The height, h to which the package of mass m bounces to depends on its initial velocity, v and the acceleration due to gravity, g and is given below:
<h3>What are perfectly elastic collision?</h3>
Perfectly elastic collisions are collisions in which the momentum as well as the energy of the colliding bodies is conserved.
In perfectly elastic collisions, the sum of momentum before collision is equal to the momentum after collision.
Also, the sum of kinetic energy before collision is equal to the sum of kinetic energy after collision.
Since some of the Kinetic energy is converted to potential energy of the body;
Therefore, the height to which the package m bounces to depends on its initial velocity and the acceleration due to gravity.
Learn more about elastic collisions at: brainly.com/question/7694106
In short, and in general:
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Answer:
The answer to your question is a = 0.25 m/s²
Explanation:
Data
mass = m = 400 kg
Force = F = 100 N
acceleration = a = ? m/s²
Process
To solve this problem use Newton's second law that states that the force applied to an object is directly proportional to the mass of the body times its acceleration.
Formula
F = ma
solve for a
a = 
Substitution
Simplification and result
a = 0.25 m/s²
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Answer:
vf = 3.27[m/s]
Explanation:
In order to solve this problem we must analyze each body individually and find the respective equations. The free body diagram of each body (box and bucket) should be made, in the attached image we can see the free body diagrams and the respective equations.
With the first free body diagram, we determine that the tension T should be equal to the product of the mass of the box by the acceleration of this.
With the second free body diagram we determine another equation that relates the tension to the acceleration of the bucket and the mass of the bucket.
Then we equalize the two stress equations and we can clear the acceleration.
a = 3.58 [m/s^2]
As we know that the bucket descends 1.5 [m], this same distance is traveled by the box, as they are connected by the same rope.
![x = \frac{1}{2} *a*t^{2}\\1.5 = \frac{1}{2}*(3.58) *t^{2} \\t = 0.91 [s]](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%2Aa%2At%5E%7B2%7D%5C%5C1.5%20%3D%20%5Cfrac%7B1%7D%7B2%7D%2A%283.58%29%20%2At%5E%7B2%7D%20%5C%5Ct%20%3D%200.91%20%5Bs%5D)
And the speed can be calculated as follows:
![v_{f}=v_{o}+a*t\\v_{f}=0+(3.58*0.915)\\v_{f}= 3.27[m/s]](https://tex.z-dn.net/?f=v_%7Bf%7D%3Dv_%7Bo%7D%2Ba%2At%5C%5Cv_%7Bf%7D%3D0%2B%283.58%2A0.915%29%5C%5Cv_%7Bf%7D%3D%203.27%5Bm%2Fs%5D)
