Question

Problem 8.14 26 of 36 Review | Constants A 750 kg car is stalled on an icy road during a snowstorm. A 1000 kg car traveling eastbound at 12 m/s collides with the rear of the stalled car. After being hit, the 750 kg car slides on the ice at 4 m/s in a direction 30° north of east. Part A What is the magnitude of the velocity of the 1000 kg car after the collision? Express your answer in meters per seconds to two significant figures. ΠΙ ΑΣΦ ? m/s Submit Request Answer Part B
PLEASE ANSWER A B AND C !! INDICATE EACH ANSWER PLEASE

Answer 1

(A) Using conservation law m Vi + m2 v2 = mvi +m,va cos 30° (1000)(12)+(750)(0)=(1000) vi +(750)(4) cos 30° (1000)(12)–(750)(4)cos 30 V1 1000 = 10.5 m/s Similarly momentum equation y-direction is, m_v; + m2 V2 = m,ví + m2v, sin 30° 1000(0)+750(0)=(1000)vi +(750)(4)sin 30° 750(4) sin 30° 1000 = -1.5 m/s The magnitude of the velocity is, 1 = = V(10.5)² + (-1.5) =10.61 m/s (B) The direction is, -1.5 tan = 10.61 -1.5 = tan 10.61 - m_v? = -8.05° south of east (C) The kinetic energy is, K; -m,v + 2 2 =(0.5)(1000)(12)? +(0.5)(750)(0)? = 72000 J =72 kJ The kinetic energy expression after collision is, 24V1? +m_v? 2 = (0.5)(1000)(10.61)° +(0.5)(750)(4)? = 62.286 kJ 1 1 KE

The ratio of kinetic energy is, KE, _62.286 KE 72 = 0.865 = 0.87