Answer:
Explanation:
Given
Initial angular velocity [tex]\omega_0=\frac{3v_0}{R}[/tex]
Initial speed [tex]u=v_0[/tex]
Coefficient of kinetic friction [tex]\mu _k=0.08[/tex]
As No external torque is applied therefore angular momentum is conserved
[tex]L_i=L_f[/tex]
[tex]I_{cm}v_i+Mv_{cm}_iR=I_{cm}\omega _f+Mv_{cm}_f\cdot R[/tex]
For rolling without slipping [tex]v_f=\omega _fR[/tex]
[tex]\frac{2}{5}MR^2\cdot (\frac{3v_0}{R})+Mv_0\cdot R=\frac{2}{5}MR^2\cdot \frac{v_f}{R}+Mv_f\cdot R[/tex]
[tex]\frac{6}{5}Mv_0R+Mv_0R=\frac{2}{5}Mv_fR+Mv_f\cdot R[/tex]
[tex]v_f=\frac{11}{7}v_0[/tex]
acceleration provided by surface
[tex]a=\mu _kg[/tex]
[tex]a=0.08\times 9.8[/tex]
[tex]a=0.784\ m/s^2[/tex]
Speed of ball after [tex]t=2\ s[/tex]
using [tex]v=u+at[/tex]
v=final velocity
u=initial velocity
[tex]\frac{11}{7}v_0=v_0+0.784\times 2[/tex]
[tex]\frac{11}{7}v_0-v_0=0.784\times 2[/tex]
[tex]v_0=2.744\ m/s[/tex]
thus [tex]v_f=\frac{11}{7}\times 2.74[/tex]
[tex]v_f=4.312\ m/s[/tex]
