(7.5) p, =Pr This is called Conservation of Momentum. Conservation of momentum s

Question

(7.5) p, =Pr This is called Conservation of Momentum. Conservation of momentum states that the total initial momentum equals the total final momentum. If the two particles move along one dimension (the X axis say), then the vector equation for conservation of momentum may be replaced with a scalar equation. Let particle 1 have a mass of m, and particle 2 have a mass of m,. Particle 1 has an initial velocity of V, and a final velocity of VF . Particle 2 has an initial velocity of U, and a final velocity of Uf (see Figure 7.3-1). For today's experiment, we will let these two particles collide by moving horizontally on a "frictionless" track. As a result, there are no net external forces acting on both particles. We may apply conservation of momentum to the collision: (7.6) Before Collision After Collision V. Figure 7.3-1: Elastic Collision. We will study momentum conservation for elastic collisions. The description elastic means that conservation of energy also holds. In the collision of these two masses, this means conservation of kinetic energy. Conservation of kinetic energy for our two-body head on elastic collision may be summed up as:

Explanation / Answer

we know that the in elastic collision  

conservation of momentum and kinetic energy takes place that is  

the total momentum before and after the collision is same and  

total kinetic energy before and after the collision is same .

here mass m1 is moving in +x direction with initial velocity v1 , and  

mass m2 is moving in -x direction with initial velocity U1  

we take the velocity in the +x direction +ve and -ve in the -ve x direction

so the momentum of the system before collision is  

m1*v1 - m2*U1

and after the collision

mass m1 is moving in -x direction with initial velocity VF , and  

mass m2 is moving in +x direction with initial velocity UF

  

the momentum of the system before collision is  

-m1*vF + m2*UF

by conservation of momentum

m1*v1 - m2*U1 = -m1*vF + m2*UF

m1(V1+VF) = m2(UF+U1)

same way the kinetic energy is  

before collision

0.5( m1*V1^2 -m2*U1^2)

and after the collision

0.5(m2*UF^2 - m1*VF^2)

equating the energies before and after the collision  

0.5( m1*V1^2 -m2*U1^2) = 0.5(m2*UF^2 - m1*VF^2)

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