State and explain law of conservation of momentum?
Answers
Explanation:
The conservation of momentum states that, within some problem domain, the amount of momentum remains constant; momentum is neither created nor destroyed, but only changed through the action of forces as described by Newton's laws of motion. ... Momentum is conserved in all three physical directions at the same time.
Answer:
Momentum Conservation Principle
Law of conservation of momentum states that
For two or more bodies in an isolated system acting upon each other, their total momentum remains constant unless an external force is applied. Therefore, momentum can neither be created nor destroyed.
Derivation of Conservation of Momentum
Newton’s third law states that for a force applied by an object A on object B, object B exerts back an equal force in magnitude, but opposite in direction. This idea was used by Newton to derive the law of conservation of momentum.
Consider two colliding particles A and B whose masses are m1 and m2 with initial and final velocities as u1 and v1 of A and u2 and v2 of B. The time of contact between two particles is given as t.
A=m1(v1−u1) (change in momentum of particle A)
B=m2(v2−u2) (change in momentum of particle B)
FBA=−FAB (from third law of motion)
FBA=m2∗a2=m2(v2−u2)t
FAB=m1∗a1=m1(v1−u1)t
m2(v2−u2)t=−m1(v1−u1)t
m1u1+m2u2=m1v1+m2v2
Therefore, above is the equation of law of conservation of momentum where m1u1+m2u2 is the representation of total momentum of particles A and B before the collision and m1v1+m2v2 is the representation of total momentum of particles A and B after the collision.
Related Articles:
Law of Conservation of Energy
Law Of Conservation Of Angular Momentum
Law of Conservation of Momentum Examples
Following are the examples of law of conversation of momentum:
Air-filled balloons
System of gun and bullet
Motion of rockets
Law of Conservation of Momentum Problems
Q1. There are cars with masses 4 kg and 10 kg respectively that are at rest. A car having the mass 10 kg moves towards the east with a velocity of 5 m.s-1. Find the velocity of the car with mass 4 kg with respect to ground.
Ans: Given,
m1 = 4 kg
m2 = 10 kg
v1 = ?
v2 = 5 m.s-1
We know from the law of conservation of momentum that,
Pinitial = 0, as the cars are at rest
Pfinal = p1 + p2
Pfinal = m1.v1 + m2.v2
= 4 kg.v1 + 10 kg.5 m.s-1
Pi = Pf
0=4 kg.v1+50 kg.m.s-1
v1 = 12.5 m.s-1
Q2. Find the velocity of bullet of mass 5 gram which is fired from a pistol of mass 1.5 kg. The recoil velocity of pistol is 1.5 m.s-1.
Ans: Given,
Mass of bullet, m1 = 5 gram = 0.005 kg
Mass of pistol, m2 = 1.5 kg
The velocity of a bullet, v1 = ?
Recoil velocity of pistol, v2 = 1.5 m.s-1
Using law of conservation of momentum,
m1u1 + m2u2 = m1v1 + m2v2
Here, Initial velocity of the bullet, u1 = 0
Initial recoil velocity of a pistol, u2 = 0
∴ (0.005 kg)(0) + (1.5 kg)(0) = (0.005 kg)(v1) + (1.5 kg)(1.5 m.s-1)
0 = (0.005 kg)(v1)+(2.25 kg.m.s-1)
v1=-450 m.s-1
Hence, the recoil velocity of pistol is 450 m.s-1.