Question

A ball of mass \(300\;g\) is dropped from a height \(10\;m\) above a sandy ground. On reaching the ground, it penetrates through a distance \(1.5\;m\) in sand and finally stops. The average resistance offered by the sand to oppose the motion is
Take \(g=10\;\text{m s}^{-2}\).

• A. \(35\;N\)
• B. \(23\;N\)
• C. \(34\;N\)
• D. \(28\;N\)

Hint:

When an object penetrates a surface and stops, use the work-energy theorem. Include both the work done by gravity and the resisting force during penetration.

Answer 1

The Correct Option is B

Step 1: Gather the data in SI units.
Mass $m = 300\ g = 0.3\ kg$, drop height $h = 10\ m$, penetration depth $s = 1.5\ m$, and $g = 10\ \text{m s}^{-2}$. We want the average sand resistance $R$.
Step 2: Picture the whole journey with energy.
The ball starts at rest $10\ m$ above the sand and finally stops $1.5\ m$ deep inside it. The total drop is $h + s = 11.5\ m$. Over the entire trip the kinetic energy starts and ends at zero.
Step 3: Apply the work-energy theorem across the full path.
Gravity does positive work over the whole $11.5\ m$, while the sand resistance acts only over the $1.5\ m$ inside. Since net kinetic energy change is zero, \[ mg(h+s) - R\,s = 0 \]
Step 4: Plug in the gravity work.
\[ mg(h+s) = 0.3 \times 10 \times 11.5 = 34.5\ J \]
Step 5: Solve for the resistance.
\[ R = \frac{mg(h+s)}{s} = \frac{34.5}{1.5} = 23\ N \]
Step 6: State the conclusion.
The average resistance offered by the sand is about $23\ N$, which matches option (2). \[ \boxed{23\ N} \]