Consider the collision of 2 identicalparticles with m 1 =m 2 = .1kg. The initial
ID: 1754269 • Letter: C
Question
Consider the collision of 2 identicalparticles with m1=m2= .1kg. The initialvelocity of particle 1 is v1 and particle 2 is initiallyat rest, v2= 0 m/s. After an elastic head-on collision ,the final velocity of particle 2 is givenby: 1. v2=(v1)/2 2. v2=0 3. v2=v1 4.v2=(5v1)/3 5. v2=v1/4 6. v2=(3v1)/4 7. v2=2v1 8. v2=v1/3 9. v2=(2v1)/3 10. v2=(4v1)/3 Consider the collision of 2 identicalparticles with m1=m2= .1kg. The initialvelocity of particle 1 is v1 and particle 2 is initiallyat rest, v2= 0 m/s. After an elastic head-on collision ,the final velocity of particle 2 is givenby: 1. v2=(v1)/2 2. v2=0 3. v2=v1 4.v2=(5v1)/3 5. v2=v1/4 6. v2=(3v1)/4 7. v2=2v1 8. v2=v1/3 9. v2=(2v1)/3 10. v2=(4v1)/3Explanation / Answer
Given that then masses are m1 = m2 = 0.1 kg initial velocity of the particle is v1 and v2 = 0 ------------------------------------------------------------------------------------------------------ Let the masses of the two bodies are m1 and m2the initial velocitiesare U1,U2 respectively
the finalvelocities are V1,V2 respectively. from the law of conservation of momentum m1*U1+m2*U2 = m1*V1+m2*V2 m1(U1-V1) = m2*(V2 - U2) ----------(1) from the law of conservation of Kineticenergy (1/2)m1*U12+(1/2)m2*U22 =(1/2)m1*V12+ (1/2)m2*V22 m1(U12-V12) =m2*(V22 - U22) ----------(2) devide the equation (1) by (2) weget U1+V1 = V2 +U2 V1 =V2 + U2 -U1 V2 = U1+V1 -U2 substitude the value of V2 value in equation (1) we get V1 = (m1-m2)*U1/(m1+m2) + (2m2)U2 /(m1+m2)-----------(3) substitude the valueof V1 value in equation (1) weget V2 = (m2-m1)U2 /(m1+m2) + (2m1)U1 /(m1+m2) -----------(4) ----------------------------------------------------------------------------------------------------------- From theequation (4) the final velocity of secone mass m2 is V2= 0 + (2m1)v1 /(m1+m2) =v1 m1(U12-V12) =m2*(V22 - U22) ----------(2) devide the equation (1) by (2) weget U1+V1 = V2 +U2 V1 =V2 + U2 -U1 V2 = U1+V1 -U2 substitude the value of V2 value in equation (1) we get V1 = (m1-m2)*U1/(m1+m2) + (2m2)U2 /(m1+m2)-----------(3) substitude the valueof V1 value in equation (1) weget V2 = (m2-m1)U2 /(m1+m2) + (2m1)U1 /(m1+m2) -----------(4) ----------------------------------------------------------------------------------------------------------- From theequation (4) the final velocity of secone mass m2 is V2= 0 + (2m1)v1 /(m1+m2) =v1 V1 =V2 + U2 -U1 V2 = U1+V1 -U2 substitude the value of V2 value in equation (1) we get V1 = (m1-m2)*U1/(m1+m2) + (2m2)U2 /(m1+m2)-----------(3) substitude the valueof V1 value in equation (1) weget V2 = (m2-m1)U2 /(m1+m2) + (2m1)U1 /(m1+m2) -----------(4) ----------------------------------------------------------------------------------------------------------- From theequation (4) the final velocity of secone mass m2 is V2= 0 + (2m1)v1 /(m1+m2) =v1 V2 = U1+V1 -U2 substitude the value of V2 value in equation (1) we get V1 = (m1-m2)*U1/(m1+m2) + (2m2)U2 /(m1+m2)-----------(3) substitude the valueof V1 value in equation (1) weget V2 = (m2-m1)U2 /(m1+m2) + (2m1)U1 /(m1+m2) -----------(4) ----------------------------------------------------------------------------------------------------------- From theequation (4) the final velocity of secone mass m2 is V2= 0 + (2m1)v1 /(m1+m2) =v1 substitude the value of V2 value in equation (1) we get V1 = (m1-m2)*U1/(m1+m2) + (2m2)U2 /(m1+m2)-----------(3) substitude the valueof V1 value in equation (1) weget V2 = (m2-m1)U2 /(m1+m2) + (2m1)U1 /(m1+m2) -----------(4) ----------------------------------------------------------------------------------------------------------- From theequation (4) the final velocity of secone mass m2 is V2= 0 + (2m1)v1 /(m1+m2) =v1