Conservation of Momentum problem

In summary, two blocks with masses of 1.17 kg and 2.73 kg, respectively, are moving on a frictionless surface with initial velocities of 5.50 m/s and 2.15 m/s. After a collision, the second block has a velocity of 5.26 m/s. Using the equation Vfinal = (m1 x v1initial) + (m2 x v2initial) / (m1+m2), the final velocity of the 1.17 kg block is approximately 3.16 m/s. However, this approach does not account for the fact that the blocks do not stick together after the collision. To solve this, the general law of conservation of momentum should be
  • #1
Lannie
11
0
I'm not sure what I've done wrong on this problem, so anyone who sees my mistake, please let me know!

Two blocks are moving on a frictionless surface. The first block's mass is 1.17 kg and its initial velocity is 5.50 m/s. The second block's mass is 2.73 kg and its initial velocity is 2.15 m/s.

What is the speed of the 1.17 kg block after the collision if the speed of the second block is 5.26 m/s after the collision?


(For clarification, before the collision the blocks are moving to the right, which I stated as the positive direction, and after the collision they are both still moving to the right.)

To solve, I used the equation Vfinal= (m1 x v1initial) + (m2 x v2initial) / (m1+m2)

so my work shows Vfinal = (1.17x5.50) + (2.73x2.15) / (1.17+2.73)
which worked out to approx. 3.16 m/s

But this approach didn't get me the right answer, which I don't understand since it seemed really straightforward.

The second part of the question asks,

If the initial velocity of the 2.73 kg block is the same magnitude but in the opposite direction, what is the velocity of the 1.17 kg block after the collision?

I used the exact same approach for this, but made the initial velocity of the second block negative. So,

Vfinal= (1.17x5.50) + (2.73x - 2.15) / (1.17+2.73)

which worked out to 0.145 m/s

this wasn't the right answer either.

Any help would be greatly appreciated!
 
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  • #2
you must remember...that after the collision they do not stick together so their final velocity is not the same...remember how you have the velocity of the second afterwards...

it should be more like...
vf = [ (m1 x v1 initial) + (m2 v v2 initial) - (m2 x v2 final) ] / m1
 
  • #3
You are confusing the general conservation law of momentum with the special case of INELASTIC COLLISION.
Remember: At the end of an inelastic collision, the two objects stick together with the same velocity.
That is what you've used, rather than the general law of conservation of momentum:
[tex]m_{1}v_{1,init.}+m_{2}v_{2,init.}=m_{1}v_{1,final}+m_{2}v_{2,final}[/tex]
(The notation ought to be self-explanatory)
Use this approach and see if you get better results..
 
  • #4
actually yeah, WOW, I just looked at my work and was hoping to scramble here and edit this... totally missed a step there. I'm sorry! I'm a bonehead.
thanks though :)
 

Related to Conservation of Momentum problem

1. What is conservation of momentum?

Conservation of momentum is a fundamental principle in physics that states that the total momentum of a system of objects remains constant if there are no external forces acting on the system.

2. How is conservation of momentum related to Newton's laws of motion?

Conservation of momentum is a consequence of Newton's third law of motion, which states that for every action, there is an equal and opposite reaction. This means that in a closed system, the total momentum of all objects before and after a collision or interaction remains the same.

3. Can you provide an example of a conservation of momentum problem?

One example of a conservation of momentum problem is a collision between two billiard balls on a pool table. Before the collision, the two balls have a certain momentum, and after the collision, the momentum is transferred between the two balls, but the total momentum of the system remains the same.

4. How is the conservation of momentum used in real-world applications?

The conservation of momentum is used in various real-world applications, such as rocket propulsion, car crashes, and sports like ice skating and hockey. It is also used in the design and engineering of machines and structures, such as roller coasters and bridges.

5. What are the limitations of conservation of momentum?

Conservation of momentum only applies to closed systems where there are no external forces acting on the system. In real-world situations, there are often external forces at play, such as air resistance or friction, which can affect the momentum of objects and make it difficult to apply the principle accurately.

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