A metal sphere with a mass of 2.00 kg rolls along a frictionless surface at 12.0 m/s?

Question

A metal sphere with a mass of 2.00 kg rolls along a frictionless surface at 12.0 m/s and stikes a stationary sphere having a mass of 6.00kg. The first sphere stops completely. At what speed does the second sphere move away from the point of impact?

Answer 1

The momentum must be conserved. 2*12 = 24. The second sphere would have to roll at 24/6 to conserve momentum, so 4 m/s.

Answer 2

The quinetic energy of the first sphere goes completely to the second So
1/2*2 *(12)^2 = 1/2 *6*v^2 so v^2= 144/3 and v= sqrt48=4*sqrt3=6.93 m/s

Answer 3

This is not an answer to your question per se, but I would like to point out that on a frictionless surface the sphere will not roll but slide. That is it will have only linear motion and not rotational. So you consider only linear momentum, not angular momentum.
The answer of course is got by conservation of momentum. As someone has pointed out, energy is conserved only in perfectly elastic collisions. When in doubt, conservation of momentum always takes precedence.

Answer 4

Momentum is always conserved in a collision. Guaranteed. Energy is also conserved, but it may, or may not, all continue to be kinetic energy. This depends on whether it's an elastic collision or not.

This collision is not completely elastic, because if you try to make all kinetic energy transfer to the 2nd sphere, then conservation of momentum is violated .

So you can't use conservation of kinetic energy. Go with the answers that used conservation of momentum.

Answer 5

asecond thought 12/2 = 6 m/s .

Answer 6

m1v1=m2v2 (Inertia)

v2 = m1v1/m2 = 24/6 = 4 m/s

Answer 7

it doesnt,shes usually laying under me.