When you hit a 0.27kg volleyball the contact time is 50ms and the average force is 125N.
If you serve the volleyball (from rest) sending it up at 30 degrees above the horizontal, what is the horizontal component of the momentum of the volleyball in kgm/s ?
To solve this problem, we need to use the conservation of momentum principle, which states that the total momentum before the collision is equal to the total momentum after the collision. In this case, we have a single object (the volleyball) that is initially at rest, so its initial momentum is zero.
After the serve, the volleyball will have some momentum in the horizontal direction due to its initial velocity. To find this momentum, we need to use the following formula:
p = m*v*cos(theta)
where p is the momentum in kgm/s, m is the mass of the volleyball in kg, v is the initial velocity in m/s, and theta is the angle above the horizontal in radians.
We can start by finding the initial velocity. We know that the volleyball is served up at 30 degrees above the horizontal, so we can use the following formula to find the vertical component of the velocity:
v_y = v*sin(theta)
where v is the initial velocity and theta is the angle in radians. Since the volleyball starts from rest, its initial velocity is zero, so we can set this equation equal to zero and solve for v:
0 = v*sin(30)
0 = v*(1/2)
v = 0 m/s
Therefore, the initial velocity of the volleyball is zero.
Now we can use the momentum formula to find the momentum in the horizontal direction:
p = m*v*cos(theta)
p = 0.27 kg * 0 m/s * cos(30)
p = 0 kgm/s
So the horizontal component of the momentum of the volleyball is zero.