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
Answer:
Energy needed = 1100 kJ
Explanation:
Energy needed = Change in kinetic energy
Initial velocity = 15 m/s
Mass, m = 1600 kg

Final velocity = 40 m/s

Energy needed = Change in kinetic energy = 1280000-180000 = 1100000J
Energy needed = 1100 kJ
Answer:
14.7 m/s
Explanation:
a = acceleration experienced by driver's head = 50 g = 50 x 9.8 m/s² = 490 m/s²
v₀ = initial speed of the driver = 0 m/s
v = final speed of the driver after 30 ms
t = time interval for which the acceleration is experienced = 30 ms = 0.030 s
Using the equation
v = v₀ + a t
Inserting the values
v = 0 + (490) (0.030)
v = 14.7 m/s
Answer
The answer and procedures of the exercise are attached in the following archives.
Explanation
You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.