I WILL ONLY RATE THE ANSWER that HAS THOSE NUMBERs ! DONT ANSWER A EXAMPLE BECAU
ID: 2104825 • Letter: I
Question
I WILL ONLY RATE THE ANSWER that HAS THOSE NUMBERs ! DONT ANSWER A EXAMPLE BECAUSE I HAVE TRY TONS OF THOSE AND THEY DON'T WORK.
5 STARS TO THE RIGHT ANSWER!
A 4.60 kg ball is dropped from a height of 13.0 m above one end of a uniform bar that pivots at its center. The bar has mass 7.00kg and is 8.60m in length. At the other end of the bar sits another 5.10 kg ball, unattached to the bar. The dropped ball sticks to the bar after the collision.
How high will the other ball go after the collision?
Explanation / Answer
velocity of ball before collision
V_ball =sqrt(2gh)=sqrt(2*9.8*13)
V_ball =15.96 m/s
R=L/2 =8.6/2 =4.3 m
total moment of inertia is
It =[mL^2/12 +(M1+M2)R^2] =[7*8.6^2/12 +(4.6+5.1)*4.3^2]
It=222.5 Kg-m^2
By conservation of momentum
L_before =L_after
M1VR=ItW
4.6*15.96*4.3 =222.5*W
W=1.42 rad/s
Velocity of ball after collision
V=RW =4.3*1.42
V=6.1 m/s
h= V^2/2g =6.1^2/2*9.8
h=1.9 m
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