Nov 14, 2019 · Consider a ball that strikes a stationary box and has momentum m v. If the ball rebounds with the same speed it had before the collision, its momentum is − m v. Therefore, the box must have a momentum of 2 m v in order for momentum to be conserved. However this would mean the kinetic energy after the collision is more than the kinetic energy ...
some of the energy is wasted as heat. In stage 1 energy is lost because of air resistance. When the ball collides with the air molecules, both the ball and the air are made a little warmer. In stage 2 energy is lost when the ball changes its shape and the ball becomes warmer. Those of you who play squash will be familiar with this. Stage 2 also loses energy as sound when the ball hits the …
Now when the ball hits the sand, due to the collisions between the particles of sand and that of the ball, most of the kinetic energy of the ball is transferred to the particles of the sand. During this process, the air molecules also get some little push/pull and some of the kinetic energy is transferred to the air molecules which appears as ...
Change in kinetic energy =mg (h-h'). Percentage change in kinetic energy = [mg (h-h')/mgh]100= ( 7.5/10) (100)=75%. A ball is dropped from a height of 10m. It loses 10 % of its initial energy due to air resistance and 10% when the ball comes in contact with the ground.
loses energy as sound when the ball hits the ground. Stage 3 loses energy for the same reason as stage 2. Stage 4 loses energy for the same reason as stage 1. If the energy transfers were perfect and there were. no energy losses , the bouncing ball would always return. to the same height at which it started.
some of the energy is wasted as heat. In stage 1 energy is lost because of air resistance. When the ball collides with the air molecules, both the ball and the air are made a little warmer. In stage 2 energy is lost when the ball changes its shape. and the ball becomes warmer. Those of you.
If we want to drop a ball from a height, we calculated that potential energy at bottom is zero and we say it is converted into kinetic energy . At that movement, if it is a kind of sand, we find it will be at rest. So, what happened to its mechanical energy over there?
1. Just before the ball reaches the ground, all of its molecules are coming down with almost an equal speed that is the speed of the ball. (Although, due to the non-zero temperature of the ball, the molecules are also vibrating about their mean position wrt COM frame of the ball).And thus the ball possesses a systematic macroscopic kinetic energy.
Now when the ball hits the sand, due to the collisions between the particles of sand and that of the ball, most of the kinetic energy of the ball is transferred to the particles of the sand. During this process, the air molecules also get some little push/pull and some of the kinetic energy is transferred to the air molecules which appears as sound.
If you're dropping a projectile into sand, the potential energy that you began with ends up being converted into kinetic energy (from the sand thrown out from the collision), sound energy, and thermal energy. Ultimately, the thermal energy is the only surviving energy after any appreciable time though.
If you're dropping a projectile into sand, the potential energy that you began with ends up being converted into kinetic energy (from the sand thrown out from the collision), sound energy, and thermal energy. Ultimately, the thermal energy is the only surviving energy after any appreciable time though. Share.
So as any object falls in a fluid (in this case air) it firstly accelerates and finally reaches constant velocity termed as terminal velocity, at this time the gravitational force equals the viscous force (applied by air). For feather the terminal velocity is very less and all the way down the potential energy is dissipated as heat and sound.
During this process, the air molecules also get some little push/pull and some of the kinetic energy is transferred to the air molecules which appears as sound.
Ball loses 25% of (final K.E.), so initial energy with which it bounces back = 75% of (final K.E) = (3/4) × 120M = 90M.
There was a kinetic energy loss because you know, when the ball will bounce it'll create some sound and heat and vibrations on the floor, so the 25% of kinetic energy is lost there.
If it falls from a state of rest, to the ground (where it is inherently also at rest) then there is no change in kinetic energy : All of the potential energy at 55m which became kinetic energy as it fell, was turned into heat and vibrational energies (sound), and some into moving the earth (but not very far).
Normally, however, a reasonable percentage of the energy will be transformed into other forms. As the GPE of a falling body is given by GPE = mgh, then if 40% of the total energy is turned into other forms, only 60% will remain as GPE when it bounces to its maximum height, and its rebound height will be 60% of its original height. If it was droppped from an original height of 10m, it will rebound to 6m.
Same will be the energy of the body just before touching the ground as the entire potential energy has been converted into kinetic energy now in your case after hitting the ground the body has lost 25 % of its kinetic energy. Now the energy losses after hitting the ground will be in the form of sound, friction of the ball hitting the ground which would generate heat. The energy left in the body after losing the 25 % is 90M which you can find out using simple math. Now equating this to the potential energy equation will give you the new height till which it would rise that being 9 metres.
It it retains 64% of its energy per bounce, it will never run out of energy, no matter how many bounces. But, if you think about what a ball does when it bounces, you can get a pretty good idea about how many bounces might be possible before it stops leaving the ground. If requiring to go airborne means bouncing, you can get an idea. If h = 1 m, then it will bound up to 0.64 m after one bounce and 0.64^2 after two bounces and 0.64^3 after three bounces, etc. Each bounce results in a conversion of 36% of the remaining KE into heat. Using this, you would find that after 10 bounces, the ball would rise to a height of 11 mm and after 12 bounces, the ball would rise to a height of 5 mm. At around this number, the “height” would no longer exceed the compression of the ball (the ball is an elastic system, like a spring). The ball would still be compressing a bit, but that would be diminishing quickly and be undetectable soon. And, since it not longer goes airborne, is it still “bouncing”? The ball you describe might bounce a dozen times at one m and bounce an additional time if released from 1/0.64 = 1.6 m. But, given the constraints of the problem, it will never stop vibrating, at least given the rules of Newtonian mechanics.
So to find the height to which the ball will bounce back we equate the initial energy (that is 90M) with the final energy (that is potential energy at the maximum height to which the ball has bounced).
This is because the COR is higher so it rebound off the wall with less speed loss. Mar 25, 2009. #9. bernard08.
However the coefficient of restitution shows how elestic or plastic a collision is. It is a ratio of velocityin/velocity out and is between 0 and 1 in most practical cases.
As a ball rises kinetic energy decreases and potential energy increases. AT the highest point, the ball stops moving and no longer passes kinetic energy, only potential. As the ball falls, energy conversion is reversed. The kinetic energy increases and the potential energy decreases.
The ball doesn't bounce as high because some energy escapes as thermal or sound when the ball hits the ground.
The law of conversion of energy states that when one form of energy is converted to another, no energy is destroyed in the process. Energy cannot be created or destroyed.