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°
