A ball is attached to one end of a wire, the other end being fastened to the ceiling. The wire is held horizontal, and the ball is released from rest (see the drawing). It swings downward and strikes a block initially at rest on a horizontal frictionless surface. Air resistance is negligible, and the collision is elastic. The masses of the ball and block are, respectively, 1.7 kg and 2.4 kg, and the length of the wire is 1.29 m. Find the velocity (magnitude and direction) of the ball (a) just before the collision, and (b) just after the collision.

m1=1.7 kg, m2 =2.4 kg, L=1.29 m.

PE=KE
m1•g•L=m1•v²/2
v=sqrt(2•g•L)
The velocities of the bodies after elastic collision are
u1=(m1-m2)v/(m1+m2),
u2=2m1v1/(m1+m2)
“-“ means the motion in opposite direction

To find the velocity of the ball just before and just after the collision, we need to analyze the conservation of energy and momentum.

First, let's find the velocity of the ball just before the collision:

(a) Just before the collision:
The potential energy of the ball at the starting position is converted into kinetic energy as it swings down. Using the law of conservation of energy:

Potential energy at starting position = Kinetic energy just before the collision

m * g * h = (1/2) * m * v^2

Where:
m = mass of the ball = 1.7 kg
g = acceleration due to gravity = 9.8 m/s^2
h = height (maximum displacement of the ball from the starting position) = length of the wire = 1.29 m
v = velocity of the ball just before the collision (to be determined)

Substituting the values:

1.7 kg * 9.8 m/s^2 * 1.29 m = (1/2) * 1.7 kg * v^2

v^2 = (2 * 1.7 kg * 9.8 m/s^2 * 1.29 m) / 1.7 kg
v^2 ≈ 31.68 m^2/s^2
v ≈ √(31.68) ≈ 5.63 m/s

So, the magnitude of the velocity of the ball just before the collision is approximately 5.63 m/s. The direction is downwards because the ball is swinging down.

Now, let's find the velocity of the ball just after the collision:

(b) Just after the collision:
Since the collision is elastic, the momentum of the system is conserved.

Initial momentum = Final momentum

m1 * v1 + m2 * v2 = m1 * v1' + m2 * v2'

Where:
m1 = mass of the ball = 1.7 kg
m2 = mass of the block = 2.4 kg
v1 = velocity of the ball just before the collision = 5.63 m/s (downwards)
v2 = velocity of the block just before the collision = 0 m/s (initially at rest)
v1' = velocity of the ball just after the collision (to be determined)
v2' = velocity of the block just after the collision (to be determined)

Substituting the values:

1.7 kg * 5.63 m/s + 2.4 kg * 0 m/s = 1.7 kg * v1' + 2.4 kg * v2'

8.15 kg·m/s = 1.7 kg * v1' + 2.4 kg * v2' (Equation 1)

Since the collision is elastic, the relative velocities of the ball and block will be reversed.

v1' = -v1
v2' = -v2

Substituting these values into Equation 1:

8.15 kg·m/s = 1.7 kg * (-5.63 m/s) + 2.4 kg * (0 m/s)

8.15 kg·m/s = -9.571 kg·m/s

This equation is not satisfied, which means that momentum cannot be conserved. Therefore, we need to recheck the given data or assumptions, such as the collision being perfectly elastic.

But based on the given information, the velocity of the ball just after the collision cannot be determined.

To solve this problem, we will use the principles of conservation of mechanical energy and linear momentum.

(a) Just before the collision:
Since the wire is held horizontally, the ball is initially at a height equal to the length of the wire (1.29 m). Therefore, the initial potential energy of the ball is converted into kinetic energy as it swings downward.

The potential energy at the initial height can be calculated using the formula:

Potential Energy = mass * gravity * height

Given that the mass of the ball is 1.7 kg and the height is 1.29 m, the potential energy becomes:

Potential Energy = 1.7 kg * 9.8 m/s^2 * 1.29 m = 21.19 J

Using the principle of conservation of mechanical energy, the potential energy at the initial height is equal to the kinetic energy just before the collision:

Kinetic Energy = Potential Energy

1/2 * mass * velocity^2 = 21.19 J

Solving for velocity gives:

velocity^2 = 2 * (21.19 J) / 1.7 kg
velocity^2 = 24.9 m^2/s^2

Therefore, the magnitude of the velocity just before the collision is:

velocity = sqrt(24.9 m^2/s^2) = 4.99 m/s

Since the ball is swinging straight downward, the direction of the velocity just before the collision is downward.

(b) Just after the collision:
Given that the collision is elastic, both linear momentum and kinetic energy are conserved.

The linear momentum before the collision is given by:

momentum = mass * velocity

momentum before collision = 1.7 kg * (-4.99 m/s) = -8.48 kg*m/s

The final momentum after the collision must also be equal to -8.48 kg*m/s.

Next, we can use the principle of conservation of kinetic energy to find the velocity just after the collision.

The kinetic energy just after the collision is equal to the kinetic energy just before the collision:

1/2 * m_ball * (velocity_ball)^2 = 1/2 * m_block * (velocity_block)^2

1/2 * 1.7 kg * (velocity_ball)^2 = 1/2 * 2.4 kg * (velocity_block)^2

Since the velocities of the ball and the block are opposite in direction (one is up and the other is down), the equation can be written as:

1.7 kg * (velocity_ball) = 2.4 kg * (-velocity_block)

Substituting the value of the ball's velocity (-4.99 m/s), we can solve for the block's velocity:

1.7 kg * (-4.99 m/s) = 2.4 kg * (-velocity_block)

velocity_block = (1.7 kg * 4.99 m/s) / 2.4 kg
velocity_block = -3.53 m/s

Therefore, the magnitude of the velocity of the block just after the collision is 3.53 m/s, and its direction is to the left.

To summarize:
(a) The magnitude of the velocity of the ball just before the collision is 4.99 m/s, downward.
(b) The magnitude of the velocity of the ball just after the collision is -4.99 m/s, upward.
The magnitude of the velocity of the block just after the collision is 3.53 m/s, to the left.