Conservation of (Energy vs Momentum)

In summary: The second approach, using the conservation of momentum, is correct as it takes into account the change in mass of the system after the collision. In summary, the two approaches result in different initial speeds for the bullet, but the second approach using the conservation of momentum is correct due to the change in mass of the system after the collision.
  • #1
Ronnie1303
2
0
1. A bullet (m1 = 0,01 kg) hits a ball hanging on a thread (m2 = 1kg) and stays in he ball (therefore new system = m1 + m2) and pushes the whole system into the height of 0,2m (max potential energy, Kinetic energy = 0) Calculate the speed initial speed of bullet (v) and the initial speed of whole system (v').



2. Law of Conservation of Energy: Ek + Ep = const.
Law of Conservation of momentum: p1 + p2 = const.




3. So I've been doing the calculations with 2 different approaches and each gives a different result and I don't know why, so here they are:

Approach #1: Let the whole system be an isolated system. Therefore the kinetic energy of bullet should equal the potential energy of the whole bullet+ball system at the max Ep state. Therefore:


1/2 m1 v(bullet)^2 = (m1 + m2)gh

(Note that for symplification we consider g = 10m/s^2). So I get

v = sqrt{[2(m1+m2)gh]/m1}

which gives the speed of bullet approximately 20,1 m/s. Now for calculation the speed of the system (at Ep of system = 0 and Ek is max) I used similar approach (Ek1 = Ek2) and got v' = 2m/s . Energy is equal at all times.

However here comes the second approach:

Conservation of momentum says that: m1v1 = (m1+m2)v' so let's use the law of conservation of energy once more, however let's start with the initial kinetic energy of system (ball + bullet) should equal the potential energy at it's max state. Therefore:

1/2(m1+m2)v'^2 = (m1+m2)gh

which gives the same result of speed of the system as v' = 2m/s.
However now I use this speed in the law of conservation of momentum and get:

v(bullet) = ([m1+m2]v')/m1 which gives the speed of bullet as 202 m/s.

So here's the conflict: Working with the conservation of energy purely the speed of bullet is aproximately 20,1 m/s. Working with the conservation of momentum the speed is 202 m/s + when I start calculating the energy values, it differs! So I don't know which solution is correct and why. Thanks for help in advance!
 
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  • #2
The first method doesn't make sense because mechanical energy isn't conserved during the inelastic collision.
 

FAQ: Conservation of (Energy vs Momentum)

What is conservation of energy?

Conservation of energy is a fundamental principle in physics that states that energy cannot be created or destroyed, but can only be transformed from one form to another.

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.

What is the difference between conservation of energy and conservation of momentum?

Conservation of energy deals with the total energy of a system, while conservation of momentum deals with the total momentum of a system. While energy can be transformed from one form to another, momentum can only be transferred between objects.

Why is conservation of energy important?

Conservation of energy is important because it allows us to predict and understand the behavior of physical systems and processes. It also helps us to develop efficient and sustainable energy sources.

How is conservation of energy and momentum applied in real-life situations?

Conservation of energy and momentum are applied in various real-life situations, such as in the design and functioning of machines, understanding the motion of objects in collisions, and in the study of celestial bodies and their interactions.

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