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 2100 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 5.30 m west and 6.49 m south of the impact point.

a-How fast was sedan traveling just before the collision?

b-How fast was SUV traveling just before the collision?

Explanation / Answer

Total energy (E) lost by the friction after the impact = mu*Mt*g*d

where mu is friction coefficient, Mt is the total mass, g is the acceleration due to gravity and d is the total distance travelled after impact till halt.

d = sqrt(5.3^2 + 6.49^2) = 8.379m

E = 0.75 * (2100+1400) * 9.81 * 8.379 = 215769.7 Joules

Now, as E = 0.5 * Mt *Vc^2

Putting values in above equation we can get the combined velocity (Vc) after the impact.

We get Vc = 11.1 m/sec

Direction of this velocity = arctan(5.3/6.49) = 39.2 degrees west of the north south line

Now we can use conservation of momentum in NS and EW direction

Denoting sedan velocity by Vse and SUV velocity by Vsu

For NS direction, 1400*Vse = (1400+2100)*Vc*cos(39.2)

Vse = 21.5 m/sec

For EW direction, 2100*Vsu = (1400+2100)*Vc*sin(39.2)

Vsu = 11.6 m/sec

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