rol Q A 5. A clay ball with mass m has speed ve as it approaches a clay ball with mass 3m. The 3m ball is initially at r

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Rol Q A 5 A Clay Ball With Mass M Has Speed Ve As It Approaches A Clay Ball With Mass 3m The 3m Ball Is Initially At R 1 (29.34 KiB) Viewed 8 times

Rol Q A 5 A Clay Ball With Mass M Has Speed Ve As It Approaches A Clay Ball With Mass 3m The 3m Ball Is Initially At R 2 (32.82 KiB) Viewed 8 times

Rol Q A 5 A Clay Ball With Mass M Has Speed Ve As It Approaches A Clay Ball With Mass 3m The 3m Ball Is Initially At R 3 (24.46 KiB) Viewed 8 times

rol Q A 5. A clay ball with mass m has speed ve as it approaches a clay ball with mass 3m. The 3m ball is initially at rest and hanging from a string with unknown length L. See figure. Then, the two balls have an inelastic collision, so that they stick together. After colliding, both balls move upward along a circular path, reaching a maximum height, max, at angle y. Find y. If the magnitude of the tension force in the string immediately 16 after the collision is 7-mg, what is L? The knowns are m, g, and vo. (20 points) 3 center of circle gravity direction. string, L=? ²1. Ww? 3m at rest before collision. hax?

Constant Acceleration Equations (1) x(1) = x₂ + vol + ar² (2) v(1) = vo + at (3) v²=v² + 2a(x-x) W=F,Ar, +F,Ar, (constant force) W = |F|AF|cose (constant force) p=mv System Momentum Conservation m₁v₁ + m₂v, m₂v₁ + m₂v₂ = Inelastic Collision Elastic Collision V₂ - V₁ = V₁ V₂1 ΣΕ = ma, ΣF, = ma, ΣF, = ma a = r fx = μ₂n Energy Equations [_mv² + m²y₁ + = kx² + [W₁ = =_ mv² + n 2 10² +mgya+W = 10 + Io+mgycs,2 K=-=-mv² Ug = mgy 1 U₁ -kx² 2 no slip condition: v=cr + +=kx

Static Equilibrium EF=0 ΣF-0 Σr-o Universal Gravity Gm,m, Simple Harmonic Motion m T=2m, -±|||F|sin T=xF, -yF r>0=CCW r<0=CW 1=-=-=- 2π @= T V =0A = 0² A F=KA= mamax max в роли ротика = mr L=10 L =Lat c (hyp) sin 4 b (opp) cos tan c²=a² + b² a (adj) ax² +bx+c=0 - opp/hyp-b/c - adj/hyp-a/c - opp/adj-b/a Quadratic Equation x= -b± √b²-4ac 2a