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Question

Physics Question on System of Particles & Rotational Motion

A solid uniform sphere resting on a rough horizontal plane is given a horizontal impulse directed through its center so that it starts sliding with an initial velocity v0v_0. When it finally starts rolling without slipping the speed of its center is

A

27v0\frac{2}{7}v_{0}

B

37v0\frac{3}{7}v_{0}

C

57v0\frac{5}{7}v_{0}

D

67v0\frac{6}{7}v_{0}

Answer

57v0\frac{5}{7}v_{0}

Explanation

Solution

Let, the final velocity be vv

So, angular momentum will remain conserved along point of contact
By conservation of angular momentum
Angular momentum will remain conserved along point of contact
Iω=I \omega= constant
mv0r=mvr+25mr2×ω(ω=vr)m v_{0} r =m v r+\frac{2}{5} m r^{2} \times \omega\,\,\, \left(\because \omega=\frac{v}{r}\right)
mv0r=mvr+25mr2(vr)m v_{0} r =m v r+\frac{2}{5} m r^{2}\left(\frac{v}{r}\right)
v0=v+25vv_{0} =v+\frac{2}{5} v
v0=75vv_{0} =\frac{7}{5} v
v=57v0\Rightarrow v=\frac{5}{7} v_{0}