A 1400 kg sedan goes through a wide intersection traveling from north to south w

Question

A 1400 kg sedan goes through a wide intersection traveling from north to south when it is hit by a 2300 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 0.75, and the cars slide to a halt at a point 5.60 m west and 6.13 m south of the impact point.

Part A

How fast was sedan traveling just before the collision?

Part B

How fast was SUV traveling just before the collision?

Explanation / Answer

Accleration after collision = friction_cofficient*g = 0.75*9.8 = 7.35 m/s2

component of accleration in N-S direction = 7.35*cos = 7.35*6.13/(sqrt(6.13*6.13 + 5.6*5.6)) = 5.426 m/s2

So V in N-S direction just before collision = sqrt(2*a*s) = sqrt(2*5.426*6.13) = 8.156 m/s

component of accleration in E-W direction = 7.35*sin = 7.35*5.6/(sqrt(6.13*6.13 + 5.6*5.6)) = 4.957 m/s2

So V in E-W direction just before collision = sqrt(2*a*s) = sqrt(2*4.957*5.6) = 7.451 m/s

Momentum will be conserved before and after collision.

So momentum conservation in N-S direction

1400*Vsedan = (1400+2300)*8.156   => Vsedan = (1400+2300)*8.156/1400 = 21.555 m/s Answer (A)

and momentum conservation in E-W direction

2300*Vsuv = (1400+2300)*7.451   => Vsedan = (1400+2300)*7.451/2300 = 11.986 m/s Answer (B)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.