Question

A bullet of mass 20 g moving with $500 \text{ ms}^{-1}$ is pierd 1 cm into a wooden block, then the retarding for experiend by the bullet is

• A. $125 \times 10^3 \text{ N}$
• B. $750 \times 10^3 \text{ N}$
• C. $500 \times 10^3 \text{ N}$
• D. $250 \times 10^3 \text{ N}$

Hint:

Tip: Use energy conservation when force is constant but time is unknown. Work done equals change in kinetic energy.

Answer 1

The Correct Option is D

The bullet has mass $m = 20 \text{ g} = 0.02 \text{ kg}$ and initial velocity $u = 500 \text{ m/s}$. It is stopped after traveling a distance $s = 1 \text{ cm} = 0.01 \text{ m}$. Since final velocity $v = 0$, we can use work-energy theorem or kinematics. The work done by the retarding force brings the bullet to rest, and is equal to its loss in kinetic energy. Using the work-energy theorem: \[ F \cdot s = \frac{1}{2} m u^2 \Rightarrow F = \frac{mu^2}{2s} \] \[ F = \frac{0.02 \cdot (500)^2}{2 \cdot 0.01} = \frac{0.02 \cdot 250000}{0.02} = 250000 \text{ N} \] So, the retarding force is $250 \times 10^3 \text{ N}$.

Answer 2

Step 1: Identify given data
Mass of bullet, \(m = 20 \text{ g} = 0.02 \text{ kg}\)
Initial velocity, \(u = 500 \text{ ms}^{-1}\)
Distance penetrated, \(s = 1 \text{ cm} = 0.01 \text{ m}\)
Final velocity, \(v = 0 \text{ ms}^{-1}\) (since bullet comes to rest)

Step 2: Use the equation of motion to find acceleration \(a\)
Using \(v^2 = u^2 + 2as\),
\[ 0 = (500)^2 + 2 \times a \times 0.01 \]
\[ a = -\frac{(500)^2}{2 \times 0.01} = -\frac{250000}{0.02} = -12,500,000 \text{ ms}^{-2} \]

Step 3: Calculate retarding force \(F\)
Using Newton's second law,
\[ F = m \times |a| = 0.02 \times 12,500,000 = 250,000 \text{ N} \]

Step 4: Conclusion
The retarding force experienced by the bullet is \(250 \times 10^3 \text{ N}\).