Posted by Hale on .
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)?
 Physics  Count Iblis,

Physics 
steven lou,
1.68?

Physics 
Help please,
i input your formula and false

Physics 
Elena,
The angular momentum of the door ‘p’= the change in amgular momentum of the ball Δp.
In vector form:
Δp⃗ =p₂⃗p₁⃗ =mv₂⃗mv₁⃗,
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. 
Physics 
hemant,
please write right answer