106the answer i have given is for elastic collision only. is it correct bhaiyya?
the amont of energy transferred to a stationary body of mass m2 by a collision witha moving body of mass m1 can be increased by interposing another stationarybody of mass m between m1 and m2 such that m1 strikes m and m strikes m2 For maximum transfer of energy what is mPlease solve it
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106are the collisions elastic? or COR = e (say) further if collisions are not elastic, are COR's for both pairs equal or e1 and e2?
if the collisions are elastic then m=√m1m2
62yes I think karan wants to ask over all colissions.. so it should be for e=1!
106106nishant bhaiyya i am getting m=√m1m2 for all cases where COR is equal for both collisions.
1yes your ans is absolutely correct
Please give the method to solve it
106consider a block of mass m1 moving with speed v which collides with stationary block of mass m. Let coefficient of restitution be e.
Applying conservation of momentum,
m1v = m1v1 + mv' ...... (i)
and COR equation
v'-v1=ev ....... (ii)
Solving (i) and (ii)
v'=m1v(1+e)/(m+m1) ...... (iii)
Now block of mass m collides with block of mass m2
Applying conservation of momentum again,
mv' = mv'' + m2v2 ...... (iv)
COR again,
v2 - v'' = ev' ....... (v)
Solving (iv) and (v),
v2 = mv'(1+e)/(m+m2)
= mm1v(1+e)2/[(m+m1)(m+m2)
now, for max. KE transfer,
KE of m2 should be max.
Take derivative of KE of B wrt m is the same as derivative of v2 wrt. x
So, the ans u will get after differentiation is m = √m1m2
Nishant bhaiyya, is there any shorter method?
62no dude.. i cant think of any..
This is partly shorter than mine.. :)