# 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?

### 2 Answers

- nyphdinmdLv 72 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

- billrussell42Lv 72 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