Two hockey pucks approach each other as shown in the figure below. Puck 1 has an initial speed of 22 m/s, and puck 2 has an initial speed of 14 m/s. They collide and stick together.

Find the components along x and y of the initial velocities of both particles.
v1xi = 19.1 m/s
v1yi = -11 m/s
v2ix = 0 m/s
v2iy = 14 m/s

I don't understand how to..

Find the final velocity of the two pucks after the collision and angle

What fraction of the initial kinetic energy is lost in the collision

To find the final velocity of the two pucks after the collision and the angle, we need to use the principle of conservation of momentum and conservation of kinetic energy.

1. Conservation of Momentum:
Since the pucks stick together after the collision, the total momentum before the collision should be equal to the total momentum after the collision. The momentum of an object is defined as the product of its mass and velocity.

Assuming both pucks have the same mass, let's denote the mass of each puck as "m."

The total momentum before the collision (p1i + p2i) is given by:
p1i = m * v1i
p2i = m * v2i

The total momentum after the collision (p1f + p2f) is given by:
p1f = m * v1f
p2f = m * v2f

Using the principle of conservation of momentum, we can equate the total momentum before and after the collision:
p1i + p2i = p1f + p2f

Substituting the given initial velocities:
m * v1i + m * v2i = m * v1f + m * v2f

Simplifying the equation:
v1i + v2i = v1f + v2f

2. Conservation of Kinetic Energy:
In an elastic collision (where no kinetic energy is lost), the total kinetic energy before the collision should be equal to the total kinetic energy after the collision.

The kinetic energy of an object is given by:
KE = (1/2) * mass * velocity^2

The total kinetic energy before the collision (KE1i + KE2i) is given by:
KE1i = (1/2) * m * v1i^2
KE2i = (1/2) * m * v2i^2

The total kinetic energy after the collision (KE1f + KE2f) is given by:
KE1f = (1/2) * m * v1f^2
KE2f = (1/2) * m * v2f^2

Using the principle of conservation of kinetic energy, we can equate the total kinetic energy before and after the collision:
KE1i + KE2i = KE1f + KE2f

Substituting the given initial velocities:
(1/2) * m * v1i^2 + (1/2) * m * v2i^2 = (1/2) * m * v1f^2 + (1/2) * m * v2f^2

Simplifying the equation:
v1i^2 + v2i^2 = v1f^2 + v2f^2

Now, we have two equations:
1. v1i + v2i = v1f + v2f (from momentum conservation)
2. v1i^2 + v2i^2 = v1f^2 + v2f^2 (from kinetic energy conservation)

Solving these two equations simultaneously will give us the final velocity of the pucks after the collision and the desired angle:

1. Subtract v2f from both sides of equation 1:
v1i - v2i = v1f + v2f - v2f

2. Square both sides of equation 1:
(v1i - v2i)^2 = (v1f + v2f - v2f)^2

3. Multiply out the squared term:
v1i^2 + v2i^2 - 2 * v1i * v2i = v1f^2 + v2f^2 - 2 * v1f * v2f + 2 * v1f * v2f - v2f^2

4. Simplify equation 3:
v1i^2 + v2i^2 - 2 * v1i * v2i = v1f^2 + v2f^2 - v2f^2

5. Cancel out the common terms:
v1i^2 + v2i^2 - 2 * v1i * v2i = v1f^2 + v2f^2

6. Substitute equation 2 into equation 5:
v1i^2 + v2i^2 - 2 * v1i * v2i = v1i^2 + v2i^2

7. Simplify equation 6:
-2 * v1i * v2i = 0

8. Divide equation 7 by -2:
v1i * v2i = 0

9. Since neither v1i nor v2i can be zero, the only possibility is that either v1i or v2i must be zero. This implies that one of the pucks is initially at rest, which contradicts the given information. Therefore, there seems to be an error in the given initial velocities. Please recheck the values.

Regarding the fraction of initial kinetic energy lost in the collision, we cannot determine that unless we know the mass of the pucks and the final velocity after the collision.