A 1600 kg sedan goes through a wide intersection traveling from north to south w
ID: 1340540 • Letter: A
Question
A 1600 kg sedan goes through a wide intersection traveling from north to south when it is hit by a 2500 kg SUV traveling from east to west. The two cars become enmeshed due to the impact and slide as one thereafter. On-the-scene measurements show that the coefficient of kinetic friction between the tires of these cars and the pavement is 0.750, and the cars slide to a halt at a point 5.54 m west and 6.43 m south of the impact point.
What's the velocity of
a. SUV
b. Sedan.
I got V(suv) = 12m/s and V(sedan)=21.7m/s. but apparently it's not correct
Explanation / Answer
Given that
The resultant displcament is given by s =Sqrt((5.54)2+(6.43)2) =8.487m
The direction is given by theta =tan-1(6.43/5.54) =49.252degrees
From the work energy therorem
W =change in kinetic energy
Ff*s =(1/2)[m1+m2]v2
µk*(m1+m2)*s =(1/2)[m1+m2]v2
Therefore v =Sqrt(2uks) =Sqrt(2*0.750*8.487) =3.567m/s
From the law of conservation of momentum in the direction of south
m1*vo =(m1+m2)vsintheta
1600vo=(1600+2500)(3.567)sin49.252
vo =6.924m/s for sedan
From the law of conservation of momentum for the SUV moving in the west direction
m2*vo =(m1+m2)vcostheta
2500*vo=(1600+2500)(3.567)cos49.252
vo =3.818m/s for SUV
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.