Answer:
Initial velocity, U = 28.73m/s
Explanation:
Given the following data;
Final velocity, V = 35m/s
Acceleration, a = 5m/s²
Distance, S = 40m
To find the initial velocity (U), we would use the third equation of motion.
V² = U² + 2aS
Where;
V represents the final velocity measured in meter per seconds.
U represents the initial velocity measured in meter per seconds.
a represents acceleration measured in meters per seconds square.
S represents the displacement measured in meters.
Substituting into the equation, we have;
35² = U + 2*5*40
1225 = U² + 400
U² = 1225 - 400
U² = 825
Taking the square root of both sides, we have;
Initial velocity, U = 28.73m/s
Answer:
280 N
Explanation:
Applying Newton's third second law of motion,
F = m(v-u)/t................... Equation 1
Where F = Magnitude of the average force on the ball during contact, v = final velocity of the ball, u = initial velocity of the ball, t = time of contact of the ball and the wall.
Note: Let the direction of the initial velocity of the ball be positive
Given: m = 4 kg, u = 3.0 m/s, v = -4.0 m/s (bounce off), t = 0.1 s
Substitute into equation 1
F = 4(-4-3)/0.1
F = 4(-7)/0.1
F = -28/0.1
F = -280 N.
Note: The negative sign tells that the force on the ball act in opposite direction to the initial motion of the ball
(a) The spring stiffness constant of the spring is 18,392 N/m.
(b) The time the car was in contact with the spring before it bounces off in the opposite direction is 0.23 s.
<h3>Kinetic energy of the car</h3>
The kinetic energy of the car is calculated as follows;
K.E = ¹/₂mv²
K.E = ¹/₂ x 950 x 22²
K.E = 229,900 J
<h3>Stiffness constant of the spring</h3>
The stiffness constant of the spring is calculated as follows;
K.E = U = ¹/₂kx²
k = 2U/x²
k = (2 x 229,900)/(5)²
k = 18,392 N/m
<h3>Force exerted on the spring</h3>
F = kx
F = 18,392 x 5
F = 91,960 N
<h3>Time of impact</h3>
F = mv/t
t = mv/F
t = (950 x 22)/(91960)
t = 0.23 s
Learn more about spring constant here: brainly.com/question/1968517
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Explanation:
