Question

What is the magnitude of angular velocity of the stick plus puck after the collision?

• A. $\frac{6v_i}{5l}$
• B. $\frac{5v_i}{6l}$
• C. $\frac{v_i}{l}$
• D. $\frac{v_i}{\sqrt{2}l}$

Answer 1

The correct option is A $\frac{6v_i}{5l}$

The new center of mass of the stick-puck system w.r.t. center of stick is

$r_{cm}=\frac{m_s\left(l/2\right)}{m_s+m_p}=\frac{m\left(d/2\right)}{m+m}=l/4$ (given- $m_s=m_p=m$)

Before collision , the total angular momentum of the stick-puck system ,

$L_1=rp+0$ (as stick is stationary, hence its angular momentum is zero)

or $L_1=\left(l/4\right)\left(m{v}_i\right)$ ..............eq1

After collision, angular momentum of stick-puck system ,

$L_2=I_{cm}^{\prime }\omega$ ...............eq2

by law of conservation of angular momentum ,

$L_2=L_1$

or $I_{cm}^{\prime }\omega =\left(l/4\right)\left(m{v}_i\right)$

or $\omega =\frac{\left(l/4\right)\left(m{v}_i\right)}{I_{cm}^{\prime }}$ ........................eq3

where $I_{cm}^{\prime }=$ moment of inertia of stick-puck system about new center of mass,

$I_{cm}^{\prime }=I_s+I_p$ ,

now , $I_s=I_{cm}+m(d/4{)}^2$ (by parallel axis theorem)

and $I_p=m(l/4{)}^2$

hence $I_{cm}^{\prime }=I_{cm}+m(l/4{)}^2+m(l/4{)}^2=I_{cm}+m\left(l^2/8\right)$

but $I_{cm}=m\left(l^2/12\right)$

therefore $I_{cm}^{\prime }=m\left(l^2/12\right)+m\left(l^2/8\right)=m\left(5l^2/24\right)$ ................eq4

putting the value of $I_{cm}^{\prime }$ in eq3 , we get

$\omega =\frac{l/4\left(m{v}_i\right)}{m\left(5l^2/24\right)}=\frac{6v_i}{5l}$