Question

A bullet of mass $20\,\text{g}$ moving with a velocity of $200\,\text{m/s}$ strikes a target and is brought to rest in $\left(\dfrac{1}{50}\right)^{\text{th}}$ of a second. The impulse and average force of impact are respectively

• A. $2\,\text{Ns},\,100\,\text{N}$
• B. $4\,\text{Ns},\,100\,\text{N}$
• C. $2\,\text{Ns},\,200\,\text{N}$
• D. $4\,\text{Ns},\,200\,\text{N}$

Hint:

Impulse equals change in momentum and determines average force over time.

Answer 1

The Correct Option is D

Step 1: Convert mass into SI units.
\[ m = 20\,\text{g} = 0.02\,\text{kg} \]

Step 2: Calculate impulse.
Impulse $= \Delta p = m(v - u)$
\[ = 0.02(0 - 200) = -4\,\text{Ns} \] Magnitude of impulse $= 4\,\text{Ns}$.

Step 3: Calculate average force.
\[ F_{\text{avg}} = \frac{\text{Impulse}}{\text{time}} = \frac{4}{1/50} = 200\,\text{N} \]

Step 4: Conclusion.
Impulse $= 4\,\text{Ns}$ and average force $= 200\,\text{N}$.