If a steel ball hits an iron surface it bounces back due to the Newton's third law of motion. Then why does'nt it bounce back when it hits a water surface? or for that matter even air? The molecules of water / air should apply the force of equal magnitude on the ball as the ball applies to the water molecules. Can anyone please explain this?
The third law of motion:
When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body.
Edit after revisit:
The molecules of water / air should apply the force of equal magnitude on the ball as the ball applies to the water molecules.
The basic misunderstanding here is in the notion of force. Acceleration for impacts will define the force, and it is best to think of the dp/dt definition of force. An incoming ball does not have 10 Newton's force , it has a constant velocity, so no force . Force is generated on impact by dp/dt. Impacts with solids distribute momentum and energy on the whole solid due to the cohesiveness of the electromagnetic interactions, and the total momentum energy balance can be calculated with the impact point , and dp/dt calculated to give the force of impact.
Impact on air is coming in tiny scatters and tiny dp/dt. Impact on water is similar if the ball has enough energy, molecules bounce left and right with a small dp/dt slowly dissipating the energy and dispersing the momentum.
You ask :
Then why does'nt it bounce back when it hits a water surface ?
Bouncing back depends on elasticity of surfaces deformations and dissipative motions of the surfaces which absorb the energy and momentum of impact , and have to be estimated for each individual case.
Have you estimated the force that the steel ball exerts on the water molecules? dp/dt will transfer momentum to all the column of water it penetrates and dissipate the energy until it hits the bottom. At the surface little except ripples because the liquid is displace-able with little energy/momentum needs . Look at the video for a drop falling, which has the same density as the water and small energy/momentum to transfer to see how the third law is obeyed by dissipating the impulse and still a small drop bounces back.
Air is a non thinker: individual molecules have very small mass with respect to the ball and pick up the tiniest fraction from the impact diminishing the motion of the ball very slightly at best.
One has to do the calculations to see that the third law is OK ? It is called a law because innumerable people have done the calculations and found that the law is validly describing data.
If a steel ball hits an iron surface it bounces back due to the Newton's third law of motion. Then why does'nt it bounce back when it hits a water surface ? or for that matter even air ? The molecules of water / air should apply the force of equal magnitude on the ball as the ball applies to the water molecules. Can anyone please explain this?
A simple explanation:
The forces that appear when a 100 kg iron sphere traveling at 100 km/h comes in contact with a motionless 0.1 kg plastic sphere are the same on both spheres. This is Newton's third law. However, this does not mean the 100 kg ball has to bounce back.
force exerted by the heavy steal ball on air partical is very very small. If a ball moves with even as much as 1000kmps may exert a very small force of 0.000001N on a body. Force exerted on a air particles by steel ball is very small although it has great speed. Force is rate of change of momentum . A very small particle of air cannot reduce its momentum to a larger extent. Instead the rate of change of momentum brought by it is very small. so, force due to momentum change by iron ball on air is very small, in return force on ball by air is also very small and ball experiences a very small drag which reduces speed of iron ball after sometime ( say 10-20 mins).
Although very small force it is, but on air it causes much greater acceleration and greater effect. Whereas since ball is very heavy deacceration is very small.
After reading your comment on Anna V 's answer , I think I have got your doubt.
When you throw a ball with a force of 10 N , you have applied the force and you are not continuously applying the force after the ball has left your hands. What I mean to say is that all the force that you applied on the ball (10N) has been applied by the time the ball leaves your hand. That force (10N ) = rate of change of momentum of the ball ... ie 10N=( mv-m.0)/t (where u=0) . Obviously t will be very small. Take m out and you get 10=ma. The point is that you have applied all the force and if there is no gravity (to avoid projectile motion) , the ball moves with a constant velocity which aboviously depends on a (and a obviously depends on t). So now the ball moves with a constant velocity. As it hits the wall it's velocity and hence it's momentum is changed. The rate of change in momentum (which is also the force by the wall on the ball and by the ball on the wall) is (mv2-mv1)/t (where v2 is the velocity of the ball after the collision and v1 is the velocity of the ball that it had gained after you had applied the 10 N force ) . Now as the others have said v2 depends on the material of the ball and the wall (coefficient of restitution) . Hence v2 will be different for a steel wall , a brick wall and a water wall. Since v2 will be different for different types of walls , the force on ball by the wall will be different and hence the acceleration of the ball would be different. Obviously for a water surface wall elastic properties will also matter and the ball will penetrate into the water surface.
So you were getting wrong the fact that the wall also applies a force on the ball which is equal to the force you had initially applied (it may be equal but that depends on the collision )
I hope you got it useful !
protected by Qmechanic♦ Sep 16 '16 at 20:29
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