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Kinematics
Kinematics is the study of how a body moves. This includes linear motion, angular motion, and rotational motion.
3,459 Questions
m=3 kg
g=9.8 m/s^2
h=2.5 m
P.E.=mgh
P.E.=3kg(9.8 m/s^2)(2.5m)
P.E.=73.5 J
When you increase the height of a ramp, the efficiency for kinetic energy decreases because you are doing work against gravity to lift the object higher. This means less of the initial potential energy is converted into kinetic energy compared to when the ramp is lower.
Is velocity a scalar quantity or vector quantity?
Velocity is a vector.
Its magnitude is called 'speed'.
In an elastic collision between two objects do both objects have the same kinetic energy after the collision as before?
Let's do the problem and see
The problem below will prove that in an elastic collision between two objects the objects do not have the same kinetic energy after the collision as before?
But the KE before collision = KE after collision
A 2 kg object with velocity of 8 m/s E hits a 5 kg object with a velocity of 5m/s W. What is the velocity of both objects after colision?
East is positive, West is negative
Draw a picture showing the 2 objects before and after collisions.
…..8 m/s……………....-5 m/s………….V2…………….V5
...2kg object..-->….<--5 kg object…..2kg object…….5kg object W……………………………………………………………….E
Mom = +16………..Mom = -25………..2V2……………5V5
Momentum = mass * velocity
Since the 5 kg object has more momentum, both objects will continue to move West. They should have negative velocities!!
Momentum = mass * velocity
Total momentum before collision = (2 * 8) + (5 * -5) =
Total momentum before collision = 16 + -25 = -9
Total momentum before collision = -9
Total momentum after collision = (2 * V2) + (5 * V5)
Momentum is always conserved.
Total momentum after collision = Total momentum before collision
2V2 + 5V5 = -9 = (add - 2V2 and +9 to both sides)
2V2 = -9 - 5V5 , (divide by -2)
V2 = -4.5 - 2.5V5
Eq. momentum = V2 = -4.5 - 2.5V5
Since the collision is elastic, kinetic energy is conserved
KE before collision = KE after collision
KE ½ mass * velocity ^2
KE before collision = ½ * 2* 8^2 + ½ * 5 * 5^2 =
KE before collision = 64 + 62.5 =
KE before collision = 126.5
KE after collision = ½ * 2 * V2^2 + ½ *5 *V5^2
KE before collision = KE after collision
126.5 = ½ * 2 * V2^2 + ½ *5 *V5^2
126.5 = V2^2 + 2.5 *V5^2
Eq. energy = V2^2 + 2.5 *V5^2 -126.5 =0
Now we have 2 equations in 2 unknowns
Substitute the value of V2 from Eq.m into Eq.e
V2 = -4.5 - 2.5V5
(-4.5 - 2.5V5)^2 + 2.5V5^2 -126.5 = 0
(20.25 + 22.5V5 + 6.25V5^2) + 2.5V5^2 -126.5 =0
Add (6.25V5^2 + 2.5V5^2) = 8.75V5^2
Subtract (20.25 - 126.5) = -106.25
8.75V5^2 + 22.5V5 -106.5 =0
Use quadratic equation to solve for V5
V5 =[ -22.5 ± [22.5^2-(4 * 8.75 * -106.25)]^0.5] ÷ (2 * 8.75)
V5 = [-22.5 ± [506.25 + 3718.75]^0.5] ÷ 17.5
V5 = [-22.5 ± [4225]^0.5] ÷ 17.5
The square root of 4225 = ± 65, I will try +65 first and try add +65 first.
V5 = [-22.5 + 65] ÷ 17.5
V5 = +42.5 ÷ 17.5
V5 = +2.42857
This means the 5 kg object is going East. I made this statement at the beginning. "Since the 5 kg object has more momentum, both objects will continue to move West." V5 = +2.4286 is wrong
I will try subtract +65.
V5 = [-22.5 - 65] ÷ 17.5
V5 = -87.5 ÷ 17.5
V5 = -5
That can not be true, because that was the velocity of the 5 kg in the beginning.
I will try using - 65 for the square root of 4225, and add.
V5 = [-22.5 + -65] ÷ 17.5
V5 = -87.5 ÷ 17.5
V5 = -5
That can not be true, because that was the velocity of the 5 kg in the beginning.
Last but not least, I will using - 65 for the square root of 4225, and subtract.
V5 = [-22.5 - -65] ÷ 17.5
V5 = +42.5 ÷ 17.5
V5 = +2.42857
I know the answer has to be V5 = -2.42857.
If the -22 was +22.5, I would get the correct answer..
If you find the mistake, email me at morrison60957@yahoo.com
V5 = [+22.5 - 65] ÷ 17.5
V5 = -42.5 ÷ 17.5
V5 = -2.42857
I will copy, paste the area where I believe my mistake is at the bottom of this work. If you find my mistake let me know!!
V5 = -2.42857
Eq. momentum = V2 = -4.5 - 2.5V5
V2 = -4.5 - (2.5* -2.42857)
V2 = -10.5714
Let's see if the momentum = -9
Total momentum after collision = (2 * -10.5714) + (5 * -2.42857)
Total momentum after collision = -21.142853 + 12.14285 = -9.000003
Now let's see if the Kinetic energy is conserved
KE before collision = 126.5
KE after collision = ½ * 2 * V2^2 + ½ *5 *V5^2
126.5 = ½ * 2 * (-10.5714)^2 + ½ *5 *(-2.42857)^2
126.5 = 111.754498 + 14.75588061
126.5 ≈ 126.49993786 OK
KE before collision = 64 + 62.5
KE of 2 kg object = 64 J
KE of 5 kg object = 62.5 J
Sum of KE = 126.5
KE after collision = 64 + 62.5
KE of 2 kg object after collision = 111.754498 J
KE of 5 kg object after collision = 14.75588061 J
Sum of KE after collision = 126.49993786
I have proved that in an elastic collision between two objects the objects do not have the same kinetic energy after the collision as before?
