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One was to write Newton's laws is:

$$F=\frac{dp}{dt}.$$

I don't understand what is the force there. I believe that $F$ is the net external force on the system. So supposedly I have a mass which is moving right and then it collides with another mass which is hanging on a rope from the ceiling.

Supposedly my system is the mass, mass on rope and the Earth. This would make forces of gravity internal. The only external force is the tension on rope(assume massless rope). Now, would the tension of the rope be before the collision, or after the collision? The mass on rope swings up obviously. At that particular instant when it is at max angle, the $T$ is obviously not equal to $T$ when before collision. So does the $T= dp/dt$, is the $T$ before or after collision?

Ok edit. In this system the momentum is not conserved right? Since there is a net external force $T$. So I supposed taking ceiling as part of the system would make $T$ an internal force.

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  • $\begingroup$ Newton's laws are valid at all times. The way you define your system (the mass), the force is the sum of all forces acting on it (transmitted through rope tension, gravity, and, during any collision, contact forces), and the impulse $p$ includes its instantaneous velocity $v$ via $p=mv$. $\endgroup$ – pyramids Feb 27 '15 at 9:22
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$F = \frac{dp}{dt}$ means that force is the rate of momentum transfer per unit time.

Lets say we have mass $m_1$ moving to the right, and mass $m_2$ is on the left side of $m_1$ with zero velocity. If $m_1$ put a force to pull $m_2$, that force will create the acceleration on $m_2$ and increase its velocity, this also means the change in momentum. At the same time the reaction force will also slow down the mass $m_1$ and decrease its momentum. It you think of it that way, you can see that the force between these two masses is just the rate of transfer of momentum from $m_1$ to $m_2$.

$$ F = ma = m\frac{dv}{dt} = \frac{d(mv)}{dt} = \frac{dp}{dt} $$

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The $d$ in front of momentum and in front of time means infinitesimal change of time

$$dt = t_{final} - t_{initial}$$

Therefore the change in momentum over the change in time equals the force. Also momentum is equal to $m\cdot u$ , where $u = \text{velocity}$.

So, the change in momentum is equal to

$$ dp = m\cdot u_{final} - m\cdot u_{initial}$$

We also know from $\sum{F} = m\cdot a$ which is equal to $\sum{F} = m \cdot \dfrac{du}{dt}$

Then you solve!

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  • $\begingroup$ I also think that the tension is not an external force(so the system is isolated) $\endgroup$ – C.A Feb 27 '15 at 8:58
  • $\begingroup$ What is wrong with my answer $\endgroup$ – C.A Feb 27 '15 at 14:45

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