A 1400 k g sedan goes through a wide intersection traveling from north to south

Question

A 1400kg sedan goes through a wide intersection traveling from north to south when it is hit by a 2400kg 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 5.51m west and 6.19m 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

a) first conservation of energy to find speed after impact
1/2 mv^2 = u m g d

0.5*v^2 = 0.75*9.81*sqrt(5.51^2 + 6.19^2)
v=11.04 m/s

now we need direction
angle = arctan( 6.19/5.51)= 48.33 degrees

now conservation of momentum in the y

1400*v = (1400+2400)*11.04*sin(48.33)

v=22.38 m/s

b) now consrvation in the x

2400*v = (1400+2400)*11.04*cos(48.33)
v=11.62 m/s

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