Anonymous
Anonymous asked in Science & MathematicsPhysics · 2 months ago

Physics question help pls! ?

Car A collides elastically with the rear of car B. Car A has an initial speed of 5.00 m/s and a mass of 400 kg while car B has an initial speed of 2.50 m/s and a mass of 500 kg.

(a) Find the velocities of the two cars after the collision. (b) What is the change in the momentum of each car?

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  • 2 months ago

    vA0 = 5 m/s  vB0 = 2.5 m/s  mA = 400 kg   mB = 500 kg

    ELastic collision -->  momentum and kinetic energy are conserved

    Initial momentum = Pi = mA*vA0 + mB*vB0 = PAi + PBi

    Initial KE = Ki = PAi^2/(2mA) + PBi^2/(2mB)

    Final momentum = Pf = mA*vAf + mB*vBf

    Final KE = KEf = PAf^2/(2mA) + PBf^2/(2mB)   PXf =  mX vX  X = A or B

    Let solve for PAf.  Use

    PI = Pf --->  PAi + PBi = PAf + PBf  --->> PAf = PAi + PBi - PBf

    Now use conservation of energy to find PBf

    PAi^2/(2mA) + PBi^2/(2mB) = PAf^2/(2mA) + PBf^2/(2mB)

    Here's where it gets messy.  Plug in for PAf and expand

    PAf^2/(2mA) = (PAi + PBi - PBf)^2 = PAi^2 +PBi^2 +PBf^2 - 2*(PAi*PBi -PBf*(PAi+PBi))

    So

    PAi^2/(2mA) + PBi^2/(2mB) =

             {PAi^2 +PBi^2 +PBf^2 - 2*(PAi*PBi -PBf*(PAi+PBi))}/(2mA) +PBf^2/(2mB)

    PBi^2/(2mB) - PBi^2/(2mA) +2*PAi*PBi/(2mA) = 2PBf*(PAi + PBi)/(2mA) + PBf^2/(2mB)

    Multiply through by 2mB

    PBf^2 + (2mB/mA)*PBf*(PAi + PBi) = PBi^2(1 -(mB/mA))+ (2mB/mA)*PAi*PBi

    THe right  hand side is just a number, call it C for now so we have

    PBf^2 + 2*D*PBf - C = 0  where B = (mB/mA)*(PAi + PB1)

    whe can use the quadratic formula to find PBf

    PBf = -D +/- sqrt(D^2 - C)

    THere will be two roots, one is likely to be negative which can be ignored since B cannot go backwards when hit from the rear.

    Using the positive root  PBf = mB*vBf  -->   vBf = PBf/mB

    Then since  Pi = mA vAf + PBf  -->  vAf = (Pi - PBf)/mA

    You can plug in the numbers

    for part b

    deltaPA = PAi - PAf

    deltaPB = PBi = PBf

  • 2 months ago

    Elastic collisions

    v is velocity after the collision, u before

    v₁ = (u₁(m₁–m₂) + 2m₂u₂) / (m₁ + m₂)

    v₂ = (u₂(m₂–m₁) + 2m₁u₁) / (m₁ + m₂)

    directions are unknown. Assume both are moving in the same direction, so both u's are positive. see below for alternative

    va = (5(400–500) + 2•500•2.5) / (400 + 500)vb = (2.5(500–400) + 2•400•5) / (400 + 500)

    va = (5(–100) + 2500) / (900)vb = (2.5(100) + 4000) / (900)

    va = (–500 + 2500) / (900)vb = (250 + 4000) / (900)

    va = (2000) / (900) = 2.22 m/svb = (4250) / (900) = 4.72 m/s

    both in the original direction.

    momentum is above multiplied by mass

    Pa = 2.22•400 = 888 kgm/s

    Pb = 4.72•500 = 2360 kgm/s

    total = 3248 kgm/s

    initial momentum is

    Pai = 5•400 = 2000

    Pbi = 2.5•500 = 1250

    total = 3250 kgm/s

    should be equal, diff is rounding errors.

    If they are in a head-on collision, this becomes

      (pick the first car's direction as +)

    va = (5(400–500) + 2•500(–2.5)) / (400 + 500)vb = (–2.5(500–400) + 2•400•5) / (400 + 500)

    va = (–500 – 2500) / (400 + 500)vb = (–250 + 4000) / (400 + 500)

    va = (– 3000) / (900) = –3.33 m/svb = (3750) / (900) = +4.17 m/s

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