A collision occurs between two pucks on an air-hockey table. Puck A has a mass of 0.020 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 0.040 kg and is initially at rest. The collision is not head-on. After the collision, the two pucks fly apart with puck A 65º above the x-axis and puck B 37º below the x-axis.

(a) Find the final speed of puck A.

(b) Find the final speed of puck B.

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Well, well, we have a little game of air-hockey going on here! Let's see if I can provide you with some humorously accurate answers.

(a) To find the final speed of puck A, we need to consider the conservation of momentum. The initial momentum is given by the mass of puck A multiplied by its initial velocity: 0.020 kg * 5.5 m/s = 0.11 kg·m/s. After the collision, puck A flies off at an angle of 65º above the x-axis. To find the x-component of its velocity, we multiply the final speed by the cosine of 65º. So, the final velocity along the x-axis is 0.11 kg·m/s * cos(65º). To find the final speed, we divide the x-component of velocity by the cosine of the angle, which gives us 0.11 kg·m/s * cos(65º) / cos(65º) = 0.11 kg·m/s. Phew, that was a lot of math for a clown!

(b) Now let's move on to puck B. After the collision, it flies off at an angle of 37º below the x-axis. Similar to before, we need to find the x-component of its velocity. Since puck B was initially at rest, its initial momentum is zero. So, we don't need to calculate anything here, just enjoy the fact that it didn't go anywhere fast. Hooray for lazy pucks!

So, to sum it up:
(a) The final speed of puck A is approximately 0.11 kg·m/s.
(b) The final speed of puck B is approximately 0 m/s, because it was too cool to move quickly.

I hope that brought a bit of laughter to your air-hockey match!

To solve this problem, we need to use the principles of conservation of momentum and conservation of kinetic energy.

(a) Find the final speed of puck A:
Let's assume the final velocity of puck A is v_A and the final velocity of puck B is v_B. Since the collision is not head-on, the x and y components of momentum and kinetic energy will be conserved separately.

1. Conservation of momentum in the x-direction:
Before the collision, the momentum in the x-direction is given by:
p_x_initial = (mass_A * velocity_A) + (mass_B * velocity_B)
Since puck B is initially at rest (velocity_B = 0), we can simplify this to:
p_x_initial = mass_A * velocity_A

After the collision, the momentum in the x-direction is given by:
p_x_final = (mass_A * v_A * cos(65°)) + (mass_B * v_B * cos(37°))

Using the principle of conservation of momentum, we have:
p_x_initial = p_x_final
mass_A * velocity_A = (mass_A * v_A * cos(65°)) + (mass_B * v_B * cos(37°))

2. Conservation of kinetic energy in the y-direction:
Before the collision, the kinetic energy in the y-direction is zero for both pucks, as they are only moving along the x-axis.

After the collision, the kinetic energy in the y-direction is given by:
KE_y_final = (1/2 * mass_A * v_A^2 * sin^2(65°)) + (1/2 * mass_B * v_B^2 * sin^2(37°))

Using the principle of conservation of kinetic energy, we have:
KE_y_final = 0

Now we have two equations to solve for v_A and v_B.

(b) Find the final speed of puck B:
Using the equation we derived from conservation of momentum:
mass_A * velocity_A = (mass_A * v_A * cos(65°)) + (mass_B * v_B * cos(37°))

Substituting the known values:
0.020 kg * 5.5 m/s = (0.020 kg * v_A * cos(65°)) + (0.040 kg * v_B * cos(37°))

Simplifying and rearranging the equation, we can solve for v_B:
0.11 = v_A * cos(65°) + 2 * v_B * cos(37°) ... (Equation 1)

Next, let's use the conservation of kinetic energy equation:
KE_y_final = (1/2 * mass_A * v_A^2 * sin^2(65°)) + (1/2 * mass_B * v_B^2 * sin^2(37°))

Since KE_y_final is zero:
0 = (1/2 * 0.020 kg * v_A^2 * sin^2(65°)) + (1/2 * 0.040 kg * v_B^2 * sin^2(37°))

Simplifying and rearranging the equation, we can solve for v_A^2:
v_A^2 = [-0.020 kg * sin^2(65°) / (0.040 kg * sin^2(37°))] * v_B^2 ... (Equation 2)

Now, we have two equations with two unknowns (v_A and v_B), which we can solve simultaneously to find the values.

Please note that you will need a scientific calculator to evaluate the trigonometric functions and perform the calculations.

Use conservation of momentum.

In the x direction:

0.02*5.5 = 0.02*vA*cos(65) + 0.04*vB*cos(37)

In the y direction:

0 = 0.02*vA*sin(65) - 0.02*vB*sin(37)

You should be able to use algebra to solve for these equations, since you have two equations with 2 unknowns; vA is the final speed of puck A; vB is the final speed of puck B