The drawing shows a collision between two pucks on an air-hockey table. Puck A has a mass of 0.021 kg and is moving along the x axis with a velocity of +5.5 m/s. It makes a collision with puck B, which has a mass of 0.041 kg and is initially at rest. The collision is not head-on. After the collision, the two pucks fly apart with the angles shown in the drawing.

| (A)
|
|
(+5.5m/s) | 65 deg.
__(A)__________________(B)__________
|
| 37 deg.
| (B)
|
(-----before collision-)(after

A(left)advance toward B(at rest) at a velocity of +5.5m/s.
B in the middle is @ rest.
A hits B and is diverted 65 degrees north east of B
B is knocked 37 degrees south east of it's initial position B.

(a) Find the final speed of puck A.

(b) Find the final speed of puck B.

You have the principle of conservation of momentum (conserved in x and y direction). Write those equations in the x and y directions. You also have the conservation of energy equation, although you may not need it. That is three equations. I am not certain you need the energy equation (two equations, two unknowns, both from momentum)

To solve this problem, we will first apply the principle of conservation of momentum in both the x and y directions. Then we will use the given angles to find the x and y components of the velocities after the collision. Finally, we will combine the x and y components to find the magnitudes of the final velocities of both pucks.

Let's start by considering the conservation of momentum in the x direction. Before the collision, only Puck A is moving along the x-axis, so its momentum is given by:

Puck A's initial momentum = mass of A * velocity of A

= 0.021 kg * 5.5 m/s

= 0.1155 kg·m/s

After the collision, the two pucks fly apart in different directions, so we need to find the x-component of the velocity for both pucks. The x-component of a velocity can be calculated by multiplying the magnitude of the velocity by the cosine of the angle. Let's denote the final speed of Puck A in the x direction as VAx, and the final speed of Puck B in the x direction as VBx.

Now, applying the conservation of momentum in the x direction, we have:

Puck A's initial momentum = Puck A's final momentum in x direction + Puck B's final momentum in x direction

0.1155 kg·m/s = (0.021 kg * VAx) + (0.041 kg * VBx)

Next, let's consider the conservation of momentum in the y direction. Since the starting y-velocity of both pucks is zero, the conservation of momentum equation in the y direction becomes:

0 = (mass of A * VAy) + (mass of B * VBy)

where VAy and VBy represent the y-components of the final velocities for Puck A and Puck B, respectively.

Given that Puck A is deflected 65 degrees northeast of B and B is deflected 37 degrees southeast of its initial position, we can find the x and y components of the final velocities for both pucks using the given angles.

For Puck A:
VAx = magnitude of VA * cos(65 degrees)
VAy = magnitude of VA * sin(65 degrees)

For Puck B:
VBx = magnitude of VB * cos(37 degrees)
VBy = magnitude of VB * sin(37 degrees)

Now, we have three equations:

1. 0.1155 kg·m/s = (0.021 kg * VAx) + (0.041 kg * VBx)
2. 0 = (0.021 kg * VAy) + (0.041 kg * VBy)

We can solve these two equations using substitution or elimination to find VAx and VAy.

Finally, to find the magnitudes of the final velocities of both pucks, we can use the Pythagorean theorem:

Final velocity of A = sqrt(VAx^2 + VAy^2)
Final velocity of B = sqrt(VBx^2 + VBy^2)

This will give us the answers to parts (a) and (b) of the question.