Car A was traveling east at high speed when it collided at Point O with car B, w

Question

Car A was traveling east at high speed when it collided at Point O with car B, which was traveling north at mi/h. Car C which was traveling west at 55 mi/h. was 32 ft cast and 10 ft north of Point O at the time of the collision. Because the pavement was wet. the driver of car C could not prevent his car from sliding into the other two cars, and the three cars, stuck together, kept sliding until they hit the utility pole P. Knowing that the weights of cars A. B, and C are. respectively. 3000 lb. 2600 lb. and 3000 lb. and neglecting the forces exerted on the cars by the wet pavement, solve the problems indicated. Knowing that the coordinates of the utility pole are x_p = 52 ft and y_p = 61 ft. Determine: (a) the time elapsed from the first collision to the stop at P. (b) the speed of car A. V_A = V_B = 50 mi/hr or 73.33 ft/s v_c = 55 mi/hr or 80.67 ft/s x_p = 52 ft y_p = 61 ft Let t = time elapsed first collision (m_A + m + m_c) (x_o i + y_o j) = M_A (0) + m_B (o) + m_c (x_c i + y_c j) x_o = 0.3 X_c = (0.3) (32) = 9.6 ft y_o = 0.3 y_c = (0.3) (10) -3ft

Explanation / Answer

Given that,

vB = 50 mi/h = 3.33 ft/s

vC = 55 mi/h = 80.6 ft/s

xc = 32 ft   ,   yc = 10 ft

coordinates of utility pole,

xp , yp = 52 ft , 61 ft

(a)

final position of center of mass of system is same as previous.

(mA + mB + mC)*rp = (mA*vA*t) i + (mB*vB*t) j + mC*(xc i + yc j + vC*t i)

(3000 + 2600 + 3000)*rp = 3000*vA*t i + 2600*73.33*t j + 3000*(32i + 10j + 80.6t i)

8600*rp = 3000vAt i + 190658t j + 96000i + 30000 j + 241800t i

on simplifying,

rp = (0.348vA - 28.11)t i + 11.16 i + 22.169t j + 3.48 j

xp = (0.348vA - 28.11)t + 11.16   ......eq1

yp = 22.169t + 3.48         .........eq2

61 = 22.169t + 3.48

by solving,

t = 2.595 s

(b)

put the value of t in eq(1),

52 = (0.348vA - 28.11)*2.595 + 11.16

by solving,

vA = 126 ft/s

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