# another doubt............

A block of mass m and a pan of equal mass are connected by a string going over a smooth light pulley. Initially the system is at rest when a particle of mass m falls on the pan and sticks to it. If the particle strikes the pan with a speed v find the speed with which the system moves just after the collision.

33
Abhishek Priyam ·

1
Great Dreams ·

HINT:1.neglect gravitational impulse
2.impulsive tension =mÎ”v
3.constraint on length of string leads to the fact velocities
at the ends are equal
4.if at all really tired ref.=HCverma pg.no 155 ex.24

1
sumit_kumar ·

hey gr8dreams , plzzzzzz give the equations

29
govind ·

u may refer to HC Verma.....it is given as an example...

3
msp ·

i dunno whether this method is rite or rong.

since we have to find the velocity just after the inelastic collision we can assume that the collision is between a pan of mass 2m and another mass m.

conserving momentum
we have mv=(2m+m)V' implies V'=v/3

3
msp ·

if anyone finds that the above post is rong,pls put your reply soon,as i myself not fully convinced,ur reply will help me,dont hesitate to correct me

1
lalit ·

msp ur answer and explanation both are correct.

1
sumit_kumar ·

msp , even i thought of using that eqn............... but i m confused about 1 thing........... the momentum due to the block and the pan-mass system after the collision are opp. in direction............ so can u add them ???

1
lalit ·

the pan, particle and block forms one system so overall momentum is in

same direction However individual momentum of block and pan are in

opposite direction.

1
alok ·

i am not convinced with msp...
it may be that its v/3 just by coincidence. because the same logic does not apply to many other impulse based questions...........

1
lalit ·

concept is same but the logic depends upon oreintation of system

1
Great Dreams ·

1
binod ·

hello hello hello hello hello

1
binod ·

Let the required speed is V.Further, let J1 = impulse between particle and panand J2 = impulse imparted to the block and the pan by the stringUsing, impulse = change in momentumFor particle J1 = mv â€“ mV ....(i)For pan J1 â€“ J2 = mV .....(ii)For block J2 = mV .....(iii)Solving, these three equation, we get V = 3vAns.

1
bubai bubai ·

Ans.V/2?