Momentum and Collision help needed (2 questions)

In summary: In an elastic collision, the momentum of the two objects is conserved. That means the total momentum (mass*velocity) of the objects is the same after the collision.
  • #1
John Comtoi
6
0

Homework Statement


A car braking will decelerate at .8g. If the car's mass is 1200kg, how much horizontal force does each tire exert on the road? What is the coefficient of friction between the tire and the road?

A 2kg steel ball rolling at 4m/s strikes a stationary 5kg steel ball elastically. What is the resultant motion of each ball? How and why would the collision change if it were elastic?

Homework Equations


F=m*a
Ff=μFN

P1+P2=P3+P4 ?
KE=1/2mv^2 ?

The Attempt at a Solution


I am having a very hard time even understanding where to start here. I started off by multiplying (1200kg)(.8g)(-9.8m/s). I am unfamiliar with the measurement being in g's. How would I factor that into the question? I then divided the number by 4 to reference the tires in the work.

The work i have for the second question is basically me attempting to do the momentum equation in the relevant equations. There is two velocities you have to solve for though and I am unsure on how to progress in this problem. I am fairly sure that I started the question wrong and I should be using a different equation. Some guidance would be appreciated!
 
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  • #2
John Comtoi said:
I am having a very hard time even understanding where to start here. I started off by multiplying (1200kg)(.8g)(-9.8m/s). I am unfamiliar with the measurement being in g's. How would I factor that into the question?
You did so already, just with two mistakes. Just replace g by its value (and correct units) and you are done.
John Comtoi said:
The work i have for the second question is basically me attempting to do the momentum equation in the relevant equations. There is two velocities you have to solve for though and I am unsure on how to progress in this problem.
Your textbook probably has a formula for both velocities. Otherwise, think which quantities are conserved, and use those.
 
  • #3
mfb said:
You did so already, just with two mistakes. Just replace g by its value (and correct units) and you are done.Your textbook probably has a formula for both velocities. Otherwise, think which quantities are conserved, and use those.

for question #1, what do u mean by done. Am i done with the problem or am i done with referencing g's in the question? If not, what is the next step?

for question #2, could you elaborate? My knowledge on collisions is iffy at best. What formulas are you talking of for velocity?
 
  • #4
John Comtoi said:
or question #1, what do u mean by done.
You have multiplied the mass by two accelerations, 0.8g and 9.8m/s2. Does that seem right?
John Comtoi said:
for question #2, could you elaborate?
What else is conserved in an elastic collision?
By the way, the question did not mention any friction in this question, so presumably you should ignore it.
 
  • #5
John Comtoi said:
for question #1, what do u mean by done. Am i done with the problem or am i done with referencing g's in the question? If not, what is the next step?
F=m*a. You know m and a, I don't see the problem.
Divide it by 4 and you are done.
 

1. What is momentum?

Momentum is a physical quantity that describes the amount of motion an object has. It is calculated by multiplying an object's mass by its velocity. In simpler terms, momentum is the product of an object's mass and how fast it is moving.

2. How does momentum affect collisions?

Momentum plays a crucial role in collisions because it is always conserved. This means that the total momentum of a system before a collision is equal to the total momentum after the collision. In other words, the sum of the momentums of all objects involved in the collision will remain the same, even if the objects interact and change direction or speed.

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