Respuesta :
Answer
given,
mass of car = 1200 Kg
initial velocity in north = 5.8 m/s
90° right-hand turn in time = 3.3 s
car stop in, time = 270 ms
a) impulse = change in momentum
Initial momentum = ( m x v) j
= ( 5.8 x 1200 ) j
= ( 6960) j
final momentum after turn is 90°
= 6960 i
impulse = 6960 i - 6960 j
impulse = 6960 ( i - j )
b) due to collision
initial momentum after turning = 6960 i
final momentum is equal to zero
impulse = 0 - 6960 i
impulse = - 6960 i
c) during turn
I = F Δ t
[tex]F = \dfrac{I}{t}[/tex]
[tex]F = \dfrac{6960 ( \hat{i} -\hat{j})}{3.3}[/tex]
[tex]F =2109.1 ( \hat{i} -\hat{j})[/tex]
magnitude of force
[tex]F =2109.1\sqrt{2}[/tex]
[tex]F =2982.72\ N[/tex]
d) during collision
[tex]F = \dfrac{I}{t}[/tex]
[tex]F = \dfrac{-6960 \hat{i} }{0.27}[/tex]
[tex]F =-25777.78\hat{i}\ N[/tex]
magnitude of average force is equal to
[tex]F =25777.78\ N[/tex]
e) angle between average force
[tex]F =2109.1 ( \hat{i} -\hat{j})[/tex]
[tex]\theta =tan^{-1}(\dfrac{-1}{1})[/tex]
[tex]\theta =-45^0[/tex]
