Question

State and prove the law of conservation of linear momentum using Newton's third law.

Answer 1

The law of conservation of linear momentum easily follows from the third law of motion.

Consider a system of two particles, A and B. Let's say they interact with each other and 'A' experts a Force Fba on 'B' and in reaction, 'B' exerts a force Fab on 'A'.

Now, according to the second law of motion,

Fba= Mb x d/dt(Vb) = d/dt(Mb x Vb)

Fab= Ma x d/dt(Va) = d/dt(Ma x Va)

where Mb and Ma are masses of 'B' and 'A' respectively, and similarly Vb and Va their velocities in that order. d/dt denotes the derivative.

Now, adding the above two equations,

Fab + Fba = d/dt (Ma x Va) + d/dt(Mb x Vb)

Note that according to the third law of motion, these forces Fab and Fba are equal and opposite, so Fab= -Fba

Hence, 0= d/dt(Ma x Va) + d/dt(Mb x Vb) = d/dt(Pa + Pb)

where Pa and Pb are the linear momenta of 'A' and 'B' respectively.

Hence, Pa + Pb= constant (since the derivative is zero)

Therefore, The sum of the linear momenta of the bodies is constant.

The above calculations assume no external forces act upon the system. Hence, if there is zero net external force, the linear momentum of a system remains conserved, i.e. unchanged.

Answer 2

Conservation of linear momentum is a consequence of the third law.Imagine two bodies interacting with each other and forming an isolated system. The change in momentum of one body is the time integral of the force on it. By the third law, the forces are equal and opposite so the two time integrals and the changes in momenta have the same magnitude but opposite signs. That is the sum of the changes in momenta is zero.It is easy to generalize the above derivation to any number of particles or bodies. I.e., given the second law, Newton’s third law implies conservation of momentum.
Consider two bodies A and B of masses mA and mB, respectively ,moving in the same direction along a straight line. The bodies are moving with velocities uA and uB, respectively , such that
uA > uB.
They collide for time, t.After the collision let their velocities be vA and vB.
F (AB ) = rate of change of momentum of A
={mA (vA - uA)}
---------------------
t
F (BA)= rate of change of momentum of B
={mB (vB - uB )}
--------------------
t
Now, according to Newton’s third law, the forces are equal and opposite.
F (AB) = -F (BA)
{mA(vA-uA)} -{mB(vB-uB)}
-----------------= ------------------
t t mAvA-mAuA=mBuB-mBvB

mAuA+mBuB=mAvB
+mBvB
That is,
total momentum before collision = total momentum after collision.
Thus, total momentum of the body is conserved provided no external force acts on them. This proves the law of conservation of mass.

Hope, this may help you, buddy....
Mark as Brainliest, plz. . . .