A 14.6-g marble is dropped from rest onto the floor 1.23 m below. If the marble bounces straight upward to a height of 0.665 m, what is the magnitude of the impulse delivered to the marble by the floor.
If the marble had bounced to a height of 0.884 m, what would the magnitude of the impulse delivered to the marble by the floor be?
Well, when a marble bounces, it's like the floor saying, "Hey, I don't want you here anymore, go up! Oh, and take some momentum with you!" So, the impulse delivered to the marble by the floor can be calculated using the principle of conservation of momentum.
Let's start with the first scenario. The marble is dropped from rest, so its initial velocity is zero. When it bounces back, its final velocity is also zero because it momentarily comes to a stop at the highest point. We can use the equation: Impulse = Change in Momentum.
Now, the momentum of an object is the product of its mass and velocity. The mass of the marble is 14.6 g or 0.0146 kg. So, the momentum before the bounce is 0 kg∙m/s since the initial velocity is zero. After the bounce, the momentum is still 0 kg∙m/s because the final velocity is also zero.
Therefore, the change in momentum, and hence the magnitude of the impulse, is 0 kg∙m/s.
For the second scenario, where the marble bounces to a height of 0.884 m, we can follow the same logic. The initial momentum is still 0 kg∙m/s, and the final velocity is zero at the highest point. However, this time, since the marble reaches a greater height, it has a greater final velocity on the bounce back.
So, the change in momentum and the magnitude of the impulse would not be zero. Unfortunately, I don't have enough information to calculate the exact values. You'll need to provide the velocity of the marble when it reaches a height of 0.884 m for a more precise answer.
Remember, when it comes to physics, bouncing marbles are no jokes!
To find the magnitude of the impulse delivered to the marble by the floor, we can use the principle of conservation of momentum.
1) When the marble bounces to a height of 0.665 m:
- The initial velocity of the marble just before hitting the floor can be found using the equation v^2 = u^2 + 2as, where v is the final velocity (0 m/s), u is the initial velocity, a is the acceleration (9.8 m/s^2), and s is the distance (1.23 m).
- Solving for u, we have u^2 = v^2 - 2as -> u^2 = 0 - 2 * 9.8 * 1.23 -> u^2 = -24.0628 -> u ≈ -4.906 m/s.
- The change in velocity, Δv, is given by Δv = final velocity - initial velocity = 0 - (-4.906) = 4.906 m/s.
- The impulse, J, is given by J = m * Δv, where m is the mass of the marble (14.6 g = 0.0146 kg).
- Plugging in the values, we have J = 0.0146 kg * 4.906 m/s = 0.0715 N s (to 4 significant figures).
Therefore, the magnitude of the impulse delivered to the marble by the floor when it bounces to a height of 0.665 m is 0.0715 N s.
2) When the marble bounces to a height of 0.884 m:
- Similarly, we can find the initial velocity u using v^2 = u^2 + 2as, where v is the final velocity (0 m/s), a is the acceleration (9.8 m/s^2), and s is the distance (1.23 m).
- Solving for u, we have u^2 = v^2 - 2as -> u^2 = 0 - 2 * 9.8 * 1.23 -> u^2 = -24.0628 -> u ≈ -4.906 m/s.
- The change in velocity Δv is the same as before, Δv = 4.906 m/s.
- The impulse J can be calculated as J = m * Δv, where m is still 0.0146 kg.
- Plugging in the values, we have J = 0.0146 kg * 4.906 m/s = 0.0715 N s (to 4 significant figures).
Therefore, the magnitude of the impulse delivered to the marble by the floor when it bounces to a height of 0.884 m is also 0.0715 N s.
To find the magnitude of the impulse delivered to the marble by the floor, we can use the principle of conservation of momentum.
The impulse experienced by an object is equal to the change in momentum it undergoes. In this case, since the marble is dropped from rest, its initial momentum is zero. The final momentum of the marble after bouncing can be calculated using the formula:
p = m * v
where p is the momentum, m is the mass of the marble, and v is the velocity of the marble.
Since the marble bounces straight upward, its final velocity (v) can be found using the equation:
v = sqrt(2 * g * h)
where g is the acceleration due to gravity (approximately 9.8 m/s^2) and h is the height to which the marble bounces.
Now, let's calculate the impulse for the given height of 0.665 m:
1. Calculating the velocity:
v = sqrt(2 * 9.8 * 0.665)
v ≈ 3.13 m/s
2. Calculating the final momentum:
p = 0.0146 kg * 3.13 m/s
p ≈ 0.0455 kg·m/s
Therefore, the magnitude of the impulse delivered to the marble by the floor for a bounce height of 0.665 m is approximately 0.0455 kg·m/s.
To find the magnitude of the impulse for a bounce height of 0.884 m, follow the same steps as above:
1. Calculating the velocity:
v = sqrt(2 * 9.8 * 0.884)
v ≈ 3.71 m/s
2. Calculating the final momentum:
p = 0.0146 kg * 3.71 m/s
p ≈ 0.054 kg·m/s
Therefore, the magnitude of the impulse delivered to the marble by the floor for a bounce height of 0.884 m is approximately 0.054 kg·m/s.
m1v1 = (0.0146)(v1) (vertically downward is neg direction)
v1 = √2gh (where h = 1.23)
v1 = 4.91 m/s
m1v1 = (0.0146)(4.91) = - 0.0717 kg-m/s (Momentum Before Floor)
m2v2 = (0.0146)v2 (vertically upward)
v2 = √2gh (where h=.665)
v2 = 3.61 m/s
m2v2 = (0.0146)(3.61) = 0.0527 kg-m/s (Momentum After Floor)
Change in Momentum is:
m2v2 - m1v1 = 0.0527 - (- 0.0717) = 0.124 kg-m/s (upward direction)
Impulse = 0.124 N s (upward direction)
Do the same thing with .884 instead of .665 for the second question.