The drawing shows a collision between two pucks on an air-hockey table. Puck A has a mass of 0.0280 kg and is moving along the x axis with a velocity of +6.64 m/s. It makes a collision with puck B, which has a mass of 0.0560 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. Find the speed of (a) puck A and (b) puck B.
Do you think we might need to know the angles?
To find the speed of puck A and puck B after the collision, we can use the principles of conservation of momentum and conservation of kinetic energy.
1. Conservation of Momentum:
The total momentum before the collision is equal to the total momentum after the collision.
Mathematically, we can represent this as:
(mass of A * velocity of A before) + (mass of B * velocity of B before) = (mass of A * velocity of A after) + (mass of B * velocity of B after)
2. Conservation of Kinetic Energy:
The total kinetic energy before the collision is equal to the total kinetic energy after the collision.
Mathematically, we can represent this as:
(1/2 * mass of A * (velocity of A before)^2) + (1/2 * mass of B * (velocity of B before)^2) = (1/2 * mass of A * (velocity of A after)^2) + (1/2 * mass of B * (velocity of B after)^2)
Given information:
Mass of A (mA) = 0.0280 kg
Velocity of A before (vA before) = +6.64 m/s
Mass of B (mB) = 0.0560 kg
Velocity of B before (vB before) = 0 m/s
Now, let's calculate the velocities of A and B after the collision.
Using the conservation of momentum:
(mA * vA before) + (mB * vB before) = (mA * vA after) + (mB * vB after)
(0.0280 kg * 6.64 m/s) + (0.0560 kg * 0 m/s) = (0.0280 kg * vA after) + (0.0560 kg * vB after)
0.1859 kg·m/s = 0.0280 kg * vA after + 0.0560 kg * vB after
0.1859 kg·m/s - 0.0280 kg * vA after = 0.0560 kg * vB after
Next, let's calculate the velocities of A and B using the conservation of kinetic energy.
(1/2 * mA * (vA before)^2) + (1/2 * mB * (vB before)^2) = (1/2 * mA * (vA after)^2) + (1/2 * mB * (vB after)^2)
(1/2 * 0.0280 kg * (6.64 m/s)^2) + (1/2 * 0.0560 kg * (0 m/s)^2) = (1/2 * 0.0280 kg * (vA after)^2) + (1/2 * 0.0560 kg * (vB after)^2)
0.629 m²/s² = 0.01498 kg * (vA after)^2 + 0.0280 kg * (vB after)^2
Now we have two equations:
0.1859 kg·m/s - 0.0280 kg * vA after = 0.0560 kg * vB after
0.629 m²/s² = 0.01498 kg * (vA after)^2 + 0.0280 kg * (vB after)^2
We can solve this system of equations to find the velocities of A and B after the collision.