a rod Mis lying on a horizontal frictionless table.a small mass m travelling along the surface comes with speed v and hits the rod elasticallyly at one end perpendicular to the rod thereby coming to rest.we have to calculate the ratio m/M

 

first we conserve linear momentum mv=MV'

:v'=(m/M)v eq(i)

 

since collision is elastic,e=1 ;[v'+w(omega)l/2]/v=1 eq(ii)

 

for eq(iii)we need to conserve angular momentum.but about which point will we conserve itand why???

a rod Mis lying on a horizontal frictionless table.a small mass m travelling along the surface comes with speed v and hits the rod elasticallyly at one end perpendicular to the rod thereby coming to rest.we have to calculate the ratio m/M

 

first we conserve linear momentum mv=MV'

:v'=(m/M)v eq(i)

 

since collision is elastic,e=1 ;[v'+w(omega)l/2]/v=1 eq(ii)

 

for eq(iii)we need to conserve angular momentum.but about which point will we conserve itand why???

 

Angular momentum is conserved if there is no net external torque on the system, just like linear momentum is conserved if there is no net external force.

 

You can pick any point. So pick a convenient one, e.g. the axis of rotation the rod.

actually the teacher told us that we are supposed to conserve angular momentum about the point on the surface just below the centre of mass. any idea why?i am confused as to why that specific point

actually the teacher told us that we are supposed to conserve angular momentum about the point on the surface just below the centre of mass. any idea why?i am confused as to why that specific point

 

Maybe to ensure that you use the various formulas for moment of inertia, like the parallel axis theorem.

why not the centre of mass itself???

why not the centre of mass itself???

 

One reason a teacher may have you do an easy problem a harder (or less obvious) way is to give you practice at the methods you might have to use in a more complicated situation. Or maybe there's something about that choice that will work out in an interesting way.