Rolling Motion and Angular Momentum

Elastic Collision – One Dimension
Assume a particles moving with velocities v ?_1 and v ?_2 before the collision
Particles move with velocities v ?_1 and v ?_2 after the collision
Conservation of momentum p ?_i=p ?_f
m_1 v ?_1+m_2 v ?_2=m_1 v ? ?_1+m_2 v ? ?_2 (1)
Conservation of energy K_i=K_f
1/2 m_1 v_1^2+1/2 m_2 v_2^2=1/2 m_1 v ?_1^2+1/2 m_2 v ?_2^2 (2)
Rewrite Eq. (1) m_1 (v ?_1-v ? ?_1 )=m_2 (v ?_2-v ? ?_2 ) (3)
Rewrite Eq. (2) m_1 (v_1^2-v ?_1^2 )=m_2 (v_2^2-v ?_2^2 ) (4)
((4))/((3))=v ?_1+v ? ?_1=v ? ?_2+v ?_2 (5)
Or v ?_1-v ?_2=v ? ?_2-v ? ?_1
Now, if we substitute Eq. (1) in Eq. (5), we obtain
v ?_2=v_1 ((2m_1)/(m_1+m_2 ))+v_2 ((m_2-m_1)/(m_1+m_2 ))
v ?_1=v_1 ((m_1-m_2)/(m_1+m_2 ))+v_2 ((2m_1)/(m_1+m_2 ))
Elastic Collision – Two Dimension
When two bodies collide and their motion is not along single axis (the collision is not head on), the collision is two dimensional
Conservational of momentum
m_1 v ?_1=m_1 v ? ?_1+m_2 v ? ?_2
m_1 v ?_1=m_1 v ? ?_1 cos?+m_2 v ? ?_2 cos?
0=m_1 v ? ?_1 sin?-m_2 v ? ?_2 sin?
Conversation of energy
1/2 m_1 v_1^2=1/2 m_1 v ?_1^2+1/2 m_2 v ?_2^2
Example (1): A 2 Kg body moving under the effect of the force, if the body at rest at t=0 and F ?=2ti ?+(3t^2-1)j ?N,
What is the momentum and the kinetic energy of the body at 2s.
What is the work done by this force from t=0 to t=2s.
Solution:
F ?=dp/dt??_0^p?dp=?_0^2?Fdt
p ?=?_0^2?[(2t)i ?+(3t^2-1)j ? ] dt=4i ?+6j ? ((Kg.m)/sec)
K.E=1/2 mv^2×m/m=(m^2 v^2)/2m=p^2/2m=(4i ?+6j ? )^2/(2×2)=13J
W=1/2 mv^2-1/2 mv_0^2=13-0=13J
Example (2): consider a system of three particle: m_1=1gm is at r ?_1=i ?+j ?(cm) and has velocity v ?_1=2i ?(cm/s), m_2=1gm is at r ?_2=j ?+k ?(cm) and has velocity v ?_2=j ?(cm/s), m_3=1gm is at r ?_3=k ?(cm) and has velocity v ?_2=i ?+j ?+k ?(cm/s)
Determine the velocity of the center of mass.
What is the total linear momentum of this system