A 66kg skater, at rest on frictionless ice, tosses a 6.0kg snowball with velocity by v=53.0i+14.0j m/s, where the x- and y- axes are in the horizontal plane.Find x-component (v_x) of the subsequent velocity of the skater.Find y-component (v_y) of the subsequent velocity of the skater.
To find the x-component (v_x) of the subsequent velocity of the skater, we need to apply the law of conservation of momentum. According to this law, the total momentum before an event is equal to the total momentum after the event, provided no external forces act on the system.
In this case, the skater and the snowball form a system. Initially, the skater is at rest, so the total momentum before the throw is zero. After the snowball is thrown, it acquires a velocity v=53.0i+14.0j m/s.
To find the x-component of the skater's subsequent velocity, we use the equation:
m1 * v1x + m2 * v2x = (m1 + m2) * vf_x
Where:
m1 = mass of the skater = 66 kg
m2 = mass of the snowball = 6.0 kg
v1x = initial x-component velocity of the skater (0 m/s because the skater is at rest)
v2x = x-component velocity of the snowball = 53.0 m/s
Substituting the values, we have:
(66 kg) * (0 m/s) + (6.0 kg) * (53.0 m/s) = (66 kg + 6.0 kg) * vf_x
0 + 318 = 72 * vf_x
318 = 72 * vf_x
Now, divide both sides of the equation by 72 to solve for vf_x:
vf_x = 318 / 72
vf_x ≈ 4.42 m/s
So, the x-component of the subsequent velocity of the skater is approximately 4.42 m/s.
To find the y-component (v_y) of the subsequent velocity of the skater, we repeat the same process using the y-component of the snowball's velocity.
Using the equation:
m1 * v1y + m2 * v2y = (m1 + m2) * vf_y
Where:
v1y = initial y-component velocity of the skater (0 m/s because the skater is at rest)
v2y = y-component velocity of the snowball = 14.0 m/s
Following the same calculations as before, we get:
vf_y = (66 kg * 0 m/s + 6.0 kg * 14.0 m/s) / (66 kg + 6.0 kg)
vf_y = 84 / 72
vf_y ≈ 1.17 m/s
So, the y-component of the subsequent velocity of the skater is approximately 1.17 m/s.