Calculating Mass in an Elastic Collision

In summary, Object 1 with a mass of 2.8 kg and a velocity of 7.7 m/s collides with a stationary object and recoils with a velocity of -2.2 m/s. To find the mass of the second object, two equations are needed: conservation of momentum and energy. However, if it is not an elastic collision, there is no solution.
  • #1
bencdavidson
4
0

Homework Statement



Object 1 has a mass of 2.8 kg and is moving eastward with a velocity of 7.7 m/s. It collides with a stationary object. After the collision object 1 recoils with a velocity of -2.2 m/s What is the mass of object 2?


Homework Equations



As far as I know:

Conservation of Momentum

Po=P

m1v1=m2v2

m=mass


The Attempt at a Solution



I keep trying to solve by using m1v1=m2v2

m2 = m1v1/v2
 
Physics news on Phys.org
  • #2
You have two unknowns: the mass and velocity of object 2 after the collision.
You will need two equations to solve this. If you know it is an elastic collision, you can use conservation of energy as your second equation. If it is not elastic, then you cannot find the solution. The result of the collision depends on how much energy is lost in the collision.
 
  • #3
but don't you need the mass of the second object to calculate using the elastic collision formula?
 
  • #4
No. With two equations, momentum and energy, you can find two unknowns.
It will not be straightforward - you would have to write the two equations, then solve one of them for either m or v, substitute into the other to eliminate one unknown, then solve for the other one.
 
  • #5
Any chance you could show me? I'm at a major loss and I have a huge exam coming up:/
 
  • #6
Glad to help. But you can't just add "elastic collision" to the question.
As it stands, there is no solution.
Take another look at it - if it says "elastic collision" in the question, then write
energy before = energy after
½mv² = ½mv² + ½mv²
and fill in the known quantities. Once you have your two equations with two unknowns, I can offer further advice.
 

Related to Calculating Mass in an Elastic Collision

What is conservation of momentum?

Conservation of momentum is a fundamental principle in physics that states that the total momentum of a closed system remains constant unless acted upon by an external force. This means that the total momentum before a collision or interaction is equal to the total momentum after the collision or interaction.

Why is conservation of momentum important?

Conservation of momentum is important because it helps us understand and predict the motion of objects in a closed system. It is also a fundamental law of nature that applies to all interactions between objects, from collisions to explosions to rocket propulsion.

How is conservation of momentum calculated?

The equation for conservation of momentum is:
m1v1 + m2v2 = m1v1' + m2v2'
where
m1 and m2 are the masses of the objects involved
v1 and v2 are the initial velocities of the objects
v1' and v2' are the final velocities of the objects

What are some real-world examples of conservation of momentum?

Some examples of conservation of momentum in action include billiard balls colliding, a rocket launching into space, and a car crash. In each of these scenarios, the total momentum of the system remains constant before and after the interaction takes place.

Can conservation of momentum be violated?

No, conservation of momentum is a fundamental law of nature and cannot be violated. However, it may appear to be violated in certain situations due to external forces or factors that are not accounted for in the calculations.

Similar threads

  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
2K
  • Introductory Physics Homework Help
Replies
16
Views
733
  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
1K
Back
Top