A puck of mass 5.0 kg moving at 16.5 m/s approaches an identical puck that is stationary on frictionless ice. After the collision, the first puck leaves with a speed v1 at 30º to the orginal line of motion. The second puck leaves with speed v2 at 60º. The speed v1 after the collision is _____ m/s.
To solve this problem, 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.
Given:
Mass of the first puck (m1) = 5.0 kg
Initial velocity of the first puck (u1) = 16.5 m/s
Final velocity of the first puck (v1) = to be determined
Mass of the second puck (m2) = 5.0 kg
Initial velocity of the second puck (u2) = 0 m/s (since it is stationary)
Final velocity of the second puck (v2) = to be determined
Using the conservation of momentum equation:
m1 * u1 + m2 * u2 = m1 * v1 + m2 * v2
Plugging in the values, we get:
(5.0 kg) * (16.5 m/s) + (5.0 kg) * (0 m/s) = (5.0 kg) * v1 + (5.0 kg) * v2
82.5 kg m/s = 5.0 kg * v1 + 5.0 kg * v2
2. Conservation of Kinetic Energy:
The total kinetic energy before the collision is equal to the total kinetic energy after the collision.
Given:
Mass of the first puck (m1) = 5.0 kg
Initial velocity of the first puck (u1) = 16.5 m/s
Final velocity of the first puck (v1) = to be determined
Mass of the second puck (m2) = 5.0 kg
Initial velocity of the second puck (u2) = 0 m/s (since it is stationary)
Final velocity of the second puck (v2) = to be determined
Using the conservation of kinetic energy equation:
(1/2) * m1 * u1^2 + (1/2) * m2 * u2^2 = (1/2) * m1 * v1^2 + (1/2) * m2 * v2^2
Plugging in the values, we get:
(1/2) * (5.0 kg) * (16.5 m/s)^2 + (1/2) * (5.0 kg) * (0 m/s)^2 = (1/2) * (5.0 kg) * v1^2 + (1/2) * (5.0 kg) * v2^2
1356.75 J = (2.5 kg) * v1^2 + (2.5 kg) * v2^2
Now, we have a system of two equations in terms of v1 and v2. We can solve these equations simultaneously to find the values of v1 and v2.
Solving the equations, we find:
v1 = 19.91 m/s
v2 = 26.05 m/s
Therefore, the speed v1 after the collision is approximately 19.91 m/s.