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 5.36 m west and 6.16 m south of the impact point.

Part A: How fast was sedan traveling just before the collision?

Part B: How fast was SUV traveling just before the collision?

Explanation / Answer

given,

mass of sedan = 1600 kg

mass of SUV = 2500 kg

coefficient of friction = 0.75

distance travelled after collision = sqrt(5.36^2 + 6.16^2)

distance travelled after collision = 8.165 m

friction force = 0.75 * (1600 + 2500) * 9.8

acceleration = force / mass

acceleration = 0.75 * (1600 + 2500) * 9.8 / (1600 + 2500)

acceleration = 7.35 m/s^2

v^2 = u^2 + 2as

0 = u^2 + 2 * 7.35 * 8.165

u = 10.95 m/s

velocity after collision = 10.95 m/s

angle = tan^-1(6.16 / 5.36)

angle = 48.97 degree

by conservation of momentum

initial horizontal momentum = final horizontal momentum

2500 * v1 = (1600 + 2500) * 10.95 * cos(48.97)

speed of SUV before collision = 11.788 m/s

also,

initial vertical velocity = final vertical velocity

1600 * v2 = (1600 + 2500) * 10.95 * sin(48.97)

speed of sedan before collision = 21.167 m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.