Car A rear ends Car B, which has twice the mass of A, on an icyroad at a speed low enough so that the collision is essentiallyelastic. Car B is stopped at a light when it is struck. Car A hasmass m and speed v before the collision. Afterthe collision

A. each car has halfthe momentum.

B. car A stops andcar B has momentum mv.

C. car A stops andcar B has momentum 2mv.

D. the momentum ofcar B is four times as great in magnitude as that of carA.

E. each car has halfof the kinetic energy.

D. The momentum of car B is four times as great in magnitude as that of car A.

Explanation:

Hi there!

Since the collision is elastic the momentum and the kinetic energy of the system are conserved, i. e, they remain constant (initial momentum = final momentum and initial kinetic energy = final kinetic energy). The momentum of the system is given by the sum of the momenta (p) of each car:

pA + pB = pA' + pB'

Where:

pA = momentum of car A before the collision.

pB = momentum of car B before the collision.

pA' = momentum of the car A after the collision.

pB' = momentum of the car B after the collision.

The momentum is calculated as follows:

p = m · v

Where:

m = mass of the car.

v = velocity.

Then, the momentum of the system before the collision will be:

initial momentum of the system = m · v + mB · vB

Where:

m = mass of car A

v = velocity of car A

mB = mass of car B

vB = velocity of car B

We know that initially, the velocity of car B is zero and mB = 2m. Then, the initial momentum will be:

initial momentum = m · v

Let's express the final momentum of the system:

final momentum of the system = m · vA' + 2m · vB'

Where vA' and vB' are the final velocities of car A and B respectively.

Since initial momentum of the system = final momentum of the system, then:

m · v = m · vA' + 2m · vB'

dividing both sides of the equation by m:

v = vA' + 2 vB'

Solving for vA':

v - 2 vB' = vA'

Now, using the coservation of kinetic energy (KE):

intial kinetic energy = final kinetic energy

1/2 · m · (v) ² + 1/2 · 2 · m · (vB) ² = 1/2 · m · (vA') ² + 1/2 · 2 · m · (vB') ²

vB = 0, then:

1/2 · m · (v) ² = 1/2 · m · (vA') ² + 1/2 · 2 · m · (vB') ²

dividing by 1/2 m both sides of the equation:

(v) ² = (vA') ² + 2 · (vB') ²

replacing vA' = v - 2 vB'

(v) ² = (v - 2 vB') ² + 2 (vB') ²

(v) ² = (v) ² - 4 vB' v + 4 (vB') ² + 2 (vB') ²

4 vB' v = 6 (vB') ²

divide both sides by vB'

4 v = 6 vB'

2/3 v = vB'

Now, we can calculate the final velocity of car A:

vA' = v - 2 vB'

vA' = v - 2 · 2/3 v

vA' = v - 4/3 v

vA' = - 1/3 v

Now let's calculate the momentum of each car after the collision:

Car A:

pA = m · vA'

pA = - 1/3 mv

Car B:

pB = mB · vB'

pB = 2 m · 2/3 v

pB = 4/3 mv

Then:

4|pA| = |pB|

The momentum of car B is four times as great in magnitude as that of car A.