Question

A $10\text{g}$ bullet moving at $100\text{m/s}$ strikes a log. Assume that bullet undergoes uniform deacceleration and stops in $6.0\text{cm}$, find the magnitude of average force experienced by the log.

• A. $600\text{N}$
• B. $1233.33\text{N}$
• C. $833.3\text{N}$
• D. $1266.66\text{N}$

Answer 1

The correct option is C $833.3\text{N}$

Given:
Mass of bullet; $m=10\times {10}^{-3}\text{kg}$
Initial velocity of bullet; $v_i=100\text{m/s}$
As bullet is coming to rest.
$\therefore$Final velocity of bullet; $v_f=0\text{m/s}$

Using Impulse $|J|=|\Delta P|$
$=|m(v_f-v_i)|=m{v}_i$
$=10\times {10}^{-3}\times 100$
$=1\text{N-s}$
We know, |Average force| $=\frac{\text{|Impulse|}}{\text{time}}$
Given that, bullet stops after travelling $S=6.0\text{cm}$ with initial velocity of $v_i=100\text{m/s}$
From : $v_f^2=v_i^2-2aS$
.$\therefore a=\frac{v_i^2}{2S}=\frac{(100{)}^2}{2\times 0.06}=8.3\times {10}^4{\text{m/s}}^2$

From : $v_f=v_i-at$
$t=\frac{v_i}{a}=\frac{100}{8.3\times {10}^4}=1.2\times {10}^{-3}\text{s}$

$\therefore |\text{Average force}|\text{experienced by bullet}=\frac{\text{|Impulse|}}{\text{time}}=\frac{1}{1.2\times {10}^{-3}}=833.3\text{N}$

By Newton's 3rd law of motion, we can say log will also experience average force of same magnitude but in opposite direction.
Average force experienced by log = $833.3\text{N}$