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)?
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1.68?
i input your formula and false
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.
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.
please write right answer