PHYS 218 Chapter 8

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Linear Momentum and Impulse

From Newton's laws, we know that

$\sum _{objB}\mathbf {F} =m\mathbf {a} =m{\frac {d\mathbf {v} }{dt}}={\frac {d(m\mathbf {v} )}{dt}}={\frac {d\mathbf {P} }{dt}}$

Linear Momentum of object

mass × velocity

$\mathbf {P} =m\mathbf {v}$

Impulse of object

change in momentum (final − initial)

$\mathbf {J} =\int _{t_{i}}^{t_{f}}\sum _{objB}\mathbf {F} \ dt=\mathbf {P} _{f}-\mathbf {P} _{i}$

if force is constant: $\mathbf {J} =\mathbf {F} \cdot \Delta t$

Note: $\mathbf {F} _{avg}={\frac {\mathbf {J} }{\Delta {t}}}$.

Example

A "complicated apparatus" takes a 1 kg object moving into it at 20 m/s and spits it out 0.1 s later at an angle $\varphi$ at 30 m/s. What is the impulse?

${\begin{aligned}\mathbf {J} &=\mathbf {P} _{f}-\mathbf {P} _{i}\\&=(30\cos {\varphi }-20){\hat {\mathbf {x} }}+30\sin {\varphi }{\hat {\mathbf {y} }}\end{aligned}}$

Average force = ${\frac {\mathbf {J} }{0.1}}$

Conservation of Momentum

By Newton's 3rd law, we know that every force has an equal and opposite force (e.g. normal). Since momentum is the change of force over time $\left(\textstyle \sum \mathbf {F} ={\tfrac {d\mathbf {P} }{dt}}\right)$, the overall system's momentum is conserved.

Sum of forces over system: $\sum {\mathbf {F} }={\frac {d}{dt}}\left(\sum _{\mathrm {particles} }m_{i}\mathbf {v} _{i}\right)={\frac {d}{dt}}\left(\sum \mathbf {P} _{i}\right)$

If the sum of external forces in a system is 0, then the sum of momentum is 0 and therefore conserved

If the forces in a particular direction are not conserved (i.e. gravity), conservation of momentum may still apply in the other components

Example 1

Before: two equal masses are moving toward each other with equal velocities.

After: the masses collide and stick together

Initial Energy

$2\times {\frac {1}{2}}mv^{2}=mv^{2}$

Final Energy

no movement, so 0

Clearly, there is no conservation of energy (Initial energy is greater than 0, while the final energy is 0)

Work of non-conservative forces is done by internal forces (i.e. deforming the objects as they collide)

However, momentum is conserved since there are no external forces and $P_{i}=P_{f}$.

$P_{i}=mv_{1}+mv_{2}=mv_{1}-mv_{1}=0$

$P_{f}=0$

Example 2: Problem 8.18

When cars are equipped with flexible bumpers, they will bounce off each other during low-speed collisions, thus causing less damage. In one such accident, a 1750-kg car traveling to the right at 1.5 m/s collides with a 1450-kg car going to the left at 1.1 m/s. Measurements show that the heavier car's speed just after collision was 0.25 m/s in its original direction. You can ignore any road friction during the collision.

1. What was the speed of the lighter car just after the collision?
2. Calculate the change in the combined kinetic energy of the two-car system during this collision.

${\begin{aligned}P_{i}&=m_{1}v_{1}+m_{2}v_{2}=1750\cdot 1.5-1450\cdot 1.1\\P_{f}&=m_{1}v_{1}'+m_{2}v_{2}'=1750\cdot 0.25+1450v_{2}'\\v_{2}'&=0.408\ {\mbox{to the right}}\end{aligned}}$

Collisions

"Strong interactions between bodies that occur in a short amount of time"

• elastic: all kinetic energy is conserved
• inelastic: final kinetic energy is less than the initial kinetic energy (energy transformed into internal energy)
• completely inelastic: maximal loss of energy, but still maintains conservation of momentum (usually the case where objects stick together)

Center of Mass

Wednesday, March 23, 2011

position of the center of mass ($\mathbf {r} _{cm}$) is the "mass-weighted average position"

Weighted average: ${\bar {x}}={\frac {\sum {w_{i}x_{i}}}{\sum {w_{i}}}}$

$\mathbf {r} _{cm}={\frac {\sum m_{i}\mathbf {r} _{i}}{\sum m_{i}}}$

For a system of continuum particles (like a solid object):

$\mathbf {r} _{cm}={\frac {\int \rho (\mathbf {r} )\mathbf {r} \ dV}{\int \rho (\mathbf {r} )\ dV}}$
where $\rho (\mathbf {r} )$ is the density at a given point and V is the volume

If the density is homogeneous, then

$\mathbf {r} _{cm}={\frac {1}{V}}\int {\mathbf {r} \ dV}$

Example

Particle of mass 1 kg at (1, 2) and another particle of mass 2 kg at (3, 1).

${\begin{aligned}\mathbf {r} _{cm}&={\frac {1\langle 1,2\rangle +2\langle 3,1\rangle }{3}}\\&=\left\langle {\frac {7}{3}},{\frac {4}{3}}\right\rangle \end{aligned}}$

Motion of the Center of Mass

Monday, March 28, 2011

Velocity is derivative of position; can also be obtained from individual particles. Same with acceleration.

${\begin{aligned}\mathbf {v} _{cm}&={\frac {d\mathbf {r} _{cm}}{dt}}={\frac {\sum m_{i}\mathbf {v} _{i}}{\sum m_{i}}}\\\mathbf {a} _{cm}&={\frac {d\mathbf {v} _{cm}}{dt}}={\frac {\sum m_{i}\mathbf {a} _{i}}{\sum m_{i}}}\end{aligned}}$

Forces on a System

From the acceleration of the center of mass above, we can solve the following given $M=\sum {m_{i}}$ (sum of all masses):

$M\mathbf {a} _{cm}=\sum {F_{i}}$

This equation gives us the value of the external forces, which happens to be related to the change in momentum of the system:

$\sum \mathbf {F} _{ext}=M\mathbf {a} _{cm}=M{\frac {d\mathbf {p} }{dt}}$

Center of mass of a system moves just as though all masses were concentrated at the center of mass and acted upon by a net external force.

Therefore, momentum is conserved iff the sum of all external forces is 0