A 1500 sedan goes through a wide intersection traveling from north to south when
ID: 2009220 • Letter: A
Question
A 1500 sedan goes through a wide intersection traveling from north to south when it is hit by a 2300 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 5.42 west and 6.49 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
Masses of Cars: m1 = 1500 Kg m2 = 2300 Kg Distance travelled after collision S = Sqrt[(5.42)^2 + (6.49)^2] = Sqrt[71.49] = 8.45 m Coefficient of Kinetic friction = 0.75 Acceleration a = g = 0.75*9.8 = 7.35 m/s^2 From Law of conservation of momentum : m1v1 = (m1 + m2)v cos and m2v2 = (m1 + m2)v sin But Sin = 6.49/S = 0.768 and cos = 5.42/S = 0.641 So v1 = 1.624v ----------[1] v2 = 1.268v ----------[2] After collision : vt^2 = v^2 - 2aS 0 = v^2 - (2*7.35*8.45) v^2 = 124.215 v = 11.14 m/s Hence velocity before collision: v1 = 18.09 m/s for first car v2 = 14.13 m/s for second car. So v1 = 1.624v ----------[1] v2 = 1.268v ----------[2] After collision : vt^2 = v^2 - 2aS 0 = v^2 - (2*7.35*8.45) v^2 = 124.215 v = 11.14 m/s Hence velocity before collision: v1 = 18.09 m/s for first car v2 = 14.13 m/s for second car.Related Questions
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