A 950-kg Jeep and a 1300-kg SUV collide in elastically at an intersection. Befor

Question

A 950-kg Jeep and a 1300-kg SUV collide in elastically at an intersection. Before the collision, the SUV was traveling in the -y direction (or southward) and the Jeep was traveling in the +x direction (or eastward). After they collide, they initially move off together at a speed of 14.5 m/s along the direction 15.0 degree clockwise from +x direction (i.e., eastward and a little bit south). a. The speed limit at the intersection is 40 miles per hour, or 17.86 m/s. Determine which vehicle(s), if any, exceeded the speed limit just before the collision. b. Calculate how much total kinetic energy is lost in the collision.

Explanation / Answer

here,

mass of jeep , m1 = 950 kg

mass of SUV , m2 = 1300 kg

final speed , v = 14.5 * ( cos(15) i - sin(15) j)

v = 14.1 i - 3.65 j m/s

initial speed of jeep and SUV be u1 i and u2 (-j)

using conservation of momentum

m1 * u1 i - m2 * u2 j = (m1 + m2 ) * v

950 * u1 - 1300 * u2 j = (1300 + 950) * ( 14.1 i - 3.65 j)

u1 = 33.4 m/s

u2 = 6.32 m/s

so, the jeep is exceeding speed limit before the colison

b)

the kinetic energy lost , KE = 0.5 * m1 * u1^2 + 0.5 * m2 * u2^2 - 0.5 * (m1 + m2) * v^2

KE = 0.5 * ( 950 * 33.4^2 + 1300 * 6.32^2 - (1300 + 950) * 14.5^2) J

KE = 3.19 * 10^5 J

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