How to derive Newton's 3rd Law from the 2nd Law?

Question

Class XI Physics...

Answer 1

let the net momentum of a system of two bodies of mass m1 and m2 be p where p=p1+p2.from the second law net external force on the body =dp1/dt+dp2/dt. if the total momentum in any direction is constant then dp1/dt+dp2/dt=0(derivative of a constant is zero.)let the velocities of the bodies change from u1 to v1 and u2 to v2 in time t due to their mutual interaction. then m1(v1-u1)/t+m2(v2-u2)/t=0 (since dp/dt is time rate of change of momentum.) again we know (v-u)/t=acceleration. so m1f1+m2f2=0. again from the second law we get force =mass*acceleration. so f1+f2=0 and so f1= -f2. which shows thr force applied by the first body on the second is equal in magnitude but opposite in direction to the force applied by the second on the first. this is stated in the third law.thus it is derieved from the second law.

Answer 2

you know
v=ds/dt;
thev;
x=x0+vt+(at*at)(ie.,at2)
here x is dependent on v & x is a variable we can also write
x=f(v,t,a)
and v=ds/dt can be written as
v=ds/dv*dv/dt
if the variable depends on more than 1 variable this could be written as ds/dv*dv/dt(dv/dt is acceleration)
then
v=a*dv/dt;
cross multiply
u will get
vdu=ads..............................1
integerate eq 1
u will get
v2-u2/2=as
therefore
v2-u2=2as
thus proved



by
vijaya raghavan.vv in balaji sir class adyar(this derivation is for class xi physics not for ix)