Question

Find the angular momentum of the particle and of the rod about the centre of mass $C$ before the collision.

• A. $\frac{M^{2}mvl}{4{(m+M)}^2},\frac{m^{2}Mvl}{4{(m+M)}^2}$
• B. $\frac{M^{2}mvl}{2{(m+M)}^2},\frac{m^{2}Mvl}{2{(m+M)}^2}$
• C. $\frac{M^{2}mvl}{{(m+M)}^2},\frac{m^{2}Mvl}{{(m+M)}^2}$
• D. $\frac{{2M}^{2}mvl}{{(m+M)}^2},\frac{{2m}^{2}Mvl}{{(m+M)}^2}$

Answer 1

Answer: (B)

$\frac{M^{2}mvl}{2{(m+M)}^2},\frac{m^{2}Mvl}{2{(m+M)}^2}$

Center of mass of the rod only lies at point O and that of whole system after collision lies at C.

$x=\frac{M\left(0\right)+M\frac{L}{2}}{M+m}$

$⟹x=(\frac{m}{M+m})\frac{L}{2}$ and $BC=(\frac{M}{M+m})\frac{L}{2}$

Also, velocity of particle w.r.t C before collision $u=\frac{Mv}{M+m}$ (as solved earlier)

Thus angular momentum of particle about C before collision, $L_p=mu\left(BC\right)=m\times \frac{Mv}{M+m}\times (\frac{M}{M+m})\frac{L}{2}$

$L_p=\frac{M^{2}mvL}{2(M+m{)}^2}$

Also, velocity of rod w.r.t C before collision $u_r=\frac{mv}{M+m}$ (as solved earlier)

Thus angular momentum of particle about C before collision, $L_p=M{u}_r\left(x\right)=M\times \frac{mv}{M+m}\times (\frac{m}{M+m})\frac{L}{2}$

$L_p=\frac{M{m}^{2}vL}{2(M+m{)}^2}$