Question

A stick of length L and mass M lies on a frictionless horizontal table on which it is free to move in any way A small particle of mass m moving on the table and perpendicular to the rod collides elastically with the stick with speed v.What must be the mass m in term of M.L and d of the particle so that it is at rest immediately after the collision.

• A. $\frac{M{L}^2}{L^2+12d^2}$
• B. $\frac{2M{L}^2}{L^2+12d^2}$
• C. $\frac{3M{L}^2}{L^2+12d^2}$
• D. $\frac{4M{L}^2}{L^2+12d^2}$

Answer 1

The correct option is A $\frac{M{L}^2}{L^2+12d^2}$

$(M,L)$
by COLM system $(m+m)$
$MV=M{V}_{cm}+m\left(0\right)$........(i)
$MV=M{V}_{cm}$...........(ii)
By COAM about a point On ground
$Li=MV.d$
$L_f=Iw+M(\overline{Rcm}\times \overline{Vcm})$
$=Iw+0$
$L_f=\frac{M{L}^2}{12}w$
$Mvd=\frac{M{L}^2}{12}w$......(2)
Velocity of Pt P of pad
$=U_p=V_{c}m+dw$
For elastic collision
Vel of seg=vel pf grouch
$V_p-0=U$
$=v_{cm}+dw=u$
$\frac{mv}{m}+d\left(\frac{12mud}{M{L}^2}\right)=V$
$m=M\left(\frac{L^2}{L^2+12d^2}\right)$