Two pucks collide on an air hockey table. Puck A has a mass of 16 g and is initially traveling in the +x direction at 6.2 m/s. Puck B has a mass of 64 g and is initially at rest. After the pucks collide, puck A moves away at an angle of 58 above the +x axis, while puck B travels at an angle of 41 below the +x axis.
Calculate puck A's final speed.
To calculate puck A's final speed, we can use the principles of conservation of momentum and conservation of kinetic energy.
1. Conservation of momentum: In a collision, the total momentum before the collision is equal to the total momentum after the collision. Mathematically, we can write it as:
momentum before collision = momentum after collision
The momentum (p) of an object is given by the product of its mass (m) and velocity (v): p = m * v.
Therefore, we can write the equation as:
(m1 * v1) + (m2 * v2) = (m1 * v1f) + (m2 * v2f)
Here, m1 and m2 are the masses of pucks A and B, v1 and v2 are their initial velocities, and v1f and v2f are their final velocities.
2. Conservation of kinetic energy: In this case, we assume that the collision is elastic, meaning that kinetic energy is conserved during the collision. Mathematically, we can write the equation as:
(1/2 * m1 * v1^2) + (1/2 * m2 * v2^2) = (1/2 * m1 * v1f^2) + (1/2 * m2 * v2f^2)
Now, let's calculate puck A's final speed using these equations:
Given:
- m1 = 16 g = 0.016 kg
- v1 = 6.2 m/s
- m2 = 64 g = 0.064 kg
- v2 = 0 m/s (initially at rest)
- v1f = final speed of puck A (to be found)
1. Conservation of momentum equation:
(m1 * v1) + (m2 * v2) = (m1 * v1f) + (m2 * v2f)
(0.016 kg * 6.2 m/s) + (0.064 kg * 0 m/s) = (0.016 kg * v1f) + (0.064 kg * v2f)
0.0992 kg·m/s = 0.016 kg·v1f + 0
Since puck B is initially at rest (v2 = 0 m/s), the equation simplifies to:
0.0992 kg·m/s = 0.016 kg·v1f
2. Conservation of kinetic energy equation:
(1/2 * m1 * v1^2) + (1/2 * m2 * v2^2) = (1/2 * m1 * v1f^2) + (1/2 * m2 * v2f^2)
(1/2 * 0.016 kg * (6.2 m/s)^2) + (1/2 * 0.064 kg * (0 m/s)^2) = (1/2 * 0.016 kg * v1f^2) + (1/2 * 0.064 kg * v2f^2)
0.3032 Joules = (1/2 * 0.016 kg * v1f^2) + 0
Since puck B is initially at rest (v2 = 0 m/s), the equation simplifies to:
0.3032 Joules = (1/2 * 0.016 kg * v1f^2)
Now, we have two equations:
1) 0.0992 kg·m/s = 0.016 kg·v1f
2) 0.3032 Joules = (1/2 * 0.016 kg * v1f^2)
Solving equation 1 for v1f:
v1f = (0.0992 kg·m/s) / (0.016 kg) = 6.2 m/s
Therefore, puck A's final speed is 6.2 m/s.