Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
A body of mass m moving with velocity u collides with another ball of mass m2 at rest. the ratio m1/m2 for maximum energy transfer is: ans:1 in the worked out solution: ΔK.E/K.E has been worked out as 4n/(1+n)^2 where n=m1/m2 and ΔK.E is decrease in K.E. then, d/dn(4n/(1+n)^2) = 0 so n=1 can you please explain the differentiation part? Is there any other method?
A body of mass m moving with velocity u collides with another ball of mass m2 at rest. the ratio m1/m2 for maximum energy transfer is:
ans:1
in the worked out solution:
ΔK.E/K.E has been worked out as 4n/(1+n)^2
where n=m1/m2 and ΔK.E is decrease in K.E.
then, d/dn(4n/(1+n)^2) = 0
so n=1
can you please explain the differentiation part?
Is there any other method?
change of KE /KE = 4n/(n+1)2 or change in KE depend upon 4n/(n+1)2 for a particular value of initial KE... this means dKE is a function of n (where n is variable...) now we have to maximise this change for this we can use concept of maxima & minima ... dKE = f(n) = 4n/(n+1)2 differentiating this eq wrt n d/dn [f(n)] = d/dn [4n/(n+1)2 ] =4[-n2 + 1]/(n+1)4 (using quotient rule) now putting the above expression equal to 0 -n2+1 = 0 n=+1 or -1 now we have two values for n ,on substituting these values we will see that function is maximum at n=1 & minimum at n=-1 so n=1 is the value for which kinetic energy transferred is maximum...
change of KE /KE = 4n/(n+1)2 or
change in KE depend upon 4n/(n+1)2 for a particular value of initial KE...
this means dKE is a function of n (where n is variable...)
now we have to maximise this change for this we can use concept of maxima & minima ...
dKE = f(n) = 4n/(n+1)2
differentiating this eq wrt n
d/dn [f(n)] = d/dn [4n/(n+1)2 ]
=4[-n2 + 1]/(n+1)4 (using quotient rule)
now putting the above expression equal to 0
-n2+1 = 0
n=+1 or -1
now we have two values for n ,on substituting these values we will see that function
is maximum at n=1 & minimum at n=-1 so n=1 is the value for which kinetic energy transferred is maximum...
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -