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Conservation of Momentum

Aug 29, 2022
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Key Concepts:

  • Conservation of Momentum

Introduction:

In our daily life we make many observations such as a fast bowler taking a run up before bowling, a tennis player moving her racket backwards before hitting the tennis ball and a batsman moving his bat backwards before hitting the cricket ball. All these activities are performed in order to make the ball move with a great speed when hit/thrown. The balls would rather move slowly if these activities are not done. But did you know in all of these the momentum is conserved. In this session we are going to discuss about law of conservation of momentum in detail. 

Explanation: 

Law of Conservation of Momentum 

The sum of momenta of the two objects before collision is equal to the sum of momenta after the collision provided there is no external unbalanced force acting on them.  

This is known as the law of conservation of momentum. 

Conservation of momentum

Derivation of law 

collision of objects

 

Consider an object A has mass ‘ mA,’ and object B has mass ‘ mB.’ 

parallel

The initial velocity of an object A is vB 

and initial velocity of an object B is vB 

We know the formula of momentum, 

Initial momentum (A) = mA vA 

& Initial momentum (B) =  mB vB 

parallel

Both collide such that,  

The final velocity of an object A is vB  

and final velocity of an object B is vB 

Final momentum after the collision:  

Final momentum of an object A is mA vA  

Final momentum of an object B is mB vB 

The force on A due to B is applied and thus an equal and opposite force will act on B due to A according to Newton’s third law of motion. Thus,  

FAB = -FBA 

Since F = mass x acceleration  

Thus, mBaB = -mAvA 

Since a = (v – u )/ t, so 

mb ( vB – uB ) = -mA (vA – uA

mBvB – mB uB = -mA vA + mA uA 

mBvB + mA vA = mA uA + mB uB 

Such that,  

Magnitude of the final momenta = magnitude of the initial momenta  

Applications of conservation of momenta: 

A few of the applications of conservation of momenta are: 

  1. When a gun is fired the gun recoils back so that the sum of the initial momenta is equal to the sum of the final momenta. 
  2. When a rocket is launched the hot gases are released such that the final momenta is equal to the initial momenta. 
  3. In an air-filled balloon, the particles inside the balloon collide with each other but the net momentum of the balloon is conserved. 

Questions and Answers 

Question 1: There are two balls of masses 3 kg and 4 kg respectively at rest. The 2nd ball of mass 4 kg moves with a velocity 5 m/s. Find the velocity of the ball of mass 3 kg with respect to ground. 

Answer: 

Mass of ball , m1 = 3 kg 

m2 = 4 kg 

Initial speed of the ball, u1 = om/s 

u2 = om/s 

Final speed of the ball, v1 = ? 

v2 = 5m/s  

According to law of conversation of momentum. 

Initial momentum = 3 (0) + 4 (0) = 0 kgm/s 

Final momentum = 3 (v) + 4 (5) = (3v + 20) kgm/s 

Equating both, 

3v + 20/3 = 0 

v = -20 = -6.66m/s 

Velocity of ball after collision is 6.66 m/s in the opposite direction (because of the negative sign). 

Question 2: There are two cars, A and B. Car A with a mass of 1500 kg, traveling at 30 m/s collides with car B of mass 3000 kg traveling at 20 m/s in same direction. After collision, the velocity of car A becomes 40 m/s. Calculate the velocity of car B after the collision? 

Answer: 

Mass of car A  = 1500 

Mass of car B = 3000 kg 

Before collision, 

Initial velocity , uA = 30m/s 

uB = 20m/s 

Initial momentum, = ma ua + mb ub    

= 1500 (30) + 3000 (20) = 105000 kgm/s 

After collision, 

Final velocity , uA = 40m/s 

uB = v 

Final momentum, = ma ua + mb ub    

= 1500 (40) + 3000 (v)  

= 60,000 + 30,000 v 

According to law of conversation of momentum. 

Initial momentum = Final momentum  

105000 = 60,000 + 30,000 v 

= 30,000 v = 45,000 

v =  45,000/30,000 = 1.5 m/s 

Velocity of car B after collision is 1.5 m/s. 

Summary:

  • Law of conservation of momentum- “The sum of momenta of the two objects before
    collision is equal to the sum of momenta after the collision provided there is no external
    unbalanced force acting on them.”
  • m1 u1 + m2 u2 = m1 v1 + m2 v2
  • Application:
  1. Firing of a gun
  2. When rocket is launched
  3. Motion of particle in an air-filled balloon

Comments:

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