
A light spring is placed inside a big tube and stretches along the whole length and no external force acts on it. A ball is placed at the top of the spring, due to which the spring is compressed to half of its length. If the tube in this inclined position falls freely, then?
A. No change will occur in the position of the spring.
B. The spring will be compressed more.
C. The spring will expand and the ball will remain in the tube.
D. The spring will expand and the ball will emerge from the tube.
Answer
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Hint: Spring initially will be compressed due to the weight of the ball and restoring force due to the spring will be exerted on the ball. When anybody freely falls under gravity it experiences weightlessness.
Complete answer:
We are given that the spring is inclined initially so we assume that it makes an angle $\theta$ with the horizontal. When we put a ball in this inclined spring, the restoring force of the spring will be equal to the cosine of the angle with the (horizontal) axis.
$mg \cos \theta = F$
Consider the case of an elevator undergoing acceleration downwards now.
W = m (g-a)
is the weight when the a is the acceleration.
When the same elevator undergoes freefall, then a becomes g so the weight of the body in it becomes zero:
W = m(g-g) = 0.
In our case, the acceleration downwards is due to gravity and the resultant from the spring’s restoring force gives us:
$ mg = mg – F \sin \theta $
This gives us directly F=0 as $\theta$ is not zero. Therefore, the spring does not feel any compression. The spring therefore expands and throws the ball out. These results were because we were already given that the uncompressed spring in the tube filled the entire tube.
So, the correct answer is “Option D”.
Note:
If a body undergoes acceleration upwards, then the acceleration is added to g and the weight increases. If there is downwards acceleration, the weight of the body decreases. We are clearly able to tell without any mathematical expressions that the ball will exert no force on the spring due to freefall. Between option (C) and (D), we chose (D) only because the spring filled the tube initially.
Complete answer:
We are given that the spring is inclined initially so we assume that it makes an angle $\theta$ with the horizontal. When we put a ball in this inclined spring, the restoring force of the spring will be equal to the cosine of the angle with the (horizontal) axis.
$mg \cos \theta = F$
Consider the case of an elevator undergoing acceleration downwards now.
W = m (g-a)
is the weight when the a is the acceleration.
When the same elevator undergoes freefall, then a becomes g so the weight of the body in it becomes zero:
W = m(g-g) = 0.
In our case, the acceleration downwards is due to gravity and the resultant from the spring’s restoring force gives us:
$ mg = mg – F \sin \theta $
This gives us directly F=0 as $\theta$ is not zero. Therefore, the spring does not feel any compression. The spring therefore expands and throws the ball out. These results were because we were already given that the uncompressed spring in the tube filled the entire tube.
So, the correct answer is “Option D”.
Note:
If a body undergoes acceleration upwards, then the acceleration is added to g and the weight increases. If there is downwards acceleration, the weight of the body decreases. We are clearly able to tell without any mathematical expressions that the ball will exert no force on the spring due to freefall. Between option (C) and (D), we chose (D) only because the spring filled the tube initially.
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