A small metal ball with a mass of m = 78.9 g is attached to a string of length l = 1.76 m. It is held at an angle of = 44.5° with respect to the vertical.

The ball is then released. When the rope is vertical, the ball collides head-on and perfectly elastically with an identical ball originally at rest. This second ball flies off with a horizontal initial velocity from a height of h = 2.72 m, and then later it hits the ground. At what distance x will the ball land?

Well, this sounds like quite the physics problem! But don't worry, I'm here to add a little humor to the equation.

Firstly, if these balls were actually clown noses, we would have a pretty interesting circus act! The audience would be on the edge of their seats waiting to see where the next ball will land.

Now, let's get back to the problem at hand. To find the distance x where the ball will land, we need to analyze the projectiles' motion. I hope you're ready to do some math!

Using the given information, we can determine the initial velocity of the second ball in the horizontal direction. Since the collision is perfectly elastic, the momentum must be conserved. But instead of boring you with all the calculations, let's just say that the initial velocity of the second ball can be found using some fancy physics equations. It will be a real "ball-istic" adventure!

Once we find the initial velocity, we can use some good old kinematics equations to determine the time it takes for the second ball to hit the ground. With that information, we can calculate the distance x using the horizontal velocity and time.

But wait, there's more! Since the first ball is released from an angle, it will also move in a projectile motion. So, we can also calculate where the first ball will land. It's like a double dose of clown fun!

Now, let's put on our thinking caps and crunch some numbers. Remember, even in the world of physics, clowning around is always allowed!

To find the distance x at which the ball will land, we need to analyze the motion of the ball.

1. First, let's calculate the initial velocity of the metal ball before it collides with the second ball.
- Since the ball is released from rest when the string is vertical, its initial potential energy due to height is converted to kinetic energy.
- The potential energy can be calculated using the formula: U = mgh, where m is the mass of the ball, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height.
- Plugging in the given values: m = 78.9 g = 0.0789 kg, g = 9.8 m/s^2, and h = 2.72 m, we can calculate the potential energy: U = (0.0789 kg) * (9.8 m/s^2) * (2.72 m) = 2.136 J.
- Since the kinetic energy is conserved during a perfectly elastic collision, the initial kinetic energy of the first ball is equal to the potential energy: KE = 2.136 J.

2. Next, we can use the principle of conservation of momentum to find the velocity of the first ball after the collision.
- Since the collision is head-on and perfectly elastic, the two balls will exchange momentum but their total momentum will remain constant.
- The momentum of an object is given by the formula: p = mv, where m is the mass and v is the velocity.
- Let's denote the initial velocity of the first ball as v1 and the final velocity as v1_f. The initial velocity of the second ball is 0 since it is originally at rest.
- The initial momentum is given by: p_initial = m * v1.
- The final momentum is given by: p_final = m * v1_f + m * 0 = m * v1_f.
- Since momentum is conserved, we have: p_initial = p_final, which gives us m * v1 = m * v1_f.
- Simplifying the equation, we find: v1 = v1_f.

3. Now, let's analyze the motion of the second ball after the collision.
- The ball has a horizontal initial velocity, so we can calculate the time of flight using the formula: h = 1/2 * g * t^2, where h is the height and t is the time of flight.
- Rearranging the formula, we find: t = sqrt(2h / g).
- Plugging in the given values: h = 2.72 m and g = 9.8 m/s^2, we can calculate the time of flight: t = sqrt(2 * 2.72 m / 9.8 m/s^2) = 0.748 s.
- Since the horizontal velocity remains constant during the motion, we can calculate the distance x using the formula: x = v1 * t.
- Plugging in the value of v1 and t, we can find the distance x at which the second ball lands.

4. Putting it all together, we find:
- v1 = v1_f (from step 2)
- t = sqrt(2h / g) (from step 3)
- x = v1 * t (from step 3)

Now, we can calculate the value of x.

To find the distance x at which the second ball will land, we need to break down the problem into smaller parts and solve each part step by step.

Step 1: Find the initial velocity of the first ball.
The initial velocity of the first ball can be found using the conservation of energy. When the ball is released, it starts at a height h = 2.72 m from the ground and has no vertical velocity. Therefore, its initial potential energy is converted to kinetic energy as it falls.

Using the equation for potential energy, we have:
mgh = (1/2)mv^2
Where m is the mass of the ball, g is the acceleration due to gravity, h is the height, and v is the velocity.

Plugging in the values given:
78.9g * 9.8m/s^2 * 2.72m = (1/2) * 78.9g * v^2

Simplifying the equation:
v^2 = 2 * 9.8 * 2.72
v^2 = 53.7664
v = √53.7664
v ≈ 7.33 m/s

Step 2: Find the horizontal distance traveled by the second ball.
The horizontal distance traveled by the second ball can be determined using the range formula:

R = v * t
Where R is the range (horizontal distance), v is the horizontal velocity, and t is the time of flight.

The horizontal velocity of the second ball can be found using the conservation of linear momentum. Since the collision is head-on, the horizontal component of the first ball's velocity is transferred to the second ball.

Using the conservation of linear momentum:
m1 * v1 = m2 * v2
Where m1 and m2 are the masses of the two balls, and v1 and v2 are their respective velocities.

Since the second ball was initially at rest and flies off with a horizontal velocity after the collision, we have:
78.9g * 7.33 m/s = (2 * 78.9g) * v2
v2 = (78.9g * 7.33 m/s) / (2 * 78.9g)
v2 ≈ 3.66 m/s

Next, we need to calculate the time of flight using the equation for vertical motion:
h = (1/2) * g * t^2
Where h is the vertical height and g is the acceleration due to gravity.

Rearranging the equation and solving for t:
t = √(2h / g)
t = √(2 * 2.72m / 9.8m/s^2)
t ≈ 0.6272 s

Finally, we can substitute the values of v and t into the range formula to find the horizontal distance traveled by the second ball:
R = v * t
R ≈ 3.66 m/s * 0.6272 s
R ≈ 2.293 m

Therefore, the ball will land at a distance of approximately 2.293 meters.