Two masses, each weighing 1.0 × 103 kilograms and moving with the same speed of 12.5 meters/second, are approaching each other. They have a head-on collision and bounce off away from each other. Assuming this is a perfectly elastic collision, what will be the approximate kinetic energy of the system after the collision?
A. 1.6 × 105 joules
B. 2.5 × 105 joules
C. 1.2 × 103 joules
D. 2.5 × 103 joules

Question

anonymous · Answer

An elastic collision is an encounter between two bodies in which the total kinetic energy of the two bodies after the encounter is equal to their total kinetic energy before the encounter.

anonymous · Answer

$$1/2 m_1 v_1^2+1/2 m_2 (v_2)^2$$

anonymous · Answer

i got 1.609375x10^4

anonymous · Answer

1/2 (1.0*10^3) (12.5)^2 +1/2 (1.0*10^3) (12.5)^2= (1.0*10^3) (12.5)^2

anonymous · Answer

okay this time i got 1.5625x10^5

anonymous · Answer

that's what I got

anonymous · Answer

alright!

turingtest · Answer

or if you want to be clever, because the masses an velocities are equal$$1/2m_1v_1^2+1/2m_2v_2^2=mv^2$$

anonymous · Answer

thanks!

anonymous · Answer

Energy is not conserved during collisions. Use momentum.

turingtest · Answer

it is in completely inelastic ones

anonymous · Answer

in elastic collision  , it is

turingtest · Answer

sorry elastic*

anonymous · Answer

Ah yes. Overlooked elastic.