But the KE before collision = KE after collision
Below is the work where I suspect I have made a mistake, If you find the mistake, email me at morrison60957@yahoo.com
Momentum = mass * velocity
Total momentum before collision = (2 * 8) + (5 * -5) =
Total momentum before collision = 16 + -25 = -9
Total momentum before collision = -9
Total momentum after collision = (2 * V2) + (5 * V5)
Momentum is always conserved.
Total momentum after collision = Total momentum before collision
2V2 + 5V5 = -9 = (add - 2V2 and +9 to both sides)
2V2 = -9 - 5V5 , (divide by -2)
V2 = -4.5 - 2.5V5
Eq. momentum = V2 = -4.5 - 2.5V5
V2 = -4.5 - 2.5V5
(-4.5 - 2.5V5)^2 + 2.5V5^2 -126.5 = 0
(20.25 + 22.5V5 + 6.25V5^2) + 2.5V5^2 -126.5 =0
Add (6.25V5^2 + 2.5V5^2) = 8.75V5^2
Subtract (20.25 - 126.5) = -106.25
8.75V5^2 + 22.5V5 -106.5 =0
What is the relationship between momentum and mass?
The more the mass, the more momentum you will need for an object to speed up more, or accelerate.
Who the Chinese acrobats in incredible balancing act are in equilibrium?
The Chinese acrobats in the incredible balancing act are skilled performers who have undergone rigorous training to develop their strength, balance, and flexibility. Their precise movements and control allow them to achieve incredible feats of balance and equilibrium that captivate audiences around the world.
What has more lift - the wing of a slow-flying bird or the wing of a fast-flying bird?
Slow flying birds
Why do you need momentum with kinetic energy?
Momentum is the product of mass and velocity.
Kinetic Energy is the product of mass and velocity squared.
As you can see, since Kinetic Energy is derived from mass and velocity, and Momentum is derived from mass and velocity, you cannot have one without the other.
The condition you are referring to is called temperature. Temperature is a measure of the average kinetic energy of the particles in a substance, such as air. When the average velocity of atmospheric molecules is not zero, it indicates that the substance has a non-zero temperature.
Energy can be looked at at the apex of the ball's bounce and is proportional to h. Therefore, the energy retained between the first two bounces = 1.3/1.6 = 81%
Therefore, it loses about 19% of its energy per bounce.
To lose 90% means to retain 10% of the energy
0.8125^x=.1
x=log(.1)/log(.8125)
round up
12 rebounds
The molecules of all samples of ideal gases have the same average kinetic energies at the same?
temperature. This is known as the kinetic theory of gases, which states that the average kinetic energy of gas molecules is directly proportional to the temperature of the gas, regardless of the type of gas.
Kinetic energy = 1/2Mass x Velocity^2
KE = 1/2(1990) x (28^2)
KE = 995 x 784
KE = 780080 Joules
Is kinetic energy at work when something is thrown?
Kinetic energy is the energy of motion; a thrown object has kinetic energy.
Kinematics are mostly concerned with the geometrically possible motion of a body or system of bodies, without consideration of the forces involved.
It is used in describing the spatial position of bodies or systems, their velocities, and their acceleration.
Mach 10 is equal to 7,612.71 miles per hour (mph) at standard atmospheric conditions.
When a sample of gas is heated at constant pressure the average kinetic energy of its molecules?
KEavg = 3/2RT
Just need to know the temperature, T. ( in Kelvin )
R is a constant.
What does v equals d divided by t mean?
Velocity = Distance / Time
Velocity is defined as the change in Distance travelled over the Time taken to travel across it at this average rate of velocity.
Therefore, average velocity and time are inversly proportional to one another, while distance is directly proportional to both time and velocity, and vice versa. At a fixed velocity, the travel time increases as the distance becomes longer; if the distance is fixed, then the velocity must become greater to make the time shorter.
How do you calculate the gain of Kinetic Energy?
The gain in kinetic energy can be calculated using the equation: ΔKE = KE_final - KE_initial, where KE is the kinetic energy. Simply subtract the initial kinetic energy from the final kinetic energy to determine the gain.
Kyle is travelling at a constant rate of 15 MILES PER HOUR! In 30 minutes , which is half of an hour, kyle would have travelled half of the 15 miles. So kyle's done 7.5 miles. Kim is travelling at a constant rate of 18 MILES PER HOUR! Now she travels 18 miles in 60 mins ....but in fact we want the time taken for her to travel 7.5 miles... I think you know what to do now :P ?
How long does it take to drive A miles at B mph?
At 'B' mph, it takes ( A/B ) hours to cover 'A' miles.
At 'B' mph, it takes ( A/B ) hours to cover 'A' miles.
What is a equation for the speed limit 75 kph in mph?
To convert 75 kilometers per hour (kph) to miles per hour (mph), you can use the conversion factor 1 kph = 0.621371 mph. Therefore, the equation would be 75 kph * 0.621371 = 46.6 mph.
How many mph is 500 miles in 2 hours 40 minutes and 3.4181 seconds?
To calculate the speed in mph, first convert the time to hours (2 hours 40 minutes and 3.4181 seconds is equivalent to 2.673394 hours). Then, divide the distance by the time: 500 miles / 2.673394 hours ≈ 186.94 mph.
