Hey there I'm having trouble with a physics problem. I'm confused why the instructions for step 5 say that for momentum, velocity = acceleration. If this is the case would I do p=1895.2kg * -8.33m/s^2? Would I remove the negative? Should I remove the m/s^2, and just do m/s? I've attached the work I've done so far, but I'm very confused as to how to get the momentum section done because of those instructions. Should I ignore those instructions and just use the final velocity from step 4 to solve for the momentum? To be clear, the focus is on question 5, I've already completed much of the preliminary work (calculating mass of objects, etc. etc)

Requirements of Final Report Primary Objective: Use the Law of Conservation of Momentum to solve for the initial velocity of the Hummer. Solve for Each- Must show work: Solve for the friction force which brought the van to rest after the collision. Solve for the acceleration of the minivan. This will be negative because the van was moving against the IIIII frictional force. Determine the length of skid marks. Solve for the velocity of the minivan. This will be your "final" velocity for the van Solve for the momentum of the minivan. Set this value equal to the momentum of the Hummer. Solve for the initial velocity of the Hummer. Final Report : . Ensure each step (except 3 & 6) of your process in solving each of the above steps is clear, including Givens, Unknowns, Equations, Set-up & Solution (box Solution) O Write a paragraph describing one other real world application of a device that deals specifically with force and momentum. Explain how the device works to reduce or increase the force acting on an object by changing the momentum RETURN Be sure you ADD ALL Equations to use / manipulate the masses per vehicle (vehicle, people, fuel) Step 1: Step 2: Step 3: Skid Friction Force Acceleration mark (m) F=UF normal Fnet = ma Review diagram *Fnet=your F value RETURN Step 4: Step 5: Step 7: Acceleration Momentum Conservation of Momentum V.2 = V.2 + 2aAX P = mv Dont forget to take the In this case v will P, = P2 square root of your answer! =acceleration mv + m,V2 = mv, + m,v2Collision Thvestigation Report 1895 ,2 kg) * blanda Odnrey . (total mass : Walking) - Car mass : 1, 70 0 keg " Driver Mass : 12 kg - Mass of 1 passengers : 35 (g 1 2 : 70 kg - Fuel load : 19 gallons might Beassigned Andand (719 gal . 1.8 kg : 91.2 kg Li WereVINAYAK#1 1.165 : 1895. 1 Kg ( ) total mass : 1 20 cup + 72# + 701 , + # - Excemal : Egravity - > F , : 18 9 5 , 1 kg . 9. RMA : 18, 52 3 N .Friction Coefficient ( H) : C.8s c [lotu, rubber tir () and asphat] Friction Force [ E . ) : C. 8SC . 19, 5 73 N = 15, 28 7 . CS N 1 12 Hummer ! ( total Mass : 6,132 . 6 kg) - Car Mass : 1, 150 ks . Driver Mass : 62 kg . Fuel load 7 gallons - 951 : 2.8 69 , 19,6 kg Ly Z gal . 2.8 kg : 19. 6 k g . Total mass : 1 150 kg + 62 kg + 19. 6 kg : 1,131, 6 kq a Step li Solve for Friction forse which bronghit the ran to cost after the collicient . True Rechose Herizeathl surface|7 5 : m . a - 218 9 5, ] kq . 9. Fang = = 18 573 N H = CESC - ( lotal mass calculated about ): M : 1895 , 2 kg FF: C. 850 . 18, 57 3 N cocking upon the real Fe : 15,782 , CSN | lis the Friction Force which branglet the van to rest " Step 2 [ solve for the acceleration of the minivan! [- 8 . 13 An! ) Fnot : FF a = ? M : 18 95 , 2 kg Lisp Fact : 15, 187. 0 49 15, 787. 05 N: Within 1895. 2 kq . a = 1845 , 145 :( Continued ) Collision Investigation *