1. CMx = M·0+3M·(L/2) + M·(L) 5M
L/2
L/2 L
CMy = M·0+3M·(L/2) + M·(0) 5M
Complex, asymetric bodies of uniform density can be simplified by breaking them into easily identified, symmetric pieces and considering all of the mass of that piece to be at it’s geometric center.
Answer= (0.5ML, 0.3ML)

2. Chop it into manageable pieces.
Locate their CMs by symmetry
Fix an origin
Find CM of whole system by finding the x and y distance to each little CM and plugging into the CM formula.
We’ve now created a 5-body problem that is easier to handle than the original whole shape by taking advantage of symmetries.

3. What patterns appear when you compare the motions before and after collisions?
What patterns appear when a wrench flips while falling?

4. The simplest way to deal with a complex, extended or rotating system is to ignore the motions and pretend all of the mass is located at the CM!! The CM will obey all of the laws of Physics you have learned so far.

6. “Suspended” means W – Fbuoyant = 0
So again Fext = macm = 0 and the CM stays at rest; no matter how fast the guy climbs, he’ll just pull the balloon and his body toward the CM faster. M u
Let’s call the speed he pulls down the balloon “u”. Since the ladder goes down with the balloon this reduces his rate of climb such that, his speed relative to the ground (let’s call it vg) such that vg = v -u.
so M moves at u and m moves at vg while vcm =0 cm v m
All this, said in 1 equation, is……….
psystem = 0 = (M+m)vcm = m vg + Mu = m(v-u) – Mu
Solve for u: u = mv/(M+m) cm
(b ): p system remains = 0, all v and u = 0 , and the balloon and man are stationary once again. The CM is in the same place, relative to earth, but the man is higher and the balloon is lower

7. On a Force vs Time graph, the impulse is the area under the curve
On a Force vs Time graph, the impulse is the area under the curve. J = FΔt =Δp