contestada

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.

Respuesta :

Kazeemsodikisola

Answer:

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

gbenewaeternity

Answer:

(a) V₂ = 5.4m/s

(b) Acceleration of car A  = -29.3m/s²

       Acceleration of car B = 54m/s²    

(c) Magnitude R =  = 50689N

Explanation:

Given data;

Weight of car A = 1730kg

Weight of car B = 935kg

Initial speed V1 = 30km/hr = 30 *1000/3600 = 8.33m/s

time duration = 0.1s

(a) Common final speed:

The common final speed can be calculated by using the principle  linear  momentum which state that;

mₐv₁ + mbv₁ = mₐv₂ + mbv₂ --------------------1

Where mₐ is the mass of car A, mb is the mass of car B, v₁ is the speed of car A and B respectively while v₂ is the common speed.

For common final speed, the equation can be written as;

mₐv₁ + mbv₁ = V₂(mₐ +mb)

Substituting into the formula, we have

(1730 * 8.33) + (935 *0) = V₂(1730 + 935)

14.41 + 0 = 2665V₂

V₂ = 14410.9/2665

      5.4m/s

(b)Average  Acceleration of Car A:

Average acceleration = change in velocity/change in time

                                    = v₂ -v₁/ Δt

                                         5.4 -8.33/0.1

                                       = -29.3m/s²

(b)Average  Acceleration of Car B:

Average acceleration = change in velocity/change in time

                                      =   V₂ - 0/Δt

                                       =5.4/0.1

                                        54m/s²    

(c) Magnitude of the average force:

The magnitude is calculated using the formula;

     R = (mvₐ₁ -mvₐ₂)/Δt

         = m(vₐ₁ -vₐ₂)/Δt

         = 1730*(8.33 - 5.4)/0.1

          = (1730*2.93)/0.1

           = 5068.9/0.1

           = 50689N

Otras preguntas