Answer:
A)x_cm = 1 / (m₁ + m₂) (x₁ m₁ + x₂ m₂)
D)v_cm = 1 / (m₁ + m₂) (m₁ v₁ + m₂ v₂)
F) p_cm = p₁ + p₂ , H) a_cm = 1 / (m₁ + m₂) (m₁ a₁ + m₂ a₂)
I) a_cm = F / (m₁ + m₂) J) a_cm = 1 / (m₁ + m₂) (F₁ + F₂), L) a_cm = 0
Step-by-step explanation:
A) The center of mass of the system is
x_cm = 1 / (m₁ + m₂) (x₁ m₁ + x₂ m₂)
D) let's find the speed
v = dx / dt
v_cm = d x_cm / dt
v_cm = 1 / (m₁ + m₂) (m₁ v₁ + m₂ v₂)
F) the moment of the center of mass
p = m v
(m₁ + m₂) v_cm = (m₁v₁ + m₂ v₂)
p_cm = p₁ + p₂
H) the acceleration of the center of mass
a = dv / dt
a_cm = d v_cm / dt
a_cm = 1 / (m₁ + m₂) (m₁ a₁ + m₂ a₂)
I) look for acceleration from Newton's second law
F = ma
a = F / m
Force is applied to the body 1
Body 2 has no applied force
a_cm = 1 / (m₁ + m₂) (m₁ F / m₁ + 0)
a_cm = 1 / (m₁ + m₂) F
J) we now have two forces each applied to a different block
a_cm = 1 / (m₁ + m₂) (F₁ + F₂)
L) as the forces are internal
F₁₂ = -F₂₁
a_cm = 1 / (m₁ + m₂) (f₁₂ - f₂₁)
a_ cm = 0