A 0.350-kg ball is dropped from rest at a point 1.10 m above the floor. The ball rebounds straight upward to a height of 0.420 m. Taking the negative direction to be downward, what is the impulse of the net force applied to the ball during the collision with the floor?
To find the impulse of the net force applied to the ball during the collision with the floor, we need to calculate the change in momentum.
The impulse is given by the formula:
Impulse = change in momentum
Momentum = mass × velocity
Since the ball is dropped from rest, the initial velocity is 0 m/s. When it rebounds, it will have an upward velocity, which we need to find.
The final velocity of the ball can be found using the principle of conservation of energy. The potential energy lost when the ball falls is equal to the kinetic energy gained when it rebounds.
Potential energy = mass × gravity × height
Initial potential energy = 0.350 kg × 9.8 m/s^2 × 1.10 m
Final kinetic energy = 0.350 kg × velocity^2 / 2
Since the ball reaches a height of 0.420 m after rebounding, we can equate the initial potential energy to the final kinetic energy:
0.350 kg × 9.8 m/s^2 × 1.10 m = 0.350 kg × velocity^2 / 2
Now, we can solve for the final velocity:
(9.8 m/s^2 × 1.10 m) = velocity^2 / 2
velocity^2 = (9.8 m/s^2 × 1.10 m) × 2
velocity^2 = 21.56 m^2/s^2
velocity ≈ √21.56 m^2/s^2
velocity ≈ 4.64 m/s
Now, we can calculate the change in momentum:
Change in momentum = final momentum - initial momentum
Since the initial velocity is 0 m/s, the initial momentum is 0:
Initial momentum = 0 kg × 0 m/s
Final momentum = mass × velocity
Final momentum = 0.350 kg × 4.64 m/s
Change in momentum = (0.350 kg × 4.64 m/s) - (0 kg × 0 m/s)
Therefore, the impulse of the net force applied to the ball during the collision with the floor is the same as the change in momentum:
Impulse = (0.350 kg × 4.64 m/s) - (0 kg × 0 m/s)
Impulse ≈ 1.624 kg·m/s
To find the impulse of the net force applied to the ball during the collision with the floor, we need to use the principle of conservation of momentum.
The impulse of a force is defined as the change in momentum it produces. Mathematically, impulse can be calculated using the formula:
Impulse (J) = change in momentum (Δp)
Since the ball rebounds straight upward, its final momentum is equal in magnitude but opposite in direction to its initial momentum before the collision.
Given:
Mass of the ball (m) = 0.350 kg
Initial height (h1) = 1.10 m
Final height (h2) = 0.420 m
Acceleration due to gravity (g) = 9.8 m/s^2 (negative)
First, let's calculate the initial and final velocities of the ball using the kinematic equation:
For the initial velocity (u) of the ball, we can use the equation:
v^2 = u^2 + 2gh
where v is the final velocity, and h is the initial height.
Rearranging the equation to solve for u:
u^2 = v^2 - 2gh
Substituting the given values:
u^2 = 0 - 2(9.8)(1.10)
u^2 = -21.56
Taking the square root of both sides (since velocity cannot be negative):
u = 4.64 m/s (upward)
Now, let's calculate the final velocity (v) after the collision using the equation:
v^2 = u^2 - 2gh
where v is the final velocity, u is the initial velocity, and h is the final height.
Substituting the given values:
v^2 = (4.64)^2 - 2(9.8)(0.420)
v^2 = 46.416 - 8.2156
v^2 = 38.2004
Taking the square root of both sides (since velocity cannot be negative):
v = 6.18 m/s (upward)
Now, we can calculate the initial momentum (before the collision) and final momentum (after the collision) using the formula:
Momentum (p) = mass (m) × velocity (v)
Initial momentum (p1) = m × u
p1 = 0.350 kg × 4.64 m/s = 1.624 kg·m/s (upward)
Final momentum (p2) = m × v
p2 = 0.350 kg × 6.18 m/s = 2.163 kg·m/s (upward)
The change in momentum (Δp) is given by the difference between the final and initial momentum:
Δp = p2 - p1
Δp = 2.163 kg·m/s - 1.624 kg·m/s
Δp = 0.539 kg·m/s (upward)
Therefore, the impulse of the net force applied to the ball during the collision with the floor is 0.539 kg·m/s (upward).