Two identical steel balls, each of mass 5.0 kg, are suspended from strings of length 29 cm so that they touch when in their equilibrium position. We pull one of the balls back until its string makes an angle θ = 25° with the vertical and let it go. It collides elastically with the other ball. How high will the other ball rise?

Suppose that instead of steel balls we use putty balls. They will collide inelastically and remain stuck together after the collision. How high will the balls rise after the collision?

To conserve both momentum and energy, the first ball stops and the second one continues with the total energy.

In part 2, momentum is conserved but the motion continues with twice the mass.

m g (.29)(1-cos 25) = .5 m v^2

solve for v
then conservation of momentum at bottom
(m+m) * new v = m v

new v = (1/2) v

then at top
g (.29)(1-cos A) = .5 (new v)^2
solve for A

To find the height the other ball rises, we can use conservation of mechanical energy. When the ball is at its highest point, all of its initial energy is converted into potential energy.

1) For the elastic collision case:
Let's calculate the speed of the first ball before the collision using the given information.

The length of the string is 29 cm = 0.29 m.
The angle θ = 25°.
Using trigonometry, we find that the vertical component of the displacement is given by:
h = 0.29 m * sin(θ)
h = 0.29 m * sin(25°)
h ≈ 0.12 m

The potential energy at the top of the swing is given by the formula:
PE = m * g * h
PE = (2 * 5.0 kg) * (9.8 m/s^2) * (0.12 m)
PE = 11.76 N*m

Since this is an elastic collision, the total mechanical energy is conserved. The maximum potential energy is equal to the maximum kinetic energy before the collision.

Assuming the initial kinetic energy is equal for both balls in the equilibrium position, the kinetic energy is given by:
KE = (1/2) * m * v^2

The total initial kinetic energy is twice the kinetic energy of one ball:
KE_initial = 2 * (1/2) * 5.0 kg * v^2
KE_initial = 5.0 kg * v^2

At the highest point, the second ball will have all the energy, which is equal to the potential energy at the top.

Therefore, we can set the initial kinetic energy equal to the potential energy to solve for the final height.

KE_initial = PE
5.0 kg * v^2 = 11.76 N*m

v^2 = 11.76 N*m / 5.0 kg
v^2 = 2.35 m^2/s^2

v ≈ 1.53 m/s (taking the positive value since speed is scalar)

At the highest point, the second ball has zero kinetic energy, so all the initial kinetic energy is converted to potential energy.

PE = m * g * h'
11.76 N*m = 5.0 kg * 9.8 m/s^2 * h'

h' = 11.76 N*m / (5.0 kg * 9.8 m/s^2)
h' ≈ 0.24 m

Therefore, the second ball will rise approximately 0.24 m after the elastic collision.

2) For the inelastic collision case:
Since the balls stick together after the collision, they will have a combined mass of 2 * 5.0 kg = 10.0 kg.

Using conservation of momentum before and after the collision:

Initial momentum = Final momentum
m1 * v1_initial = (m1 + m2) * v_final

The initial momentum of the first ball is given by:
m1 * v1_initial = 5.0 kg * 1.53 m/s (from the previous calculation)
m1 * v1_initial = 7.65 kg*m/s

The final velocity of the balls after the collision is the same, denoted as v_final.

5.0 kg * 1.53 m/s = (5.0 kg + 5.0 kg) * v_final
7.65 kg*m/s = 10.0 kg * v_final

v_final = 7.65 kg*m/s / 10.0 kg
v_final ≈ 0.76 m/s

To find the maximum height after the collision, we used the same approach as in the elastic collision case:

PE = m * g * h' (potential energy at the highest point)
PE = 10.0 kg * 9.8 m/s^2 * h'

The potential energy is equal to the initial kinetic energy:

PE = KE_initial
10.0 kg * 9.8 m/s^2 * h' = 7.65 kg*m^2/s^2 (using the velocity from the previous calculation)

h' = 7.65 kg*m^2/s^2 / (10.0 kg * 9.8 m/s^2)
h' ≈ 0.08 m

Therefore, the balls will rise approximately 0.08 m after the inelastic collision.

To answer this question, we need to analyze the conservation of energy and momentum in both scenarios.

First, let's consider the scenario with elastic collisions:

Step 1: Calculate the potential energy of the first ball when it is pulled back using the equation:

Potential Energy = mass * gravity * height
Potential Energy = 5.0 kg * 9.8 m/s^2 * h (where h is the height)

Step 2: When the ball is released, it will convert its potential energy into kinetic energy.

Step 3: As the first ball collides elastically with the second ball, momentum is conserved. Therefore, the total momentum before and after the collision will be the same.

Step 4: The second ball will gain some kinetic energy during the collision and convert it into potential energy as it rises. We need to find the height it will reach.

Now, let's consider the scenario with inelastic collisions:

Step 1: Instead of analyzing the conservation of energy and momentum separately, we can analyze them together as they are conserved in an inelastic collision. Therefore, we can apply the law of conservation of momentum and energy directly to this scenario.

Step 2: The total kinetic energy before the collision will be equal to the potential energy gained by the balls after the collision as they rise.

Now, let's calculate the heights in both scenarios:

For the scenario with elastic collisions:
Step 1: Calculate the initial potential energy of the first ball:
Potential Energy = 5.0 kg * 9.8 m/s^2 * h1

Step 2: Equate the initial potential energy to the sum of the final potential energy of both balls:
Potential Energy = 5.0 kg * 9.8 m/s^2 * h1 = 5.0 kg * 9.8 m/s^2 * h2

Step 3: Calculate the initial kinetic energy of the first ball using the angle θ:
Kinetic Energy = (1/2) * mass * velocity^2
velocity = (length of string) * angular velocity = 0.29 m * √(g / (length of string)) * sin(θ)
Kinetic Energy = (1/2) * 5.0 kg * [(0.29 m * √(9.8 m/s^2 / 0.29 m)) * sin(25°)]^2

Step 4: Equate the initial kinetic energy to the final kinetic energy of both balls:
Kinetic Energy = (1/2) * mass * (velocity)^2 = (1/2) * 5.0 kg * [(-0.29 m * √(9.8 m/s^2 / 0.29 m)) * sin(θ)]^2

Now, solve the equation system to find the value of h2, which is the height the second ball will rise to.

For the scenario with inelastic collisions:
Step 1: Apply the conservation of energy and momentum equation directly to this scenario:
(1/2) * mass * (velocity)^2 = mass * gravity * h

Now, solve the equation to find the value of h, which is the common height both balls will rise to after the collision.

By calculating these equations, we can determine the heights to answer the original question.