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 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 0.75, and the cars slide to a halt at a point 5.54 m west and 6.10 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

let West be +x axis

acceleration of the car, ax = ay = -g*mue_k

= -9.8*0.75

= -7.35 m/s^2

let vx and vy are the components of velocity of cars after the collision.s

vf^2 - vx^2 = 2*ax*dx

==> vx = sqrt(2*ax*dx)

= sqrt(2*7.35*5.54)

= 9.02 m/s

vf^2 - vy^2 = 2*ay*dy

==> vy = sqrt(2*ay*dy)

= sqrt(2*7.35*6.1)

= 9.47 m/s

A) Apply conservation of momentum in x-direction

1400*V_sedan = (1400 + 2500)*9.02

==> v_sedan = (1400 + 2500)*9.02/1400

= 25.1 m/s

B) pply conservation of momentum in y-direction

2500*V_SUV = (1400 + 2500)*9.47

==> v_SUV = (1400 + 2500)*9.47/2500

= 14.8 m/s

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