Answer:
106.7 N
Explanation:
We can solve the problem by using the impulse theorem, which states that the product between the average force applied and the duration of the collision is equal to the change in momentum of the object:
where
F is the average force
is the duration of the collision
m is the mass of the ball
v is the final velocity
u is the initial velocity
In this problem:
m = 0.200 kg
u = 20.0 m/s
v = -12.0 m/s
Solving for F,
And since we are interested in the magnitude only,
F = 106.7 N
Answer:
ELASTIC collision
kinetic energy is conservate
Explanation:
As the ball bounces to the same height, it can be stated that the impact with the floor is ELASTIC.
As the floor does not move the conservation of the moment
po = pf
-mv1 = m v2
- v1 = v2
So the speed with which it descends is equal to the speed with which it rises
Therefore the kinetic energy of the ball before and after the collision is the same
Distance, since distance represents how far something has travelled, which would be in our case 2.5m.
Answer:
The shortest distance in which you can stop the automobile by locking the brakes is 53.64 m
Explanation:
Given;
coefficient of kinetic friction, μ = 0.84
speed of the automobile, u = 29.0 m/s
To determine the the shortest distance in which you can stop an automobile by locking the brakes, we apply the following equation;
v² = u² + 2ax
where;
v is the final velocity
u is the initial velocity
a is the acceleration
x is the shortest distance
First we determine a;
From Newton's second law of motion
∑F = ma
F is the kinetic friction that opposes the motion of the car
-Fk = ma
but, -Fk = -μN
-μN = ma
-μmg = ma
-μg = a
- 0.8 x 9.8 = a
-7.84 m/s² = a
Now, substitute in the value of a in the equation above
v² = u² + 2ax
when the automobile stops, the final velocity, v = 0
0 = 29² + 2(-7.84)x
0 = 841 - 15.68x
15.68x = 841
x = 841 / 15.68
x = 53.64 m
Thus, the shortest distance in which you can stop the automobile by locking the brakes is 53.64 m