Camron Higgins

2023-03-13

A rock of mass $m$ is dropped to the ground from a height $h$. A second rock with a mass of $2m$ is dropped from the same height. When the second rock strikes the ground, what is its kinetic energy?

twice that of the first rock

four times that of first rock

the same as that of the first rock

half that of the first rock

twice that of the first rock

four times that of first rock

the same as that of the first rock

half that of the first rock

Dillan Perez

Beginner2023-03-14Added 6 answers

$K.E.=\frac{1}{2}m{v}^{2}$

Kinetic energy of first rock will be $\left(K.E\right)FR=\frac{1}{2}m{v}^{2}$ …(1)

Kinetic energy of second rock will be $\left(K.E\right)SR=\frac{1}{2}\left(2m\right){v}^{2}$ …(2)

$\frac{\left(K.E\right)}{FR}=\frac{\frac{1}{2}m{v}^{2}}{\frac{1}{2}\left(2m\right){v}^{2}}$

$\Rightarrow $$\frac{\left(K.E\right)}{FR}=\frac{1}{2}$

$\Rightarrow $ $\left(K.E\right)SR=2\left(K.E.\right)FR$

Answer is twice that of the first rock.

Kinetic energy of first rock will be $\left(K.E\right)FR=\frac{1}{2}m{v}^{2}$ …(1)

Kinetic energy of second rock will be $\left(K.E\right)SR=\frac{1}{2}\left(2m\right){v}^{2}$ …(2)

$\frac{\left(K.E\right)}{FR}=\frac{\frac{1}{2}m{v}^{2}}{\frac{1}{2}\left(2m\right){v}^{2}}$

$\Rightarrow $$\frac{\left(K.E\right)}{FR}=\frac{1}{2}$

$\Rightarrow $ $\left(K.E\right)SR=2\left(K.E.\right)FR$

Answer is twice that of the first rock.