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 2200 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 payment is .75 and the car slide to a halt at a point 5.57 meters West and 6.46 meters south of the impact point

how fast was the sedan traveling just before the Collision
how fast was the 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*6.46/(sqrt(6.46^2 + 5.57^2)) = 5.426 m/s2

So V in N-S direction just before collision = sqrt(2*a*s) = sqrt(2*5.426*6.46) = 8.156 m/s

component of accleration in E-W direction = 7.35*sin = 7.35*5.57/(sqrt(6.46^2 + 5.57^2)) = 4.957 m/s2

So V in E-W direction just before collision = sqrt(2*a*s) = sqrt(2*4.957*5.57) = 7.451 m/s

Momentum will be conserved before and after collision.

So momentum conservation in N-S direction

1400*Vsedan = (1400+2200)*8.156   => Vsedan = (1400+2200)*8.156/1400 = 21.555 m/s

and momentum conservation in E-W direction

2200*Vsuv = (1400+2200)*7.451   => Vsedan = (1400+2200)*7.451/2200 = 11.986 m/s

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