Question

Derive law of linear momentum. Without using calculas

Answer 1

Answer:

Alternatively, it states that if net external force acting on a system is zero, the total momentum of the system remains constant.

Proof:

Let us consider a particle of mass ‘m’ and acceleration ‘a’. Then, from 2nd law of motion,

Law of Conservation of Linear Momentum

If no external force acts on the body then, F=0,

Law of Conservation of Linear Momentum

Therefore, ‘P’ is constant or conserved.

(Note: If the derivative of any quantity is zero, it must be a constant quantity.)

Deduction of Law of Conservation of linear momentum for two colliding bodies

Deduction of Law of Conservation of linear momentum for two colliding bodies

Let us consider two bodies of masses m1 and m2 moving in straight line in the same direction with initial velocities u1 and u2. They collide for a short time ∆t. After collision, they move with velocities v1 and v2.

From 2nd law of motion,

Force applied by A on B = Rate of change of momentum of B

FAB = (m2v2-m2u2)/∆t

Similarly,

Force applied by B on A = Rate of change of momentum of A

FBA = (m1v1-m1u1)/∆t

From Newton’s 3rd law of motion,

FAB = -FBA

Or, (m2v2-m2u2)/∆t = -(m1v1-m1u1)/∆t

Or, m2v2-m2u2 = -m1v1+m1u1

Or, m1u1 + m2u2 = m1v1 + m2v2

This means the total momentum before collision is equal to total momentum after collision. This proves the principle of co conservation of linear momentum.

Answer 2

Answer:

Derivation of Conservation of Momentum

Suppose, a truck of mass m1, velocity u1 and its momentum = m1u1 and a car of mass m2, velocity u2 and its momentum m2u2; are moving in the same direction but with different speeds.Therefore, total momentum=m1u1 + m2u2.

Now suppose the car and truck collide for a short time t, their velocities will change. So now the velocity of the truck and car become v1 and v2 respectively. However, their mass remains the same. Hence, now the total momentum = m1v1 + m2v2

Acceleration of car (a) = (v2–u2)/t

Also, F = ma

F1 = Force exerted by truck on the car

F1 = m2(v2–u2)/t .............eq 1

Acceleration of truck =(v1–u1)/t

F2 = m1(v1–u1)/t ...............eq 2

F1 exerted by the truck is action and F2 exerted by the car is reaction.

F1 = –F2

Now substituting the values of F1 and F2 from equation 1 and 2, we get:

m2(v2– u2)/t = –m1(v1– u1)/t

m2v2–m2u2 = –m1v1+m1u1

(or)

m1u1+m2u2 = m2v2+m1v1