Question

A bullet of mass ${10}^{-3}\text{kg}$ strikes an obstacle and moves at ${60}^{\circ }$ to its original direction. If its speed also changes from $20\text{m/s}$ to $10\text{m/s}$. Find the magnitude of impulse acting on the bullet.

• A. $\sqrt{5}\times {10}^{-2}\text{Ns}$
  
• B. $\sqrt{3}\times {10}^{-2}\text{Ns}$
  
• C. $\sqrt{2}\times {10}^{-2}\text{Ns}$
  
• D. $2\times {10}^{-2}\text{Ns}$

Answer 1

The correct option is B $\sqrt{3}\times {10}^{-2}\text{Ns}$


Given that,
Mass of the bullet $m={10}^{-3}\text{kg}$


To find impulse $J_1:$

$J_1={10}^{-3}[-10\cos {60}^{\circ }-(-20\left)\right]$

$\Rightarrow J_1=15\times {10}^{-3}\text{Ns}$

Similarly, to find impulse $J_2$, we have

$J_2={10}^{-3}[10\sin {60}^{\circ }-0]$

$\Rightarrow J_2=5\sqrt{3}\times {10}^{-3}\text{Ns}$

The magnitude of resultant impulse is given by
$J=\sqrt{J_1^2+J_2^2}$

$\Rightarrow J={10}^{-3}\sqrt{(15{)}^2+(5\sqrt{3}{)}^2}$

$\therefore J=\sqrt{3}\times {10}^{-2}\text{Ns}$

Hence, option (b) is correct answer.

Why this question: This question checks the basic application of Newton`s Impulse - momentum theorem $(J=\Delta P)$