Normal bounce
A ball of mass m is dropped from some height in a (uniform) gravitational field of g . It reaches the ground and bounces directly upward, to 70% of its original height. During the bounce, the ball is in contact with the ground for a small, non-zero time interval of δ t .
What can be said about the normal force between the ball and ground during the interval δ t ?
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2 solutions
This is a pretty clever problem, since the time-averaged value of the force is proportional to the impulse. It is trivial to realize that the total normal impulse must be greater. And then it directly follows that the time-averaged normal force must exceed the time-averaged gravity force.
Is the normal force greater than mg throughout?
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Almost all the time. At contact, the falling object and/or floor will start being compressed; the restoring (elastic) force essentially provides the normal force. At first this force is N < m g : the object continues to accelerate downward yet. But soon N = m g (the object reaches maximum speed), then N > m g . This causes the falling motion to slow down; the object turns around, and starts accelerating upward. The upward acceleration stops as soon as N < m g , when the object almost (but not quite) loses contact with the ground.
Am I missing something, or are you minimizing the larger term. If dt is small, compared to the time the ball spent dropping, then N must be much larger than mg, to achieve a greater change in velocity in a much shorter time. (I'm sorry, I'm not used to creating special notation with this editor.)
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On average N must be larger than m g to accomplish the bounce. You are also correct that, if δ t is small, it must be significantly larger (if the ball is dropped from sufficiently high).
Another way of looking at it: if ⟨ N ⟩ is the average normal force during the bounce and v 0 the downward speed at which the ball enters the bounce, then the momentum-impulse relation F δ t = m Δ v implies ⟨ N ⟩ = m g + δ t ( 1 + 0 . 7 ) v 0 , from which it is obvious that the average normal force is greater as δ t decreases.
Velocity at impact = √(2gh) Momentum at impact = m√(2gh)
Velocity after impact = √(2g(0.7h)) = √(1.4gh) Momentum after impact =m√(1.4gh)
Change in momentum = m(√(2gh) + √(1.4gh)) Change in momentum = m(√gh)(√2 + √1.4)
Force = rate of change of momentum Average Force = Change in momentum/(delta T)
Average Force = m(√gh)(√2 + √1.4)/(delta T) Any solution involving an equality must include all those terms.
Enough of the junior high math, practical view, if the ball is filled with air, "delta T" will increase as its internal air pressure is reduced. The softer the ball the lower the force will be and the longer the time it will operate.
OR we can say if it's moving downward in a gravitational field and is brought to rest, the force applied must be greater than the weight of the object. The weight of the object is mg. Thus F > mg If it is accelerated upward in a gravitational field, the force applied must be greater than the weight of the object. F > mg
These are the only two cases here. Therefore the average force > mg
This is a poorly worded question. The normal force is defined as the force a surface applies to an object to keep it from falling through the surface, and is always perpendicular to the angle of that surface. In this scenario, the normal force should always be equal to the weight (N=mg*cos 90°). The only time a normal force should not equal the weight of an object is when the surface is inclined.
Normal force is not the net force acting on the ball when it strikes the surface, but it does negate the weight when doing the net force calculation. Since the normal force and the weight cancel one another, we can focus purely on momentum.
The ball has momentum A (m√(gh)) when it strikes the surface, which is passed to the surface.
Since the mass of the surface is so much larger than the ball, the surface acceleration is virtually nil, but there is a force applied (A*dt, not weight). It is important to remember that we are assuming 100% efficiency here, so when the ball strikes the surface it stops dead, imparting all its momentum to the surface. By Newton's 3rd Law, the surface applies an equal and opposite force to the ball. The ball bounces back upward.
Since gravity will still be pulling the ball down, the acceleration will be lower for the upwards bounce, causing the ball not to bounce as high.
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This is a dynamics problem, not a statics problem. Hence, the comment about the normal force equaling the weight does not apply. If that were really true, bounces would be impossible.
There are conceptual problems in your statement.
