A ball of mass 0.120 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.600 m. What impulse was given to the ball by the floor?
Impulse is change of momentum, integral of F dt
What you need to know is the speed of the ball down when it hits the floor, call that U (negative down), and the speed of the ball as it rebounds up, call that V (positive up)
then the impulse is m (V - U)
Those two speeds can be gotten from potential and kinetic energy relations.
the potential energy at 1.25 meters (mgh) is the kinetic energy at the floor (.5 m U^2)
so
9.8 * 1.25 = .5 U^2 (the mass on both sides cancels)
U^2 = 24.5
U = -4.95 (remember negative because down)
same deal for V but h in this case is .6 meters
9.8*.6 = .5 V^2
V = +3.43
V - U = 3.43 - (-4.95) = 8.38 m/s
That is the total change in velocity. Multiply by the mass to get the impulse or change in momentum
.120 * 8.38 = 1.01 kg m/s
U =
I thought I'd answered this already, but my answer seems to have disappeared. Please read Damon's response and ignore the nonsense posted by John.
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Well, let's bounce into some calculations here! To determine the impulse given to the ball by the floor, we need to find the change in momentum of the ball.
The change in momentum (Δp) is given by the formula Δp = mΔv, where m is the mass and Δv is the change in velocity.
Initially, the ball was dropped from rest, so the initial velocity (vi) is 0 m/s. As the ball rebounds and reaches a height of 0.600 m, we can use this information to find the final velocity (vf).
Using the equation for potential energy change, ΔPE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height, we can find the change in potential energy. Since the ball rebounds back, the change in potential energy is equal to twice the potential energy at the final height.
ΔPE = 2 * m * g * h
Now, the change in potential energy is equal to the kinetic energy change, which is given by the formula ΔKE = 1/2 * m * (vf^2 - vi^2).
Setting ΔKE equal to ΔPE, we have:
1/2 * m * (vf^2 - vi^2) = 2 * m * g * h
Now, we can solve for vf:
vf^2 - vi^2 = 4 * g * h
vf^2 = 4 * g * h
vf = sqrt(4 * g * h)
Substituting the known values for g (acceleration due to gravity) and h (height), you can calculate the final velocity (vf).
Once you have the final velocity, you can calculate the change in momentum (Δp = m * Δv).
Hope this helps you get to the answer, and remember, don't let gravity bring you down!
To determine the impulse given to the ball by the floor, we need to calculate the change in momentum of the ball during the collision.
Impulse is defined as the change in momentum, which can be calculated using the equation:
Impulse = m * Δv
Where:
- Impulse is the force applied over a specific time interval
- m is the mass of the object
- Δv is the change in velocity
In this scenario, the ball is dropped from rest, so its initial velocity is zero. To calculate the final velocity, we can use the conservation of energy principle.
At the highest point of its rebound, the ball is momentarily at rest before it starts falling again, so the gravitational potential energy is converted into elastic potential energy.
Using the conservation of energy:
m * g * h = 0.5 * k * Δx^2
Where:
- m is the mass of the ball
- g is the acceleration due to gravity (9.8 m/s^2)
- h is the initial height (1.25 m)
- k is the spring constant of the ball (unknown)
- Δx is the change in position (0.600 m)
We could calculate the spring constant k if we had more information about the ball or its rebound characteristics, but since it is not provided, we cannot determine it.
However, we can still calculate the change in velocity (Δv) by using the kinematic equation:
v^2 = u^2 + 2aΔx
At the topmost point, v = 0 (since it is momentarily at rest) and the acceleration (a) is equal to -g due to the opposite direction.
0 = u^2 + 2*(-g)*Δx
Simplifying, we can find u, the final velocity after the rebound:
u = sqrt(2*g*Δx)
Now, we can calculate the impulse:
Impulse = m * Δv = m * (u - 0)
Impulse = m * sqrt(2*g*Δx)
Plugging in the given values:
m = 0.120 kg
Δx = 0.600 m
g = 9.8 m/s^2 (acceleration due to gravity)
Impulse = 0.120 kg * sqrt(2*9.8 m/s^2 * 0.600 m)
Impulse = 0.120 kg * sqrt(11.76 m^2/s^2)
Impulse ≈ 0.120 kg * 3.428 m/s
Impulse ≈ 0.4116 Ns
Therefore, the impulse given to the ball by the floor is approximately 0.4116 Ns.