Please help me with the highlighted portion. Thank you very much. PRACTICE IT Us

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Please help me with the highlighted portion. Thank you very much.

PRACTICE IT Use the worked example above to help you solve this problem. An pickup truck with mass 1.75 x 103 kg is traveling eastbound at +14.8 m/s, while a compact car with mass 9.01 × 102 kg is traveling westbound at -14.8 m/s. (See figure.) The vehicles collide head-on, becoming entangled. (a) Find the speed of the entangled vehicles after the collision 4.74 m/s (b) Find the change in the velocity of each vehicle. Vtruck 10.06 m/s Vcar19.54 m/s (c) Find the change in the kinetic energy of the system consisting of both vehicles -1057559.51 X Your response differs from the correct answer by more than 100%. J EXERCISE HINTS: GETTING STARTED I'M STUCK! Use the values from PRACTICE IT to help you work this exercise. Suppose the same two vehicles are both traveling eastward, the compact car leading the pickup truck. The driver of the compact car slams on the brakes suddenly, slowing the vehicle to 6.23 m/s. If the pickup truck traveling at 17.1 m/s crashes into the compact car, find the following. (a) the speed of the system right after the collision, assuming the two vehicles become entangled m/s (b) the change in velocity for both vehicles Avtruck m/s m/s car (c) the change in kinetic energy of the system, from the instant before impact (when the compact car is traveling at 6.23 m/s) to the instant right after the collision

Explanation / Answer

m1(truck) = 1.75*10^3 kg

v1i = 17.1 m/s

m2*(car) = 9.01*10^2 kg

v2i = 6.23 m/s

linear momentum pf system before collision Pi = m1*v1i + m2*v2i

after collsion , the car and truck entangled

momentum of the system Pf = (m1 + m2)*vf

from momentum conservation

Pf = Pi

(1.75*10^3 + 9.01*10^2)*vf = (1.75*10^3*17.1) + (9.01*10^2*6.23)

speed of the system vf = 13.4 m/s

=====================

part(b)

change in velocity

Vtruck = (vf - v1i) = (13.4 - 17.1) = -3.7 m/s

dVcar = vf - v2i = 13.4 - 6.23 = 7.17 m/s

=================

part(c)

kinetic energy before imnpact KEi = (1/2)*m1*v1i^2 + (1/2)*m2*v2i^2

KEi = (1/2)*1.75*10^3*17.1^2 + (1/2)*9.01*10^2*6.23^2 = 273344 J

kinetic energy after imnpact KEf = (1/2)*(m1+m2)*vf^2

KEi = (1/2)*(1.75*10^3 + 9.01*10^2)*13.4^2 = 238006.78 J

change in KE

dKE = KEf = KEi = 273344 - 238006.78 = 35337.22 J

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