Asked by Hale

A 15 kg, 1m wide door which has frictionless hinges is closed but unlocked. A 400 g ball hits the exact middle of the door at a velocity of 35 m/s and bounces off elastically, thereby causing the door to slowly swing open. How long in seconds does it take for the door to fully open (rotate 90 degrees)?

Answers

Answered by Count Iblis
http://www.jiskha.com/display.cgi?id=1369042520
Answered by steven lou
1.68?
Answered by Help please
i input your formula and false
Answered by Elena
The angular momentum of the door ‘p’= the change in amgular momentum of the ball Δp.
In vector form:
Δp⃗ =p₂⃗-p₁⃗ =mv₂⃗-mv₁⃗,
for magnitudes
Δp =p₂-(-p₁) =mv₂+mv₁= 2mv.
p=Mu =>
Mu=2mv
The speed of the center of the door is
u=2mv/M=2•0.4•35/15 = 1.37 m/s
Its angular speed is
ω=u/r=2u/L =2. •1.87/1 = 3.73 rad/s
t= φ/ ω =π/3.73 = 0.84 s.
Answered by hemant
please write right answer
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