On a highly polished, essentially frictionless lunch counter, a 0.500 kg submarine sandwich moving 3.00 m/s to the left collides with an 0.250 kg grilled cheese sandwich moving 1.20 m/s to the right.
(a) if the two sandwiches stick together, what is their final velocity?
(b) how much mechanical energy dissipates in the collision? where did this energy go?

a) the final velocity is 1.6 m/s

b) The lost kinetic energy is: -1.47 J

Explanation:

Step 1: Data given

Mass submarine sandwich = 0.500 kg

Velocity submarine sandwich = 3.00 m/s

Mass submarine sandwich = 0.250 kg

Velocity submarine sandwich 1.20 m/s

Initial momentum = m(sub)*v(sub) - m(cheese)*v(cheese)  =  final momentum = (m(sub) + m(cheese))*vfinal)

vfinal = (v(sub)*m(sub) - v(cheese)*m(cheese)) / (m(sub) + m(cheese))

vfinal = ((3 *0.5)-(1.2*0.25) / (0.5 + 0.25) =  1.6 m/s

Step 2: The lost kinetic energy is:

ΔKE = final KE - initial KE

with final KE = 1/2 * m(sub + cheese)* vfinal²

with initial KE = 1/2*m(sub)*v(sub)² - 1/2*m(cheese)*v(cheese)²

= 1/2 * m(sub + cheese)* vfinal² -1/2*m(sub)*v(sub)² - 1/2*m(cheese)*v(cheese)²

= 1/2 *(0.5 + 0.25)*1.6²  - 1/2 * 0.5 *3² - 1/2*0.25*1.2²

ΔKE  = -1.47 J

While the total energy of a system is always conserved, the kinetic energy carried by the moving objects is not always conserved. In an inelastic collision, energy is lost to the environment, transferred into other forms such as heat.

im sorry i dont know

statement #3 on andre"s list is false.