A 0.25-kg punk, Initially at rest on a frictionless horizontal surface, is struck by a 0.20kg punk that is initially moving along the x-axis with a velocity of 2.20 m/sec. After the collision, the 0.20 kg punk has a speed of 1.5 m/s at an angle of 37 to the positive x-axis. Calculate:

A.The velocity of the .25 kg punk after collision

B.The angle of the second punk

C.The total energy before and after collision.

D.What is the percentage of energy lost in this collision?

I have had punks try to sit on my air tables, however, they didn't last long in physics. Oh, you mean pucks.

The math is too difficult to do here.
draw the figure.
write momentum equations in x direction, and y direction.
Unknowns will be velocity of .25 puck, and the angle. Two equations, two unknowns.

To solve these questions, we will use the principle of conservation of momentum and conservation of energy.

A. The velocity of the 0.25 kg punk after the collision can be found using the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.

The momentum before the collision is given by:
Initial momentum of the 0.25 kg punk = 0 (as it is initially at rest)
Initial momentum of the 0.20 kg punk = mass * velocity = 0.20 kg * 2.20 m/s = 0.44 kg m/s

Therefore, the total initial momentum is 0.44 kg m/s.

The momentum after the collision can be calculated by considering the velocity of the 0.20 kg punk after the collision. Let's denote the final velocity of the 0.20 kg punk as V2 and the final velocity of the 0.25 kg punk as V1.

Then the x-component of the final velocity of the 0.20 kg punk can be found using:
V2x = V2 * cos(37°)

The y-component of the final velocity of the 0.20 kg punk can be found using:
V2y = V2 * sin(37°)

The momentum after the collision is given by:
Final momentum of the 0.20 kg punk = 0.20 kg * V2x = 0.20 kg * V2 * cos(37°)
Final momentum of the 0.25 kg punk = 0.25 kg * V1

The total final momentum is equal to the initial momentum:
0.44 kg m/s = 0.20 kg * V2 * cos(37°) + 0.25 kg * V1

Now we can solve for V1:
V1 = (0.44 kg m/s - 0.20 kg * V2 * cos(37°)) / 0.25 kg

B. The angle of the second punk can be found using the y-component of its velocity after the collision:
sin(θ) = V2y / V2
θ = sin^(-1)(V2y / V2)

C. The total energy before the collision is given by the sum of the kinetic energies of both punks:

Initial energy = 1/2 * mass1 * velocity1^2 + 1/2 * mass2 * velocity2^2
= 1/2 * 0.25 kg * 0^2 + 1/2 * 0.20 kg * (2.20 m/s)^2

The total energy after the collision is given by the sum of the kinetic energies of both punks:

Final energy = 1/2 * mass1 * V1^2 + 1/2 * mass2 * V2^2

D. The percentage of energy lost in this collision can be calculated by comparing the initial and final energies:
Energy lost = Initial energy - Final energy
Percentage of energy lost = (Energy lost / Initial energy) * 100