Question

A car of mass m1 traveling north at a speed of v1 collides with a car of mass m2

traveling east at a speed of v2. They lock together after the collision.
> Part A: Determine expression for the distance the cars will move until they stop if the coefficient of kinetic friction μk between the cars' tires and the road is about the same for both cars. (Express your answer in terms of the variables m1, m2, v1, v2, μk and appropriate constants.)

> Part B: Determine expression for the direction of the cars' movement after the collision. Zero angle direction is due east. (Express your answer in terms of the variables m1, m2, v1 , v2)

Answer 1

Explanation: Part A:

The initial momentum of the two-car system is:

p = m1v1 + m2v2

After the collision, the two cars will be stuck together and move as one. The final momentum of the system is:

p' = (m1 + m2) v'

where v' is the final velocity of the two-car system. Since momentum is conserved, we have:

p = p'

Therefore,

m1v1 + m2v2 = (m1 + m2) v'

The force of friction opposing the motion of the car is:

Ff = μk N

where N is the normal force, which is equal to the weight of the cars:

N = (m1 + m2) g

The force of friction causes the two-car system to decelerate until it comes to a stop. The deceleration is given by:

a = Ff / (m1 + m2)

Substituting the expression for Ff, we get:

a = μk g

The distance the cars will move until they stop is given by:

d = v'² / (2a)

Substituting the expression for v' and a, we get:

d = (m1v1 + m2v2)² / (2μk g (m1 + m2)²)

Therefore, the expression for the distance the cars will move until they stop is:

d = (m1v1 + m2v2)² / (2μk g (m1 + m2)²)

Part B:

Since the two cars lock together and move as one after the collision, the direction of their movement will be determined by the vector sum of their initial velocities. To find the direction of the cars' movement after the collision, we can use the law of conservation of momentum in the x and y directions separately.

In the x-direction, the initial momentum is:

p_x = m1v1 + m2v2

After the collision, the two cars will move in the x-direction only, so the final momentum in the x-direction is:

p'_x = (m1 + m2) v'_x

where v'_x is the final velocity in the x-direction.

Since momentum is conserved in the x-direction, we have:

p_x = p'_x

Substituting the expressions for p_x and p'_x, we get:

m1v1 + m2v2 = (m1 + m2) v'_x

Therefore,

v'_x = (m1v1 + m2v2) / (m1 + m2)

In the y-direction, the initial momentum is:

p_y = 0 + 0 = 0

After the collision, the two cars will move in the y-direction only, so the final momentum in the y-direction is:

p'_y = (m1 + m2) v'_y

where v'_y is the final velocity in the y-direction.

Since momentum is conserved in the y-direction, we have:

p_y = p'_y

Substituting the expressions for p_y and p'_y, we get:

0 = (m1 + m2) v'_y

Therefore,

v'_y = 0

This means that the two cars will move in the x-direction only, and the direction of their movement after the collision is given by:

θ = tan⁻¹(v2/v1)