Honors Physics I - 3 Momentum and Kinetic Energy

Center of Mass (com)

The motion of the mass point is simple. And all other points rotate around this point.

xcom=m2m1+m2d=m1x1+m2x2m1+m2x_{com}=\frac{m_2}{m_1+m_2}d=\frac{m_1x_1+m_2x_2}{m_1+m_2}

For many particles and in higher dimensions :

r⃗com=1M∑i=1nmiri⃗\vec r_{com}=\frac 1 M \sum_{i=1}^nm_i\vec{r_i}

Usually, the ∑i=1nmiri⃗\sum_{i=1}^nm_i\vec{r_i} is called the moment

For solid bodies:

xcom=1M∫x dmx_{com}=\frac 1 M\int x\, dm

Or choose an origin, and then:

r⃗com=1M∫r⃗i dx\vec r_{com}=\frac 1 M\int \vec r_i\, dx

If we limit ourselves to uniform density.

ρ=MVxcom=1V∫x dV\rho=\frac M V\\ x_{com}=\frac 1 V\int x\, dV

Remember when add or remove things, the total mass will be changed

Newton’s Second Law for a System of Particles

Center of mass motion continues unaffected by forces internal to a system.

Because of the Newton’s third law, the internal force push to each internal object can be eliminated, only the external forces will be remained

F⃗net=Ma⃗net=Mr⃗¨\vec F_{net}=M\vec a_{net}=M\ddot{\vec r}

Kinetic Energy

Energy

  • A scalar quantity
  • Conserve in a closed system

Kinetic Energy

K=12mv2K=\frac 1 2 mv^2

The unit of kinetic energy is a joule (J) 1J=1kg⋅m2/s21J=1kg\cdot m^2/s^2

Work WW is energy transferred. Transfer to the object is positive work, while transferred from the object is negative work.

12mv2−12mv02=FxdW=Fxd\frac 1 2mv^2-\frac 1 2mv_0^2=F_xd\\ W=F_xd

For an angle ϕ\phi between force and displacement :

W=Fdcos⁡ϕW=F⃗⋅d⃗W=Fd\cos \phi\\ W=\vec F\cdot \vec d

Work by Spring Force

  • Relaxed state : no force is applied
  • Stretch and extend the spring it resists : restoring force

Hooke’s law : F⃗s=−kd⃗\vec F_s=-k\vec d, where kk is the spring constant.

Ws=∫xixfFxdx=(−12k)(xf2−xi2)=12kxi2−12kxf2W_s=\int_{x_i}^{x_f}F_xdx=(-\frac 1 2k)(x_f^2-x_i^2)=\frac 1 2k x_i^2-\frac 1 2 kx_f^2

Work by a General Variable Force

dW=F(x)dx∫ifdW=∫xixfF(x)dxdW=F(x)dx\\ \int_i^fdW=\int_{x_i}^{x_f}F(x)dx

Power

Pavg=WΔtP=dWdtP_{avg}=\frac W {\Delta t}\\[10px] P=\frac {dW}{dt}

Or :

P=Fcos⁡ϕ(dxdt)P=F⃗⋅v⃗P=F\cos \phi(\dfrac{dx}{dt})\\ P=\vec F\cdot \vec v

Linear Momentum

p⃗=mv⃗F⃗net=ma⃗=mdv⃗dt=dp⃗dt\vec p =m\vec v\\ \vec F_{net}=m\vec a=m\dfrac{d\vec v}{dt}=\frac {d\vec p}{dt}

Collision and Impulse

J⃗=∫titfF⃗(t)dt\vec J=\int_{t_i}^{t_f}\vec F(t)dt

Don’t confuse with work, the impulse is integrated by time

Δp⃗=J⃗\Delta \vec p=\vec J

If no net external force, the P⃗\vec P is a constant. This is called the law of conservation of linear momentum.

Momentum and Kinetic Energy in Collisions

  • Elastic collisions
    • Total kinetic energy is unchanged
  • Inelastic collisions : some energy is transferred
  • Completely inelastic collisions
    • The objects stick together
    • Greatest loss of kinetic energy

Elastic Collisions in One Dimension

For a stationary target :

v1f=m1−m2m1+m2v1iv2f=2m1m1+m2v1iv_{1f}=\frac{m_1-m_2}{m_1+m_2}v_{1i}\\ v_{2f}=\frac{2m_1}{m_1+m_2}v_{1i}

  • If m2>>m1m_2>>m_1, the first object will bounce back
  • If m1>>m2m_1>>m_2, then v2f≈2v1iv_{2f}\approx 2v_{1i}

Collisions in Two Dimensions

A glancing collision :

image-20260915113357685

  • Along xx
  • Along yy
  • Energy

Sometimes the θ\theta can be determined by R1,R2R_1,R_2 and DD.

Rocket

The ejection of mass from the rocket’s rear increases the rocket’s speed.

Pi=PfMv=−dM U+(M+dM)(v+dv)P_i=P_f\\ Mv=-dM\,U+(M+dM)(v+dv)

while UU is the velocity in the inertial reference frame

The first rocket equation :

−dMdt⋅vrel=Mdvdt→Rvrel=Ma-\frac {dM}{dt}\cdot v_{rel}=\frac{Mdv}{d_t} \to Rv_{rel}=Ma

The second rocket equation :

vf−vi=vrelln⁡MiMfv_f-v_i=v_{rel}\ln\frac{M_i}{M_f}