It is true that the ball imparts its momentum to the surface, and that the surface at the same time exerts an upward force on the ball. (This is precisely the normal force N .) But it won't do to explain the bounce merely from the Third Law. Theoretically this does not work, because the upward bounce happens at a different moment in time than the imparting of momentum you were talking about. Pracitcally, your statement can easily be proven wrong by dropping a chunk of clay to the floor: it does not bounce upward.
The bounce happens because the ground and/or ball undergo elastic deformation. When the ball has come to a full stop, compressional stress results in continuation of the normal force, even though this is no longer necessary to prevent it from passing into the surface. The height of the bounce depends on the percentage of kinetic energy that can be recovered through this elastic process. For a perfectly elastic ball, the bounce can reach 100% of its initial height (in contradiction of your last paragraph).
The "story" of the bouncing ball can be told in three stages, and from three different perspectives.
Force perspective : When the ball reaches the floor, the floor responds with a normal force to keep the ball from passing through the surface. This normal force is the result of compressional stresses that grow rapidly as the ball is squished. Since the normal force opposes the downward motion of the ball, the ball slows down and comes to a full stop. If the ball were not at all bouncy, the story would end here. For a bouncy ball, the normal force continues pushing upward for a while, as the ball decompresses, giving the ball upward motion. Eventually the ball reaches a highest point because the acceleration due to gravity takes away upward speed.
Momentum perspective : When the floor stops the ball, the ball's momentum is transferred to the floor. If the ball were not bouncy, the story would end here. In the case of a bouncy ball, the transfer of momentum continues, so that the ball actually turns around. Specifically, in this situation, with v 0 = 2 g h and v f = 0 . 7 v 0 , we have
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initial momentum of the ball at bounce: p 0 = − m v 0
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final momentum of the ball after bounce: p f = + m v f
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momentum transferred from ball to floor during bounce: p 0 − p f = − m ( v 0 + v f )
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momentum transferred from floor to ball during bounce: p f − p 0 = m ( v 0 + v f )
Energy perspective : When the floor stops the ball, its kinetic energy is reduced to zero. Some of this energy is lost in friction and zero-stress deformation; some of this energy is stored in elastic deformation of the ball (and/or floor). During the upward bounce, elastic energy becomes kinetic energy again. In the given situation, 30% of the initial kinetic energy is lost due to friction and 70% of it becomes final kinetic energy in the upward bounce.
"If N=mg, there would only be enough force to stop the ball from accelerating downward"
Are you saying that if a falling object just hits the ground and stops (i.e. doesn't bounce), then N=mg??
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No, "stop from accelerating downward" is not at all the same as "stop from moving downward". In order to stop the ball, one also needs N > m g on average. To be precise, ⟨ N ⟩ a v g = t 1 ∫ 0 t N d t = m ( g + t v 0 ) .
N>mg is required to reverse the momentum during the contact. Regarding the 4th alternative, N<mg is required for some period of time to maintain N=mg on average. But there are no other forces involved (at least not described) except from "N" and "mg" to bring back N>mg so that the ball can bounce up, so the 4th alternative is not possible.
Nice solution. The momentum is to be increased in the upward direction, so average net force on the ball should be upwards, so N > mg.
Relevant wiki: Newton's Second Law
Conceptual solution
If N = m g , there would only be enough force to stop the ball from accelerating downward; but there would not be enough upward force to accelerate it upward so that it can slow to a stop and turn upward again. Therefore N > m g must be true during the bounce.
Clarification: There will be a brief time, at the very beginning and the very end of the bounce, where the normal force is small yet, N < m g . However, on average this cannot be the dominant situation. See the details below.
Mathematical detail
Call the upward direction positive. During the bounce, the momentum of the ball changes from negative to positive, so that m Δ v = ( + ) − ( − ) > 0 . The change in momentum is the net impulse, which is the sum of the impulse of gravity and the impulse of the normal force: m Δ v = ∫ δ t ( N − m g ) d t = ∫ δ t N d t − m g δ t . Thus the average of N over interval δ t is δ t 1 ∫ δ t N d t = δ t 1 ( m g δ t + m Δ v ) = m g + m δ t Δ v , and since Δ v > 0 this implies ⟨ N ⟩ δ t = δ t 1 ∫ δ t N d t > m g . Therefore the average normal force during the bounce is greater than m g .