Asked by Leslie
Completely lost on what formula to use: A 50 gram ball enters a pendulum with mass 200 g. The pair then swings up to a height of 12 cm. Find the velocity at which the pair move immediately after the collision. Find the initial velocity of the ball before the collision. Appreciate anyone's guidance.
Answers
Answered by
bobpursley
ok, the masses moved up 12 cm, so the change in Potential Energy is mgh=(.250*9.8*.12)
Now that energy must have came from initial Kinetic energy, so
1/2 m vi^2=mgh
vi=sqrt(2*9.8*.12) That is the velocity of the pair after collision.
Now, at collision, conservation of momentum applies
mb*vb=(mb+mp)Vi
.05*Vball=(.250)sqrt*2*9.8*.12)
solve for vball.
Now that energy must have came from initial Kinetic energy, so
1/2 m vi^2=mgh
vi=sqrt(2*9.8*.12) That is the velocity of the pair after collision.
Now, at collision, conservation of momentum applies
mb*vb=(mb+mp)Vi
.05*Vball=(.250)sqrt*2*9.8*.12)
solve for vball.
Answered by
Leslie
Works perfectly. Thanks so much!
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