STRATEGY The total momentum of the vehicles before thecollision, Pi, equals the total momentum of the vehicles after the collision, pf, if we ignore friction andassume the two vehicles form an isolated system. (This is called the "impulse approximation.") Solve themomentum conservation equation for the final velocity of the entangled vehicles. Once the velocities arein hand, the other parts can be solved by substitution.SOLUTION(A) Find the final speed after collision.Let m₁ and V₁¡ represent the massand initial velocity of the pickup truck,while m₂ and V₂¡ pertain to thecompact. Apply conservation ofmomentum.Substitute the values and solve for thefinal velocity, Vf.Pi Pfm₁v₁; + m₂V₂i = (m₁ + m₂) vfVf(B) Find the change in velocity for each vehicle.Change in velocity of the pickup truck.Change in velocity of the compact car.Calculate the final kinetic energy of thesystem and the change in kineticenergy, AKE.(1.80 x 10³ kg)(15.0 m/s) + (9.00 × 10² kg)(-15.0 m/s)(1.80 x 10³ kg + 9.00 × 10² kg)vf+5.00 m/s==(C) Find the change in kinetic energy of the system.Calculate the initial kinetic energy ofthe system.bAV₁ = Vf - V₁iAV₂ = VfV₂i = 5.00 m/s - (-15.0 m/s) = 20.0 m/sKE₁ = 1/2m₁v₁/² + 1/2m₂v₂2²2=AKE= 5.00 m/s 15.0 m/s ==(1.80 × 10³ kg)(15.0 m/s)²+ (9.00 × 10² kg)(-15.0 m/s)²= 3.04 x 105 JVEKE₁ = 1/(m₂ + m₂) v²=(1.80 × 10³ kg + 9.00 × 10² kg)(5.00 m/s)²= 3.38 x 104 JVE-10.0 m/s2.70 x 105 1 PRACTICE ITUse the worked example above to help you solve this problem. An pickup truck with mass 1.85 x 10³ kg istraveling eastbound at +14.3 m/s, while a compact car with mass 8.61 x 10² kg is traveling westbound at-14.3 m/s. (See figure.) The vehicles collide head-on, becoming entangled.(a) Find the speed of the entangled vehicles after the collision.m/s(b) Find the change in the velocity of each vehicle.Av truckAv car=(c) Find the change in the kinetic energy of the system consisting of both vehicles.Jm/sm/sEXERCISEHINTS: GETTING STARTED I I'M STUCK!Use the values from PRACTICE IT to help you work this exercise. Suppose the same two vehicles are bothtraveling eastward, the compact car leading the pickup truck. The driver of the compact car slams on thebrakes suddenly, slowing the vehicle to 6.09 m/s. If the pickup truck traveling at 17.7 m/s crashes intothe compact car, find the following.(a) the speed of the system right after the collision, assuming the two vehicles become entangledm/s(b) the change in velocity for both vehiclesAvtruckm/sm/sAv car==(c) the change in kinetic energy of the system, from the instant before impact (when the compactcar is traveling at 6.09 m/s) to the instant right after the collisionAKE =J

Name: STRATEGY The total momentum of the vehicles before thecollision, Pi,
Uploaded: 2024-03-20T06:57:37.000Z
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Description: STRATEGY The total momentum of the vehicles before thecollision, Pi, equals the total momentum of the vehicles after the collision, pf, if we ignore friction andassume the two vehicles form an isolated system. (This is called the "impulse approximat

Note:- “Since you have asked multiple questions, we will solve the first question for you. If you want any specific question to be solved then please specify the question number or post only that question." Given:- The mass of the pickup truck is M = 1.85×103 kg is traveling eastbound at the speed of u1 = 14.3 m/s The mass of the compact car is m = 8.61×102 kg is traveling westbound at a speed of u2= -14.3 m/s The vehicle collided head-on, becoming entangled Find:- a) The speed of the entangled vehicle after the collision b) The change in the velocity of each vehicle. c) The change in the kinetic energy of both vehicles.Solution:- a) We know that the law of conservation of momentum is given as Mu1+mu2 = M+mvfvf = Mu1+mu2M+mvf = 1.85×103×14.3-8.61×102×14.31.85×103+8.61×102vf = 5.21678≈5.22vf= 5.22 m/s This is the final velocity after the collision.b) Now We know that the change velocity of the truck is given as ∆vtruck = vf-u1∆vtruck =5.22-14.3∆vtruck =-9.08 m/s Now the change in the velocity of the car is given as ∆vcar = vf-u2∆vcar =5.22--14.3∆vcar =19.52 m/sc) Now We know that the charge in kinetic energy is given as ∆K.E = K.Ei-K.Ef∆K.E =- 12Mu12+12mu22-12M+mvf2∆K.E =-12×1.85×103×14.32+12×8.61×102×-14.32-121.85×103+8.61×1025.222∆K.E =-277186.195-36935.2062∆K.E =- 240250.989 J a)The speed of the entangled vehicle after the collision is 5.22 m/s b) The change in velocity of each of the vehicles is as ∆vtruck =-9.08 m/s ∆vcar =19.52 m/s c)The change in the kinetic energy of both vehicles is -240250.989 J

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PRACTICE IT Use the worked example above to help you solve this problem. An pickup truck with mass 1.85 x 10³ kg is traveling eastbound at +14.3 m/s, while a compact car with mass 8.61 x 102 kg is traveling westbound at -14.3 m/s. (See figure.) The vehicles collide head-on, becoming entangled. (a) Find the speed of the entangled vehicles after the collision. m/s (b) Find the change in the velocity of each vehicle. m/s Av truck Av car m/s (c) Find the change in the kinetic energy of the system consisting of both vehicles. J