<span>Mass of the ball is m = 0.10kg
Initial speed of the Ball v = 15m/s
a. When the ball is at maximum height the velocity is 0
Momentum of ball = mass x velocity
Momentum = 0.10kg x 0 = 0
b. Getting the maximum height,
Using the conservation of energy equation KEinitial = mgh
1/2mVin^2 = mgh => h = v^2/2g
h = 15^2/2x9.8 = 11.48m => Half Height h = 5.96m
Applying the conservation of energy equation at halfway V^2 = 2gh
V = square root of (2x9.8x5.96) => V = square root of (116.816)
So the velocity at the half way V = 10.81 m/s
Momentum M = m x V => M = 0.10 x 10.81 => M = 1.081kg-m/s</span>
Answer:
Point A
Explanation:
The work done by stretching or compressing a spring is given by E=1/2kx²
The potential energy is numerically equal to the work done.
This means that the higher the bigger the value of the extension, x, the higher the energy contained.
In this scenario the modulus of x is considered.
Among the given values of x the modulus of -5 is the largest.
thus it gives the highest value of energy.
The elastic potential energy stored in the stretched spring is 1 J.
<h3>What is Hooke's law?</h3>
Hooke's law states that; provided the elastic limit is not exceeded, the extension of the spring is directly proportional to the force on the spring.
Given that;
Force on the spring = 350 Newton
Distance stretched = 7 centimeters or 0.07 m
Hence;
F = ke
k = F/e = 350 Newton/0.07 m = 5000 N/m
Work done in stretching a spring = 1/2ke^2
= 0.5 × 5000 × (2 × 10^-2)^2 =1 J
Learn more about elastic potential energy: brainly.com/question/156316
