Momentum and law of conversation of momentum.

AI Thread Summary
Momentum is defined as mass multiplied by velocity, emphasizing that velocity is a vector quantity. The conservation of momentum principle states that the total momentum before a collision equals the total momentum after the collision. An example illustrates this: a 5 kg ball traveling at 1 m/s has a momentum of 5 kg*m/s, and after colliding with a 10 kg ball, the total momentum remains 5 kg*m/s. To apply the momentum formula, calculate the vector momentum of each object before and after a collision, ensuring the total remains constant. Understanding momentum requires precision in distinguishing it from simply being "mass in motion."
winkle
Messages
1
Reaction score
0
can anybody please explain these two definations?
i know momentum is the mass in motion..but how do we use the formula...explain please

thx
 
Physics news on Phys.org
basic non-relativistic formulas:
momentum = mass * velocity (note: velocity is not speed, velocity is a vector quantity)
and the conservation of momentum means:
momentum (before) = momentum (after)

For example:
a 5 kg ball is traveling at 1m/s : its momentum is 5*1 = 5 kg*m/s
if it hits a different ball (which is 10 kg) directly on and bounces directly back at 0.5 m/s. (directly on so that the other ball will head in the same direction as the inital ball was.
then the total momentum after the collision will still have to be 5 kg*m/s
we know that the first ball has momentum of -0.5 m/s * 5 kg = -2.5kg*m/s. Therefore the momentum of the second ball must be 7.5 kg*m/s so that the total momentum in the system is still 5 kg*m/s. From this the velocity of the second ball can be determined.
 
No, momentum is not "the mass in motion". You have to be a lot more precise than that: momentum is "mass times velocity". It is, as dmoravec said, a vector quantity; scalar mass times vector velocity.
To use that formula, calculate the total (vector) momentum of each particle before a collision, then after the collision and set them equal. The momentum of individual particles might have changed but the total is "conserved"- remains the same.
 
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Hello everyone, Consider the problem in which a car is told to travel at 30 km/h for L kilometers and then at 60 km/h for another L kilometers. Next, you are asked to determine the average speed. My question is: although we know that the average speed in this case is the harmonic mean of the two speeds, is it also possible to state that the average speed over this 2L-kilometer stretch can be obtained as a weighted average of the two speeds? Best regards, DaTario
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
Back
Top