A 0.400-kg ball is dropped from rest at a point 1.80 m above the floor. The ball rebounds straight upward to a height of 0.730 m. What are the magnitude and direction of the impulse of the net force applied to the ball during the collision with the floor?
Get the velocity before floor impact from the height H from which it is dropped. Call it V1 (positive down)
V1 = sqrt(2H/g)
Get the velocity just after floor impact from the height H' to which it rises. Call that V2 (positive up)
V2 = sqrt(2H'/g)
The momentum change, which is the impulse, is
M*(V1 + V2)
The impulse on the ball is up, since the momentum increases in the upward direction.
To find the magnitude and direction of the impulse of the net force applied to the ball during the collision with the floor, we need to apply the principle of conservation of momentum. According to this principle, the total momentum of an isolated system remains constant before and after a collision.
First, let's calculate the initial and final velocities of the ball. We can use the kinematic equation:
v^2 = u^2 + 2as
Where:
v = final velocity
u = initial velocity (which is 0 in this case, since the ball is dropped from rest)
a = acceleration due to gravity (-9.8 m/s^2, as the ball is moving vertically)
s = displacement (1.80 m, the height from which the ball is dropped)
Substituting the given values into the equation:
v^2 = 0^2 + 2*(-9.8)*1.80
v^2 = -35.28
v ≈ -5.94 m/s
The negative sign indicates that the final velocity is in the opposite direction to the initial direction (downward).
Next, let's calculate the change in velocity (Δv), which is the difference between the final and initial velocities:
Δv = v - u
Δv = -5.94 - 0
Δv = -5.94 m/s
Now, we can calculate the magnitude of the impulse (J) using the formula:
J = m * Δv
Where:
m = mass of the ball (0.400 kg)
J = 0.400 * (-5.94)
J ≈ -2.38 kg·m/s
The magnitude of the impulse is approximately 2.38 kg·m/s.
Finally, the direction of the impulse can be determined by the direction of the change in velocity. Since the change in velocity is negative, the impulse is directed opposite to the motion of the ball. In this case, it is directed upward since the ball rebounds in that direction.
Therefore, the magnitude of the impulse is 2.38 kg·m/s, and its direction is upward.