The magnitude of the change in momentum of the stone is about 18.4 kg.m/s

<h3>Further explanation</h3>
Let's recall Impulse formula as follows:

<em>where:</em>
<em>I = impulse on the object ( kg m/s )</em>
<em>∑F = net force acting on object ( kg m /s² = Newton )</em>
<em>t = elapsed time ( s )</em>
Let us now tackle the problem!

<u>Given:</u>
mass of ball = m = 0.500 kg
initial speed of ball = vo = 20.0 m/s
final kinetic energy = Ek = 70% Eko
<u>Asked:</u>
magnitude of the change of momentum of the stone = Δp = ?
<u>Solution:</u>
<em>Firstly, we will calculate the final speed of the ball as follows:</em>



→ <em>negative sign due to ball rebounds</em>


<em>Next, we could find the magnitude of the change of momentum of the stone as follows:</em>

![\Delta p_{stone} = - [ mv - mv_o ]](https://tex.z-dn.net/?f=%5CDelta%20p_%7Bstone%7D%20%3D%20-%20%5B%20mv%20-%20mv_o%20%5D)
![\Delta p_{stone} = m[ v_o - v ]](https://tex.z-dn.net/?f=%5CDelta%20p_%7Bstone%7D%20%3D%20m%5B%20v_o%20-%20v%20%5D)
![\Delta p_{stone} = m[ v_o + v_o\sqrt{0.7} ]](https://tex.z-dn.net/?f=%5CDelta%20p_%7Bstone%7D%20%3D%20m%5B%20v_o%20%2B%20v_o%5Csqrt%7B0.7%7D%20%5D)
![\Delta p_{stone} = mv_o [ 1 + \sqrt{0.7} ]](https://tex.z-dn.net/?f=%5CDelta%20p_%7Bstone%7D%20%3D%20mv_o%20%5B%201%20%2B%20%5Csqrt%7B0.7%7D%20%5D)
![\Delta p_{stone} = 0.500 ( 20.0 ) [ 1 + \sqrt{0.7} ]](https://tex.z-dn.net/?f=%5CDelta%20p_%7Bstone%7D%20%3D%200.500%20%28%2020.0%20%29%20%5B%201%20%2B%20%5Csqrt%7B0.7%7D%20%5D)


<h3>Learn more</h3>

<h3>Answer details</h3>
Grade: High School
Subject: Physics
Chapter: Dynamics
The quantity that has a magnitude of zero when the ball is at the highest point in its trajectory is
the vertical velocity.
In fact, the motion of the ball consists of two separate motions:
- the horizontal motion, on the x-axis, which is a uniform motion with constant velocity

, where

- the vertical motion, on the y-axis, which is a uniformly accelerated motion with constant acceleration

directed downwards, and with initial velocity

. Due to the presence of the acceleration g on the vertical direction (pointing in the opposite direction of the initial vertical velocity), the vertical velocity of the ball decreases as it goes higher, up to a point where it becomes zero and it reverses its direction: when the vertical velocity becomes zero, the ball has reached its maximum height.
Impulse = Integral of F(t) dt from 0.012s to 0.062 s
Given that you do not know the function F(t) you have to make an approximation.
The integral is the area under the curve.
The problem suggest you to approximate the area to a triangle.
In this triangle the base is the time: 0.062 s - 0.012 s = 0.050 s
The height is the peak force: 35 N.
Then, the area is [1/2] (0.05s) (35N) = 0.875 N*s
Answer> 0.875 N*s
Answer:
43.58 m
Explanation:
If you travel 500 m on a straight road that slopes upward at a constant angle of 5 degrees
Using trigonometry ratio
Sin 5 = opposite/hypothenus
Where the hypothenus = 500m
Opposite = height h
Sin 5 = h/500
Cross multiply
500 × sin 5 = h
h = 500 × 0.08715
h = 43.58m
Therefore, the height above the starting point is equal to 43.58m
Answer:
1. False 2) greater than. 3) less than 4) less than
Explanation:
1)
- As the collision is perfectly elastic, kinetic energy must be conserved.
- The expression for the final velocity of the mass m₁, for a perfectly elastic collision, is as follows:

- As it can be seen, as m₁ ≠ m₂, v₁f ≠ 0.
2)
- As total momentum must be conserved, we can see that as m₂ > m₁, from the equation above the final momentum of m₁ has an opposite sign to the initial one, so the momentum of m₂ must be greater than the initial momentum of m₁, to keep both sides of the equation balanced.
3)
- The maximum energy stored in the in the spring is given by the following expression:

- where A = maximum compression of the spring.
- This energy is always the sum of the elastic potential energy and the kinetic energy of the mass (in absence of friction).
- When the spring is in a relaxed state, the speed of the mass is maximum, so, its kinetic energy is maximum too.
- Just prior to compress the spring, this kinetic energy is the kinetic energy of m₂, immediately after the collision.
- As total kinetic energy must be conserved, the following condition must be met:
- So, it is clear that KE₂f < KE₁₀
- Therefore, the maximum energy stored in the spring is less than the initial energy in m₁.
4)
- As explained above, if total kinetic energy must be conserved:

- So as kinetic energy is always positive, KEf₂ < KE₁₀.