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# Liner Momentum and Collisions

الكلية كلية التربية للعلوم الصرفة     القسم قسم الفيزياء     المرحلة 1
أستاذ المادة فؤاد عطية مجيد       29/05/2018 09:21:28
Center of Mass
7.1 System of Particles
Up to now we have studied motion of a single particle. We now want to look at a system of particles to see what we can learn about the motion. The momentum for a single particle p ?=mv ?, total momentum for a system of particle is the sum of the individual momentum
p ?=p ?_1+p ?_2+p ?_3+?+p ?_n=?_(i=1)^n?p_i
Where p ?_1=mv ?_1 , p ?_2=mv ?_2, ……..etc.
The simplest many – particle system in 2-particle system in which particles exert forces on each other by newton 3rd law
F ?_1=-F ?_2
Equation of motion
F ?_1=(dp ?_1)/dt , F ?_2=(dp ?_2)/dt
F ?_1+F ?_2=(dp ?_1)/dt+(dp ?_2)/dt=0
?d/dt (p ?_1+p ?_2 )=0
p ?=p ?_1+p ?_1
Particles exchange momentum as they interact
If only internal forces act the total linear momentum is conserved
Consider 2-particle with external forces acting on them. Then
d/dt (p ?_1+p ?_2 )=F ?_(external )
Where F ?_(external ) is total external force on system, if F ?_(external )=0
(dp ?)/dt=0?p ?=[constant]
7.2 Center of Mass
Up to now we have ignored the size of objects, we will now show that for an object of finite size. The position of the center of mass is the average position of the mass of the system.
r ?_cm=(m_1 r ?_1+m_2 r ?_2+?+m_n n)/(m_1+m_2+?+m_n )
r ?_cm=(m_1 r ?_1+m_2 r ?_2+?+m_n r_n)/M=(???m_1 r ?_i ?)/M
In terms of vector components:
x_cm=1/M [m_1 x_1+m_2 x_2+?+m_n x_n ]=1/M ?(m_i x_i )
y_cm=1/M [m_1 y_1+m_2 y_2+?+m_n y_n ]=1/M ?(m_i y_i )
z_cm=1/M [m_1 z_1+m_2 z_2+?+m_n z_n ]=1/M ?(m_i z_i )
r ?_cm=x_cm i ?+y_cm j ?+z_cm k ?
7.3 Motion of the Center of Mass
Suppose we take the time derivative of the position vector of the center of mass. Assuming m is constant (no particles center or leave system). Then we get the velocity of the center of mass.
v ?_cm=(dr ?_cm)/dt=1/M [m_1 (dr ?_1)/dt+m_2 (dr ?_2)/dt+?+m_n (dr ?_n)/dt]
=1/M [m_1 v ?_1+m_2 v ?_2+?+m_n v ?_n ]=p ?/m
p ?=mv ?_cm
(dv ?_cm)/dt=1/M [m_1 (d^2 r ?_1)/?dt?^2 +m_2 (d^2 r ?_2)/?dt?^2 +?+m_n (d^2 r ?_n)/?dt?^2 ]
Total momentum of the system is its total mass multiplied by the velocity of the center of mass. Differentiate again to get the acceleration of the center of mass
a ?_cm=(dv ?_cm)/dt=1/M ??m_i (dv ?_i)/dt=1/M ???m_i a ?_i ?
Ma ?_cm=??F ?_i ? Force of particle
The net force of the system is due only to the external force
??F ?_ext =Ma ?_cm=(dp ?)/dt

المادة المعروضة اعلاه هي مدخل الى المحاضرة المرفوعة بواسطة استاذ(ة) المادة . وقد تبدو لك غير متكاملة . حيث يضع استاذ المادة في بعض الاحيان فقط الجزء الاول من المحاضرة من اجل الاطلاع على ما ستقوم بتحميله لاحقا . في نظام التعليم الالكتروني نوفر هذه الخدمة لكي نبقيك على اطلاع حول محتوى الملف الذي ستقوم بتحميله .