A 1600 kg sedan goes through a wide intersection traveling from north to south w

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 0.75, and the cars slide to a halt at a point 5.57 m west and 6.34 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

from the law of conservation of momentum

Psedan + P Suv = P = (msedan + MsuV )V

here V^2 = ugd

d ^2 = 5.57^2 + 6.34^2

d = 8.307 m

so v^2 = 0.75 * 8.307* 9.81

V = 7.81 m/s

tan theta = y/x = 6.34/5.57

theta = 48.17 deg

so

MV) of suv = (Msuv + Msedan) * V cos Theta

Vsuv = (1600 + 2500) * 7.81 cos 48.17/2500

Vsuv = 8.54m/s -

MV) of sedan = (Msuv + Msedan) * V sin Theta

Vsuv = (1600 + 2300) * 7.81 sin 48.17/1600 = 14.18 m/s

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