1answer is v/3
HINT:1.neglect gravitational impulse
2.impulsive tension =mΔv
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
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.
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15 Answers
1
3i 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
3if 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
1msp , 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 ???
1the 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.
1i 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...........
1Let 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 – mV ....(i)For pan J1 – J2 = mV .....(ii)For block J2 = mV .....(iii)Solving, these three equation, we get V = 3vAns.