A 74.0 kg ice skater moving to the right with a velocity of 2.90 m/s throws a 0.15 kg snowball to the right with a velocity of 23.2 m/s relative to the ground.
(a) What is the velocity of the ice skater after throwing the snowball? Disregard the friction between the skates and the ice.
(b) A second skater initially at rest with a mass of 61.50 kg catches the snowball. What is the velocity of the second skater after catching the snowball in a perfectly inelastic collision?
(PLEASE HELP!)
To solve this problem, we can use the principle of conservation of linear momentum. The total momentum before throwing the snowball is equal to the total momentum after throwing the snowball.
(a) To find the velocity of the ice skater after throwing the snowball, we can use the formula for conservation of linear momentum, which states:
(m1 * v1) + (m2 * v2) = (m1 * v1') + (m2 * v2')
Where:
m1 is the mass of the ice skater before throwing the snowball (74.0 kg)
v1 is the velocity of the ice skater before throwing the snowball (2.90 m/s)
m2 is the mass of the snowball (0.15 kg)
v2 is the velocity of the snowball relative to the ground (23.2 m/s)
v1' is the velocity of the ice skater after throwing the snowball (unknown)
Plugging in the known values, we have:
(74.0 kg * 2.90 m/s) + (0.15 kg * 23.2 m/s) = (74.0 kg * v1') + (0.15 kg * 23.2 m/s)
Using this equation for conservation of linear momentum, we can solve for v1':
(74.0 kg * 2.90 m/s) + (0.15 kg * 23.2 m/s) = (74.0 kg * v1') + (0.15 kg * 23.2 m/s)
Calculating the left side:
214.6 kg·m/s = (74.0 kg * v1') + (0.15 kg * 23.2 m/s)
Simplifying the equation:
214.6 kg·m/s - (0.15 kg * 23.2 m/s) = 74.0 kg * v1'
209.9 kg·m/s = 74.0 kg * v1'
Now, we can solve for v1':
v1' = 209.9 kg·m/s / 74.0 kg
v1' ≈ 2.835 m/s
Therefore, the velocity of the ice skater after throwing the snowball is approximately 2.835 m/s to the right.
(b) To find the velocity of the second skater after catching the snowball in a perfectly inelastic collision, we again use conservation of linear momentum. The total momentum before the collision is equal to the total momentum after the collision.
(m1 * v1) + (m2 * v2) = (M * v')
Where:
m1 is the mass of the ice skater before throwing the snowball (74.0 kg)
v1 is the velocity of the ice skater before throwing the snowball (2.90 m/s)
m2 is the mass of the snowball (0.15 kg)
v2 is the velocity of the snowball relative to the ground (23.2 m/s)
M is the combined mass of the second skater and the snowball (61.50 kg + 0.15 kg)
v' is the velocity of the second skater after catching the snowball (unknown)
Plugging in the known values, we have:
(74.0 kg * 2.90 m/s) + (0.15 kg * 23.2 m/s) = ((61.50 kg + 0.15 kg) * v')
Calculating the left side:
214.6 kg·m/s = (61.65 kg * v')
Now, we can solve for v':
v' = 214.6 kg·m/s / 61.65 kg
v' ≈ 3.48 m/s
Therefore, the velocity of the second skater after catching the snowball in a perfectly inelastic collision is approximately 3.48 m/s to the right.