Question

A steel ball of mass $5g$ is thrown downward with a velocity of $10m{s}^{-1}$ from a height of $19.5m$ and it penetrates sand by $50cm$. The change in mechanical energy will be (take, $g=10m{s}^{-2}$ )

• A. $1J$
• B. $1.25J$
• C. $1.5J$
• D. $1.75J$

Answer 1

The correct option is B

$1.25J$

Step 1: Given data and assumptions.

Mass of steel ball, $m=5g=0.005kg$

Height, $h=19.5m$

Velocity, $v=10\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Penetrates under sand by distance, $x=50cm=0.5m$

Acceleration due to gravity, $g=10m{s}^{-2}$

Step 2: Finding the change in mechanical energy.

Since taking the potential energy to be zero at the final level where the particle reaches after penetration in the sand.

Then the change in mechanical energy,

$∆U=U_f-U_i=PE+KE$

Where, $U_i$ is the initial mechanical energy, $U_f$ is the final mechanical energy, $PE$ potential energy and $KE$ kinetic energy.

$∆U=mg\left(h+x\right)+\frac{1}{2}m{v}^2$

$\Rightarrow$$∆U=m\left[g\left(h+x\right)+\frac{1}{2}{v}^2\right]$

$\Rightarrow$$∆U=0.005\left[10\times \left(19.5+0.5\right)+\frac{1}{2}\times {10}^2\right]$

$\Rightarrow$$∆U=1.25J$

Hence, option B is correct.