3/164 Car B is initially stationary and is struck by car A moving with initial speed v1 = 30 km/h. The cars become entangled and move together with speed v ′ after the collision. If the time duration of the collision is 0.1 s, determine (a) the common final speed v ′, (b) the average acceleration of each car during the collision, and (c) the magnitude R of the average force exerted by each car on the other car during the impact. All brakes are released during the collision.

Incomplete question

Check attachment for the diagram of the question and the masses of the car are given in the diagram

Explanation:

Given that,

Car B is initially at rest

Ub = 0m/s

Car A is moving at

Ua = 30km/hr

Ua = 30×1000/3600 = 8.33m/s

This is an inelastic collision, after the collision the car move together at a speed of V

From the diagram

Mass of car A Ma = 1730kg

Mass of car B Mb = 935kg

Time taken during collison t = 0.1s

A. Common velocity V?

Applying conversation of momentum

Momentum before collision = momentum after collision

Momentum is given as p=mv

Now, momentum before collision

P(before) = Ma•Ua + Mb•Ub

P(before) = 1730 × 8.33 + Mb × 0

P(before) = 14,416.67

P(after) =(Ma+Mb)V

P(after) = (1730 +935)V

P(after) = 2665V

Then,

P(after) = P(before)

2665V = 14,416.67

V = 14,416.67/2665

V = 5.41m/s

To km/h

V = 5.41 ×1000./3600 = 19.47km/hr.

B. Average acceleration of each car

Car A

Acceleration is given as

a = ∆V/t

a = V-Ua/ t

a = (5.41 - 8.333) / 0.1

a = -2.924/0.1

a = -29.24m/s²

The negative sign show that car A is decelerating

Car B

a = (V - Ub) /t

a = (5.41 - 0) / 0.1

a = 5.41/0.1

a = 54.1 m/s²

This is showing that car B is accelerating, and it is reasonable because car B was initially at rest

C. Reaction of each car exerted on the other.

Using newton second second law of motion

Car A on B

F = ma

Ra = 1730 × 29.24

Ra = 50580N

Car B on A

F = ma

Rb = 935 × 54.1

Rb = 50583.5 N

Rb ≈ 50580N

As expected, the two reaction are suppose to be equal but due approximation along the way cause a slight change.

From newtons third law, for every action their is always equal and opposite reaction

So we expect reaction of car A to be equal to reaction of car B

multiply 2kg by latent heat of vap