- #36
netgypsy
- 246
- 1
I found this one for first level physics class. The essential point being that before and after the interaction the velocities of the masses (two in this case) are constant so the derivative of the velocity times the constant mass is zero. From this the F one = -F two is then concluded.
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.
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.