Two identical steel balls, each of mass 4.40 kg, are
suspended from strings of length 25.0 cm so that they touch
when in their equilibrium position. We pull one of the balls
back until its string makes an angle θ = 70.0◦ with the vertical
and let it go. It collides elastically with the other ball. How
high will the other ball rise?
since energy is conserved, and the balls are identical, the 2nd ball will rise just as high as the first ball was.
Be prepared for the next step, where the balls are not identical in mass!
To determine how high the other ball will rise after the collision, we can use the principle of conservation of energy.
1. First, let's find the initial potential energy of the system. The initial potential energy is given by the equation: PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height.
2. Since both balls are in the equilibrium position initially, the potential energy is zero for both balls.
3. When one ball is pulled back and released, it gains gravitational potential energy equal to mgh, where h is the vertical displacement of the ball from the equilibrium position.
4. The ball collides elastically with the other ball, and after the collision, the total mechanical energy of the system remains constant.
5. The energy gained by the first ball is transferred to the second ball, causing it to rise.
6. The maximum height reached by the second ball can be found by equating the initial potential energy of the first ball to the potential energy gained by the second ball.
Let's calculate the height using the above steps:
Step 1: Calculate the gravitational potential energy gained by the first ball:
PE1 = m1 * g * h1
Step 2: Assume the maximum height reached by the second ball is h2.
Step 3: Calculate the gravitational potential energy gained by the second ball:
PE2 = m2 * g * h2
Step 4: Since the collision is elastic, the total mechanical energy before and after the collision remains constant:
Initial total energy = Final total energy
PE1 + KE1 = PE2 + KE2
Step 5: The initial kinetic energy of the first ball is zero (KE1 = 0) as it is pulled back and released from rest.
This simplifies the equation to:
PE1 = PE2 + KE2
m1 * g * h1 = m2 * g * h2 + m2 * v2^2 / 2
Step 6: We know the angle θ and the length of the string. We can use trigonometry to relate the vertical displacement h1 to the horizontal displacement x:
x = L * sin(θ)
Where L is the length of the string.
Using this relationship, we can find the velocity of the first ball just before the collision (v1) using conservation of energy:
v1^2 = 2 * g * (h1 - h2)
Step 7: Substitute the expression for v1^2 into the equation from step 5:
m1 * g * h1 = m2 * g * h2 + m2 * (2 * g * (h1 - h2)) / 2
Simplifying gives:
m1 * g * h1 = m2 * g * h2 + m2 * g * (h1 - h2)
m1 * g * h1 = m2 * g * h1
Dividing both sides by m1 * g:
h1 = h2
Therefore, the maximum height reached by the second ball will be equal to the displacement of the first ball, which is h1.