Q#9
A collision experiment is done on a horizontal table kept in an elevator. Do you expect a change in the results if the elevator is accelerated up or down because of the noninertial character of the frame?Answer:
No. When we analyze the motion of a body with respect to a noninertial frame we add a pseudo force on the body opposite to the direction of acceleration of the frame. Since the elevator is accelerated only in vertical direction, the direction of the pseudo force is also vertical. It does not affect the horizontal movement of the body.
So, collision experiment done on a horizontal table in the elevator will not be affected if the elevator is accelerated up or down.
Q#10
Two bodies make an elastic head-on collision on a smooth horizontal table kept in a car. Do you expect a change in the result if the car is accelerated on a horizontal road because of the noninertial character of the frame? Does the equation "Velocity of separation = Velocity of approach" remain valid in an accelerating car? Does the equation "final momentum = initial momentum" remain valid in the accelerating car?
Answer:
In a noninertial frame both the bodies will experience equal acceleration in the horizontal direction. So, the experiment result will change in the accelerating car.
After applying pseudo forces the motion can be analyzed as if in an inertial frame. Since the sum of external forces are not zero now, hence the equations
(i) Velocity of separation = Velocity of approach and
(ii) final momentum = initial momentum do not remain valid in the accelerating car.
Q#11
If the total mechanical energy of a particle is zero, is its linear momentum necessarily zero? Is it necessarily nonzero?
Answer:
(i) No. Consider the mechanical energy of a particle under gravity. The total mechanical energy = K.E. + P.E.
K.E. is always positive because either v is positive or negative, v² will always be positive and so will ½mv² = K.E. but P.E. = mgh may be positive or negative because h is positive or negative depending upon its position with respect to the reference line. Suppose we take ground as the reference line. If the particle falls into a dry well, at some depth below ground level the magnitude of negative P.E. may be equal to K.E. thus making the total mechanical energy of the particle zero. But due to its velocity, the linear momentum is not zero.
(ii) No. It is not necessarily nonzero. Suppose the particle is on the ground at rest. K.E. = 0, and P.E. = 0. Total mechanical energy = 0. Since velocity is zero, the linear momentum is also zero.
Q#12
If the linear momentum of a particle is known, can you find its kinetic energy? If the kinetic energy of a particle is known can you find its linear momentum?
Answer:
(i) If only we know the mass of the particle, it's kinetic energy can be found out. Because both the K.E. and the magnitude of the momentum involve mass m and magnitude of the velocity v.
(ii) Even if we know the mass of the particle, only the magnitude of the momentum can be calculated. The direction of the momentum is unknown. Momentum, being a vector thus cannot be found out.
Q#13
What can be said about the center of mass of a uniform hemisphere without making any calculation? Will its distance from the center be more than r/2 or less than r/2?
Answer:
Less than r/2.
Just visualize a uniform cylinder of radius r and height r. Its center of mass will be at r/2 from the center of the base circle. Since the given hemisphere can be made by carving out masses from the upper parts the center of mass of the remaining hemisphere will be less than r/2.
Q#14
You are holding a cage containing a bird. Do you have to make less effort if the bird flies from its position in the cage and manages to stay in the middle without touching the walls of the cage? Does it make a difference whether the cage is completely closed or it has rods to let air pass?
Answer:
When the bird flies and maintains its position in the middle of the cage, it pushes air downward to get upward lift. The pushed air partly escapes down the rods of the cage, so the full downward pressure is not on the cage. So, less effort will be needed to hold the cage.
If the cage is completely closed the full downward pressure of the pushed down air (equal to the weight of the bird) will be on the base of the cage. So, same effort will be needed to hold the cage.
Q#15
A fat person is standing on a light plank floating on a calm lake. The person walks from one end to the other on the plank. His friend sitting on the shore watches him and finds that the person hardly moves any distance because the plank moves backward about the same distance as the person moves on the plank. Explain.
Answer:
Since the plank is light its mass is negligible in comparison to the fat person. Combined center of mass of the system containing the fat person and the plank can be considered near the person. Since there is no external force on this system the center of mass of the system will remain at the same place. Due to this the friend on the shore finds the fat person hardly move.
Q#16
A high-jumper successfully clears the bar. Is it possible that his center of mass crossed the bar from below it? Try it with appropriate figures.
Answer:
Nowadays high jumpers clear the rod keeping their center of mass at minimum possible height in order to put less effort. To do this they bend their body in a semicircular shape above the bar as shown in the figure below.
As we know the center of the mass of a uniform semicircular wire is at
r - 2r/π = (π-2)r/π = 0.36r below its crest.
So, the center of mass may cross below the bar during the successful clearance of the bar as shown in the figure.
Q#17
Which of the two persons shown in the figure (9-Q1) is more likely to fall down? Which external force is responsible for his falling down?
Answer:
Moving cart with an acceleration is a non-inertial frame for the passengers. So both of them will experience a pseudo force in the opposite direction of the movement. The person sitting on the left will be saved by the wall of the cart but not the person on the right. So the person on the right is more likely to fall and the force responsible for it is the pseudo force.
r - 2r/π = (π-2)r/π = 0.36r below its crest.
So, the center of mass may cross below the bar during the successful clearance of the bar as shown in the figure.
Q#17
Which of the two persons shown in the figure (9-Q1) is more likely to fall down? Which external force is responsible for his falling down?
Answer:
Moving cart with an acceleration is a non-inertial frame for the passengers. So both of them will experience a pseudo force in the opposite direction of the movement. The person sitting on the left will be saved by the wall of the cart but not the person on the right. So the person on the right is more likely to fall and the force responsible for it is the pseudo force.
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