Physics - Center of Mass Question with Solution | TestHub

Let the initial velocity be $\vec{u}$ and the final velocity be $\vec{v}$.

The horizontal component of velocity remains unchanged because the floor is smooth (no friction).

$u_x=u\cos \theta =10\cos 60^{\circ }=10\times \frac{1}{2}=5m/s$

$v_x=u_x=5m/s$

The vertical component of velocity

$v_y=e{u}_y=0.6\times 5\sqrt{3}=3\sqrt{3}m/s$ (upwards)

Now we can find the final velocity $v$ and angle $\varphi$.

$v=\sqrt{v_x^2+v_y^2}=\sqrt{5^2+(3\sqrt{3}{)}^2}=\sqrt{25+27}=\sqrt{52}\approx 7.21m/s$

So, option C is approximately correct

The angle $\varphi$ is given by $\tan \varphi =\frac{v_y}{v_x}=\frac{3\sqrt{3}}{5}$

$\varphi ={\tan }^{-1}\left(\frac{3\sqrt{3}}{5}\right)={\tan }^{-1}(0.6\sqrt{3})$

So, option B is correct.

Impulse $\vec{J}=m(\vec{v}-\vec{u})$

The horizontal impulse is $J_x=m(v_x-u_x)=1(5-5)=0$.

The vertical impulse is $J_y=m(v_y-(-u_y))=1(3\sqrt{3}-(-5\sqrt{3}))=1(3\sqrt{3}+5\sqrt{3})=8\sqrt{3}\mathrm{N}\cdot \mathrm{s}$

The magnitude of the impulse is $J=\sqrt{J_x^2+J_y^2}=\sqrt{0^2+(8\sqrt{3}{)}^2}=8\sqrt{3}\approx 8\times 1.732=13.856\mathrm{kg}\cdot m/s$

So, option A is correct ($13.86\mathrm{kg}\cdot m/s$).

Initial kinetic energy $K{E}_i=\frac{1}{2}m{u}^2=\frac{1}{2}\times 1\times (10{)}^2=50\mathrm{J}$

Final kinetic energy $K{E}_f=\frac{1}{2}m{v}^2=\frac{1}{2}\times 1\times (7.21{)}^2=\frac{1}{2}\times 52=26\mathrm{J}$

Ratio of final to initial kinetic energy = $\frac{K{E}_f}{K{E}_i}=\frac{26}{50}=0.52$

So, option D is correct.