(a)
PE=mgh=mgL
KE = mv²/2
Law of conservation of energy
PE =KE
mgL= mv²/2
v=sqrt(2gL0=sqrt(2•9.8•2.05)=6.34 m/s.
(b) Law of conservation og linear momentum
mv= (m+m1)u
u= m•v/(m+m1)=80•6.34/(80+47)=4 m/s
(c)
Law of conservation of energy
KE1=PE1
(m+m1) •u²/2 =(m+m1) •g•h1
h1=(m+m1) •u²/2 • (m+m1) •g =...
In a Broadway performance an 80.0 kg actor swings from a 2.05 m long cable that is horizontal when he starts. At the bottom of his arc he picks up his 47.0 kg costar in an inelastic collision. (a) What is the velocity of the swinging actor just before picking up the costar? (b) What is the velocity of both actors just after picking up the costar? (c) What maximum height do they reach after their upward swing?
4 answers
I'm still confused on part c.
h1=(m+m1) •u²/2 • (m+m1) •g =u²/2 •g=
=4²/2•9.8=0.82 m
=4²/2•9.8=0.82 m
Thank you!
0 answers