Question

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?

• A. twice that of the first rock
• B. four times that of first rock
• C. the same as that of the first rock
• D. half that of the first rock

Answer 1

The correct option is A

twice that of the first rock

Step 1. Given Data:

Mass of first rock $=m$

Mass of second rock $=2m$

Height of dropping the rock $=h$

Step 2. Find the kinetic energy of the second rock when it strikes the ground:

Kinetic energy will be maximum when rock reaches the ground from a certain height.

A heavy and light body when released from the same height $h$ reaches the ground simultaneously and with the same velocity $v$.

we know,

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

Step 3: Calculation

We can see from the formula that Kinetic energy is directly proportional to mass.

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)

divide equation (1) by (2), we get

$\frac{{\left(K.E\right)}_{FR}}{{\left(K.E\right)}_{SR}}=\frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}m{v}^2}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(2m\right)v^2}$

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

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

So, the K.E. of the second rock will be twice that of the first rock.

Hence, Option A is correct.