Question

g Two balls of equal size are dropped from the same height from the roof of a building. The mass of ball A is twice that of ball B, m subscript A equals 2 m subscript B. When the two balls reach the ground, how do their kinetic energies compare? Group of answer choices when reaching the ground, LaTeX: KE_A=\sqrt{2}KE_B K E A = 2 K E B when reaching the ground, LaTeX: KE_A=\frac{1}{2}KE_B K E A = 1 2 K E B when reaching the ground, LaTeX: KE_A=2KE_B K E A = 2 K E B when reaching the ground, LaTeX: KE_A=KE_B K E A = K E B when reaching the ground, LaTeX: KE_A=4KE_B

Answer 1

Answer:

$K_A = 2K_B$

Explanation:

As we know that initial height of both the balls are same

also the mass of the two balls is given as

$m_A = 2m_B$

so here we can say by mechanical energy conservation

initial total mechanical energy = final total mechanical energy

since both balls are initially at rest so initial total kinetic energy of the balls will be zero

now final total kinetic energy = initial total potential energy

now we say

$K_A = m_A gh$

$K_B = m_B gh$

so we have

$\frac{K_A}{K_B} = \frac{m_A}{m_B}$

$K_A = 2K_B$