Question

Two masses, MA and MB, with MB = 2MA, are released at the same time and allowed to fall straight down. Neglect air resistance. When we compare their kinetic energies after they have fallen for equal times, we find thata.KB =KA.

b.KB =2KA.
c.KB =4KA.
d.KA =2KB.
e.KA =4KB.

Answer 1

KB = 2KA

Step-by-step explanation:

Given

MA = Mass of object A

MB = Mass of object B

Using, K.E = ½mv², where m = mass and v = velocity

The kinetic energy of A;

KA = ½MA.V²

The kinetic energy of B;

KB = ½MB.V²

Assuming air resistance is negligible, then velocity of A and B are equal; VA = VB = V

Given that MB = 2MA

The kinetic energy of B, becomes

KB = ½(2MA).V²

KB = 2 * ½MA.V²

By substituton KA = ½MA.V²

KB becomes

KB = 2 * KA

KB = 2KA