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Question

A ball of mass $$0.2kg$$ strikes an obstacle and moves at $$60^{0}$$ to its original direction. If its speed also changes from $$20m/s$$ to $$10m/s$$, the magnitude of the impulse received by the ball is


A
27 Ns
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B
23 Ns
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C
25 Ns
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D
32 Ns
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Solution

The correct option is B $$2\sqrt{3} \ N s$$
$$\textbf{Step 1: Momentum calculation}$$   $$\textbf{{Ref. Fig.}}$$
Lets assume that initially ball is moving in positive $$x$$ direction.
$$\therefore$$ $$\text {Initial momentum}$$ $$\vec{P_i}=m\vec{u_1}$$    
              $$=0.2\times 20\hat{i}\,kg\,m/s$$
              $$=4\hat{i}\,kg\,m/s$$  

$$\text{Final momentum}$$ $$\vec{P_f}=m\vec{u_2}$$    
          $$\vec{u_2}=10\cos 60^0 \hat{i}+10\sin 60\hat{j}\,m/s$$
                $$=5\hat{i}+5\sqrt{3}\hat{j}\,m/s$$

      $$\therefore P_f=0.2(5\hat{i}+5\sqrt{3}\hat{j})\,kg\,m/s$$
                $$=(\hat{i}+\sqrt{3}\hat{j})\,kg\,m/s$$

$$\textbf{Step 2: Impulse calculation}$$ 
Impulse is given by
           $$\vec{J}=\Delta \vec{P}=\vec{P_f}-\vec{P_i}$$ 
               $$=(\hat{i}+\sqrt{3}\hat{j})-4\hat{i}\,Ns$$
           $$\vec{J}=-3\hat{j}+\sqrt{3}\hat{j}\,Ns$$

$$\large \text{Magnitude of impulse}$$ $$|\vec{J}|=\sqrt{(-3)^2+(\sqrt{3})^2}\,Ns$$
                                                      $$=\sqrt{12}$$        $$=2\sqrt{3}\,Ns$$

Hence magnitude of impulse received by the ball is $$2\sqrt{3}\,N.s.$$
Option $$B$$ is correct.

2111013_3453_ans_71151f900df04cb59508bd9be351a909.png

Physics

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