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
B
23 Ns
C
25 Ns
D
32 Ns

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.Physics

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