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 2300 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 5.60 m west and 6.55 m south of the impact point.

Part A

How fast was sedan traveling just before the collision?

v= ? m/s

Part B

How fast was SUV traveling just before the collision?

v=? m/s

Explanation / Answer

Mass m1 = 1600 Kg
Mass m2 = 2300 Kg

Initial Velocity of Sedan = v1
Initial Velocity of SUV = v2

The acceleration after impact is found from
ug(1600 + 2400) = (1600 + 2400)a

a = ug
a = 0.75 * 9.8 = 7.35 m/s^2

The distance slid = sqrt(5.6^2 + 6.55^2) = 8.62 m

After Collision,
Initial Velocity, u = ?
Final Velocity , v = 0

v^2 - u^2 = 2as
-u^2 = -2 x 7.35 x 8.62
u = 11.25 m/s

angle = tan^-1(6.55/5.6) = 49.5 deg south of west.

Resolving into south and west components and equating momentum before and after -

1600*v1 = (1600 + 2400) * 11.1 * sin(49.5)
v1 = 21.1 m/s
2400*v2 = (1600 + 2400) * 11.1 * cos(49.5)
v2 = 12.0 m/s

Part A
Speed of sedan traveling just before the collision, v1 = 21.1 m/s

Part B
Speed of SUV traveling just before the collision, v2 = 12.0 m/s

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