An ice skater of mass 55.3 kg stops to tie his shoe and is at rest. A second ice skater of mass 32.7 kg is moving at a velocity of 15.9 m/s when he strikes the first ice skater. After the collision the second ice skater has a velocity of 7.5 m/s at an angle of 45.0 degrees above the direction of the initial velocity.
What is the magnitude of the velocity of the first ice skater after the collision? Give your answer in m/s to the first decimal place.
To find the magnitude of the velocity of the first ice skater after the collision, we can use the principles of conservation of momentum and conservation of energy.
Conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision. Mathematically, this can be written as:
(m1 * v1) + (m2 * v2) = (m1 * v1') + (m2 * v2')
where m1 and m2 are the masses of the first and second ice skater respectively, v1 and v2 are their initial velocities before the collision, and v1' and v2' are their velocities after the collision.
In this case, the first ice skater is at rest before the collision (v1 = 0 m/s), so the equation simplifies to:
(m2 * v2) = (m1 * v1') + (m2 * v2')
Plugging in the given values:
m1 = 55.3 kg (mass of the first ice skater)
m2 = 32.7 kg (mass of the second ice skater)
v2 = 15.9 m/s (velocity of the second ice skater before the collision)
v2' = 7.5 m/s (velocity of the second ice skater after the collision)
(32.7 kg * 15.9 m/s) = (55.3 kg * v1') + (32.7 kg * 7.5 m/s)
520.53 kg·m/s = (55.3 kg * v1') + (245.25 kg·m/s)
Simplifying further:
520.53 kg·m/s - 245.25 kg·m/s = 55.3 kg * v1'
275.28 kg·m/s = 55.3 kg * v1'
Dividing both sides by 55.3 kg:
v1' = 275.28 kg·m/s / 55.3 kg
v1' ≈ 4.98 m/s
Therefore, the magnitude of the velocity of the first ice skater after the collision is approximately 5.0 m/s.