In a game of pool, the cue ball strikes another ball of the same mass and initially at rest. After the collision, the cue ball moves at 3.10 m/s along a line making an angle of 27.0° with its original direction of motion, and the second ball has a speed of 2.30 m/s. Find (a) the angle between the direction of motion of the second ball and the original direction of motion of the cue ball and (b) the original speed of the cue ball.

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To solve this problem, we can utilize the principles of conservation of momentum and conservation of kinetic energy.

Let's start by analyzing the conservation of momentum. In an isolated system, the total momentum before the collision is equal to the total momentum after the collision.

(a) The angle between the direction of motion of the second ball and the original direction of motion of the cue ball can be found using the concept of momentum conservation.

First, let's express the initial and final momentum in terms of their x and y components.

The initial momentum of the system (cue ball + second ball) is given by:

P_initial = m1 * v1 + m2 * v2

Where m1 and m2 are the masses of the cue ball and the second ball, respectively, and v1 and v2 are their initial velocities. Since the second ball is initially at rest, v2 = 0, and the expression simplifies to:

P_initial = m1 * v1

After the collision, the cue ball moves at a speed of 3.10 m/s along an angle of 27.0° with its original direction of motion. Let's break down this velocity into its x and y components:

v1x = v1 * cos(theta)
v1y = v1 * sin(theta)

Where theta is the angle between the cue ball's original direction of motion and the line it moves along after the collision.

Therefore, the final momentum of the system (cue ball + second ball) is:

P_final = m1 * v1x + m2 * v2x

Since the second ball moves along a line making an angle with the original direction of the cue ball, its x-component of velocity v2x is given by:

v2x = v2 * cos(phi)

Where phi is the angle between the second ball's direction of motion and the original direction of the cue ball.

Since momentum is conserved, we have:

P_initial = P_final

m1 * v1 = m1 * v1x + m2 * v2x

Substituting the expressions for v1x and v2x, we get:

m1 * v1 = m1 * v1 * cos(theta) + m2 * v2 * cos(phi)

Now, we can solve for the angle phi, which represents the angle between the direction of motion of the second ball and the original direction of motion of the cue ball.

(b) To determine the original speed of the cue ball, we can use the principle of conservation of kinetic energy. In an elastic collision, the total kinetic energy before the collision is equal to the total kinetic energy after the collision.

The initial kinetic energy of the system is given by:

KE_initial = 1/2 * m1 * v1^2

After the collision, the cue ball's velocity can be separated into its x and y components:

v1x = v1 * cos(theta)
v1y = v1 * sin(theta)

Therefore, the final kinetic energy of the system is:

KE_final = 1/2 * m1 * v1x^2 + 1/2 * m2 * v2^2

Since kinetic energy is conserved, we have:

KE_initial = KE_final

1/2 * m1 * v1^2 = 1/2 * m1 * v1x^2 + 1/2 * m2 * v2^2

Substituting the expressions for v1x and v2, we get:

1/2 * m1 * v1^2 = 1/2 * m1 * (v1 * cos(theta))^2 + 1/2 * m2 * v2^2

Now, we can solve the equation for the initial speed of the cue ball, v1, to find our answer.