Please help form the law of conservation of energy states E

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form the law of conservation of energy states: Energy cannot he created or destroyed; it may be transformed from one form to another (mechanical, thermal, electrical, nuclear, radiant, etc.), but the total energy in any isolated system never changes. Laboratory Homework Assignment (2 points) Reference: Knight\'s textbook, Chapters 9, 10 & 11. lfan object of mass m1 = 0.55 kg is sliding without friction in the +x-direction on a level surface at a speed of v1-0.72 m/s and it collides with a stationary object of mass m1 = 0.55 kg, determine the total initial and final momenta (before and after the collision) as well as the total initial and final Mechanical Energy (before and after the collision) for; 1) a perfectly elastic collision, and 2) a perfectly inelastic collision. Energy losces during an energy transformation process an e quantified as the efficieney useful energy out × 100% e= total energy in 3) State what the useful energies in and out are and then calculate the efficiency for the perfectly elastic and the efficiency for the perfectly inelastic collisions described above.

Solution

According to the concept of the collision \'

Given that

mass m1=0.55 kg

mass m2=0.55 kg

speed u1=0.72 m/s

speed u2=0 m/s

now we find the velocity after collsion

perfect elastic collsion

speed of mass m1 after collision =>V1={(0.55-0.55)0.72/(0.55+0.55)}+0

=0

speed of mass m2 after collsion =>V2=2m1u1/m1+m2+(m2-m1)u2/(m1+m2)={2*0.55*0.72/(0.55+0.55)}+0

=0.72 m/s

perfect inelastic collsion

speed after collsion V=m1u1/(m1+m2)=0.55*0.72/(0.55+0.55)=0.36 m/s

now we find the initial and finial momentum both collsion

perfect elastic collision

the initial momentum =m1u1+m2u2=0.55*0.72+0.55*0=0.396 kg.m/s

the finial momentum =m1v1+m2v2=0.55*0+0.55*0.72=0.396 kg.m/s

the initial kinectic energy =>E1=1/2*0.55*0.72^2+1/20.55*0^2=0.143 J

the finial kineticx energy =>E2=1/2*0.55*0^2+1/2*0.55*0.72^2=0.143 J

perfect inelastic collsion

the initial momentum =m1u1+m2u2=0.55*0.72+0.55*0=0.396 kg.m/s

the finial momentum =(m1+m2)v=(0.55+0.55)*0.36=0.396 m/s

the initial kinetic energy E1=1/2*0.55*0.72^2+1/20.55*0^2=0.143 J

the finial kinetic energy E2=1/2(0.55+0.55)*0.36^2=0.0196 J

therefore the energy lost in the perfect elstic collsion =>E=E2-E1=0.143-0.143=0 J

therefore the energy lost in the perfect inelastic collsion =>E=E2-E1=-0.1234 J