PLEASE GIVE THE ANSWER,I ALREADY INPUT TO YOUR FORMULA AND IT IS FALSE...
Is your answer 0.57?
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)?
The answer I posted was correct as far as I can see, the problem may well be a typo in the problem itself, e.g. a 400 gram ball slamming into the door at the speed of a speeding car is not going to "slowly" open the door.
Anyway, it's a complete waste of time to e.g. try to see if what was meant is 3.5 m/s. I would suggest forgetting about this problem and moving on to another problem, you simply say that you did the problem correctly, that the teacher made a mistake and that you demand full marks for this problem.
iblis got the problem right. for those trying to use jiskha to cheat on the brilliant problems, there was a mistake made in the acceptable answers for this week's question.
ibis ,I beg to differ,the door slowly open because the ball bounces back elastically,please don't be certain,when u are wrong
To find the time it takes for the door to fully open, we can use the principle of conservation of angular momentum.
The initial angular momentum of the system, consisting of the door and the ball, is equal to the final angular momentum. Initially, the door is at rest and the ball is moving with a velocity of 35 m/s. As the ball hits the door and bounces off, it imparts angular momentum to the door.
To calculate the initial angular momentum of the system, we need to find the moment of inertia of the door and the angular velocity of the ball after the collision.
The moment of inertia of a rectangular door rotating about an axis perpendicular to its width is given by the formula:
I = (1/3) * m * L²
where m is the mass of the door and L is its width.
Given:
Mass of the door (m) = 15 kg
Width of the door (L) = 1 m
Plugging these values into the formula, we can calculate the moment of inertia (I) of the door.
I = (1/3) * 15 kg * (1 m)²
I = 5 kg m²
Next, we need to find the angular velocity of the ball after the collision. Since the collision is elastic, we can use the conservation of kinetic energy:
(1/2) * m * v² = (1/2) * I * ω²
where m is the mass of the ball, v is its velocity, I is the moment of inertia of the door, and ω is the angular velocity of the ball.
Given:
Mass of the ball (m) = 0.4 kg
Velocity of the ball (v) = 35 m/s
Moment of inertia of the door (I) = 5 kg m²
Plugging these values into the equation, we can solve for ω.
(1/2) * 0.4 kg * (35 m/s)² = (1/2) * 5 kg m² * ω²
49 kg m²/s² = 2.5 kg m² * ω²
ω² = 19.6 rad²/s²
ω = √(19.6) rad/s
ω ≈ 4.43 rad/s
Now, we need to find the time it takes for the door to rotate 90 degrees (π/2 radians) with a constant angular velocity.
Time (t) = θ / ω
Given:
Angle to rotate (θ) = π/2 radians ≈ 1.57 radians
Angular velocity (ω) ≈ 4.43 rad/s
Plugging these values into the equation, we can calculate the time it takes for the door to fully open.
t = (1.57 radians) / (4.43 rad/s)
t ≈ 0.35 seconds
Therefore, it takes approximately 0.35 seconds for the door to fully open (rotate 90 degrees) after being hit by the ball.