves in the opposite direction compared to the heavier bob.
Step 6: Calculate the velocity of the heavier bob just after the collision.
The kinetic energy (KE) just before and just after the collision are the same for an elastic collision, therefore:
(1/2) * m2 * u2^2 = (1/2) * m2 * v2^2
Simplifying, we get:
u2^2 = v2^2
Since u2 = 0 (initial velocity of the heavier bob), then v2 = 0.
Therefore, the velocity of the heavier bob just after the collision (v2) is zero.
Step 7: Use the principle of conservation of energy to find the maximum height.
According to the principle of conservation of energy:
PE1 + KE1 = PE2 + KE2
Since KE1 = 0 and KE2 = (1/2) * m1 * v1^2, the equation becomes:
PE1 = PE2 + (1/2) * m1 * v1^2
Substituting the values we have, the equation becomes:
0 J = m1 * g * h2 + (1/2) * m1 * v1^2
Simplifying, we get:
h2 = -v1^2 / (2 * g)
Step 8: Substitute the values to find the height to which the lighter bob rebounds.
Substituting the values we have:
h2 = -((-((m2 / m1) * v2))^2) / (2 * g)
Since v2 = 0 (velocity of the heavier bob just after the collision), the equation becomes:
h2 = 0 / (2 * g) = 0 m
Therefore, the height to which the lighter bob rebounds after the collision is 0 meters.
To find the height to which the lighter bob rebounds after colliding elastically with the other bob, we need to apply conservation of energy.
Step 1: Find the initial potential energy of the lighter bob before it is released.
The potential energy of an object at a certain height h is given by the equation PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height.
The mass of the lighter bob is 65g = 0.065kg (since 1kg = 1000g), and the height is 0.75m. The acceleration due to gravity, g, is approximately 9.8 m/s^2.
Therefore, the initial potential energy of the lighter bob is PE = (0.065kg)(9.8 m/s^2)(0.75m).
Step 2: Find the kinetic energy of the lighter bob just before the collision.
The kinetic energy of an object in motion is given by the equation KE = 0.5mv^2, where m is the mass and v is the velocity.
Since the lighter bob is released from rest, its initial velocity would be zero. Hence, the initial kinetic energy of the lighter bob is KE = 0.5(0.065kg)(0)^2 = 0 Joules.
Step 3: Determine the total mechanical energy just before the collision.
The total mechanical energy of the system is the sum of the potential energy and kinetic energy. Therefore, just before the collision, the total mechanical energy is given by Total Energy = PE + KE.
Step 4: Calculate the velocity of the lighter bob after the collision.
Since the collision is elastic, both the momentum and mechanical energy are conserved.
Before the collision, as the lighter bob is at rest, it has no initial momentum.
After the collision, both the lighter bob and the heavier bob will have the same velocity as they separate.
Using the principle of conservation of mechanical energy and momentum, we can determine the velocity of the lighter bob after the collision.
Step 5: Calculate the final potential energy after the collision.
The final potential energy of the lighter bob after the collision is equal to the initial kinetic energy of both bobs just before the collision.
Using the equation PE = mgh, where m is the combined mass of both bobs (m = 0.065kg + 0.085kg) and g is the acceleration due to gravity, we can find the height to which the lighter bob rebounds.
Step 6: Calculate the height to which the lighter bob rebounds.
Using the equation PE = mgh, where PE is the final potential energy (which is equal to the initial kinetic energy of both bobs just before the collision), m is the combined mass of both bobs, g is the acceleration due to gravity, and h is the height to be determined, we can rearrange the equation to solve for h.
By following these steps and performing the necessary calculations, you will be able to find the height to which the lighter bob rebounds after colliding elastically with the other bob.