A uniform solid sphere of mass m , radius R moving with velocity v0 is rolling without slipping on a frictionless surface. It collides elastically with a frictionless vertical wall. Ratio of magnitude of angular momentum of the sphere before and after the collision about its bottommost point is

Question

A uniform solid sphere of mass m , radius R moving with velocity v0 is rolling without slipping on a frictionless surface. It collides elastically with a frictionless vertical wall. Ratio of magnitude of angular momentum of the sphere before and after the collision about its bottommost point is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-gpglhiwwxkt5.png" /> (A) (3/5) (B) (5/3) (C) (3/7) (D) (7/3)

Tardigrade Admin · Accepted Answer

Li n=mv0R+(2/5)mR2w Li n=(7/5)mv0R Lf i n a l=mv0R-(2/5)mR2ω =(3/5)mvR Ratio=(7/3)

$\frac{3}{5}$

$\frac{5}{3}$

$\frac{3}{7}$

$\frac{7}{3}$

$L_{in} = m v_{0} R + \frac{2}{5} m R^{2} w$
$L_{in} = \frac{7}{5} m v_{0} R$
$L_{f ina l} = m v_{0} R - \frac{2}{5} m R^{2} ω$
$= \frac{3}{5} m v R$
$R a t i o = \frac{7}{3}$

53​

35​

73​

37​

Lin​=mv0​R+52​mR2w Lin​=57​mv0​R Lfinal​=mv0​R−52​mR2ω =53​mvR Ratio=37​

$\frac{3}{5}$

$\frac{5}{3}$

$\frac{3}{7}$

$\frac{7}{3}$

$L_{in} = m v_{0} R + \frac{2}{5} m R^{2} w$
$L_{in} = \frac{7}{5} m v_{0} R$
$L_{f ina l} = m v_{0} R - \frac{2}{5} m R^{2} ω$
$= \frac{3}{5} m v R$
$R a t i o = \frac{7}{3}$