Does total KE after bouncing equal initial PE of two dropped balls?

[UrbanXrisis]

a baskeball and a tennis ball are dropped from the same height simultaneously. The potential energy (PE) is mgh. m=mass(kg) g=9.8m/s/s h=height(m)

after the ball hits the floor, the tennis ball flies off as the basket ball bounces.
The inital velocity of the basketball after it bounces is Vb, the inital velocity after the tennis ball bounces is Vt. Would the Kinetic Energy of the tennis ball plus the kinetic energy of the basketball right after the bounce equal the same as the original potential energy?

aka, PEtotal=KEtennis+KEbasket?

 

[vincentchan]

If no energy loss, yes... since the energy is conserve..

 

[UrbanXrisis]

If there was energy loss...for example...

The basketball bounces to a certain height, then with each consecutive bounce, it loses less and less height... Is this loss of momentum a constant? Since energy and momentum are related in that p=mv while KE=.5mv^2, related by the velocity. And since energy is loss through heat, drag force, and the friction force. Would this loss of momentum be an exponential loss or linear?

 

[UrbanXrisis]

any ideas?

 

[HallsofIvy]

A lot depends upon your assumptions. If we assume constant "elasticity" of both balls and the floor, then the total energy will be multiplied by that factor (elasticity of 1 would be an "elastic" collision and would result in conservation of energy). If the elasticity is α and the velocity just before hitting the floor is v₀ then the energy after the collision is (1/2)mv²= α (1/2)mv₀² so that v= $\sqrt{\alpha}v_0$. The (magnitude of the) momentum of the ball after the collision is mv= $\sqrt{\alpha}mv_0$. (Of course, it will have opposite sign now.)

The answer to your question , then, is the momentum reduces "exponentially" just like the energy but the factor is the square root of the factor for energy.