Asked by pyhsiclover

A 15 kg, 1 m 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)?
Answered by steven lou
1.68?
Answered by Anonymous
no. wrong answer
Answered by Elena
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 anonymous
wrong answer Elena
Answered by Anonymous
0.43 s
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