Question

A 1300 kg car moving at 5.2 m/s is initially traveling north in the positive y direction. After completing a 90° right-hand turn to the positive x direction in 4.4 s, the inattentive operator drives into a tree, which stops the car in 350 ms. (a) In unit-vector notation, what is the impulse on the car during the turn? N·s + N·s (b) In unit-vector notation, what is the impulse on the car during the collision? N·s + N·s (c) What is the average force that acts on the car during the turn? N + N (d) What is the average force that acts on the car during the collision? N + N (e) What is the angle between the average force in (c) and the positive x direction? °

Answer 1

Answer:

a. 6760(i - j) kgm/s b. -6760i kgm/s c. 2172.74 N d. 19314.29 N e. 45°

Step-by-step explanation:

Given that mass of car, m = 1300 kg

initial velocity in the y-direction, v₁ = 5.2j m/s

time taken for 90° turn to positive x-direction, t₁ = 4.4 s

time taken for collision in positive x-direction, t₂ = 350 ms

a. Impulse, J₁  on car during turn.

impulse, J =mv₂ - mv₁

v₁= initial velocity and v₂ = final velocity

v₁ = 0i + 5.2j m/s since it was initially moving in the positive y-direction.

v₂ = 5.2i + 0j m/s since it was initially moving in the positive x-direction.

So, J₁ = m(5.2i + 0j -(0i + 5.2j))

          = m(5.2i - 5.2j)

         = 1300× 5.2(i-j)

         = 6760(i-j) kgm/s

b. Impulse after collision J₂

v₁= initial velocity=5.2i + 0j m/s since it was initially moving in the positive x-direction.

v₂= 0m/s since the car stops after collision

So, J₂= mv₂ - mv₁

         = m(0 - (5.2i + 0j)) m/s

         =-5.2mi kgm/s

         = -1300 × 5.2i kgm/s

         =-6760i kgm/s

c. Average force, F₁ during turn

Impulse J = Ft

From (a) the impulse J₁ = 6760(i-j) kgm/s. The time taken for the turn t₁ = 4.4 s. So, F₁ = J₁/t₁ = 6760(i-j)/4.4 = 1536.36(i-j) N.

Magnitude of F₁ = F₁ = average force during turn=1536.36√2= 2172.74 N

d. Average force, F₂ after collision

From J=Ft, F=J/t

From (b) above, our impulse during collision is J₂ = -6760i kgm/s. The time taken for the impulse or collision to occur is t₂ = 350 ms.

So, F₂ = J₂/t₂ = -6760i /(350 × 10⁻³) N= - 19314.29i N.

So magnitude of F₂= F₂=average force during collision = 19314.29 N

e. Angle between average force in (c) and the positive x- direction.

We know that F₁= 1536.36(i-j) N = 1536.36i - 1536.36j N. The unit vector in the positive x-direction is i.

For the angle between two vectors, we have that cosθ= a.b/ab where a,b are vectors and a,b their magnitudes respectively.

So, cosθ = (1536.36i - 1536.36j).i/(1536.36√2) =1536.36/1536.36√2=1/√2

cosθ = 1/√2

θ=cos⁻¹(1/√2)

θ=45°