# Giancoli physics chapter 7 solutions

### Tranmanuscript for this Giancoli solution

So this smaller vehicle here is going to affect a larger automobile which is initially at rest at a red light and also we understand what the acceleration is going to be bereason we know friction.

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We also know just how much the linked wreck goes and so from that we can number out the initial speed of this wreck and then knowing that we’ll be able to usage momentum to figure out what should have been the rate of the incoming auto. So we’ll talk around momentum first of all. So we have actually that "m1’ times "v1’ amounts to "m1’ plus "m2’ times "v`’, so the full initial momentum which is simply of the car equates to the unified mass once it becomes a wreck times their combined velocity. And we deserve to deal with that for "v1’: that’s "m1’ plus "m2’ over "m1’ times "v`’. We know that the force of friction in this case is the net force once the two are sliding after the collision they’re sliding coming to a sheight, tright here is just one pressure acting on them which is friction. So that suggests that ‘µ’ times ‘FN’ which is a friction formula equates to ‘m’ times ‘a’ which is the net force formula and also ‘FN’ in this case is ‘m’ times ‘g’, it’s the pressure of gravity, bereason they are not increasing vertically. So we have ‘µ’ times ‘m’ times ‘g’ equates to ‘m’ times ‘a’ and the ‘m’s cancel interpretation that ‘a’ equates to ‘µ’ times ‘g’. So now we know the deceleration of the wreck we’ll use the kinematics formula "vf’ squared equals "v1’ squared plus 2 times ‘a’ times ‘d’ bereason we understand all these points except for "vi’ which is what we want to find. We know that they involved a stop, we understand what the acceleration is and we understand exactly how far they go because the question tells us. So we’ll fix for "vi’: "vi’ is the square root of two times ‘a’ times ‘d’ which is the square root negative two times ‘µ’ times ‘g’ times the distance that it slides, then substitute for "vi’ via "v`’, the "vi’ is the rate just once the wreck starts and also that’s "v`’ so much as the momentum is concerned. So "v`’ is the rate after the two cars collide and also "vi’ right here is the same speed as that rate that the wreck starts to slide at. So we have: "v1’ is "m1’ plus "m2’ over "m1’ times this expression square root of two times ‘µ’ times ‘g’ times ‘d’. ‘g’ or the acceleration is negative and so this ‘µ’ times ‘g’ must be an adverse expression so the negative authorize disappears, so we have nine hundred and also twenty kilograms plus two thousand three hundred kilograms separated by nine hundred and twenty times square root of 2 times zero point eight, the coreliable of friction, times nine allude eight which is ‘g’ times 2 allude eight meters that the wreck slides. And this provides twenty 3 meters per second. Now for the fifth Edition the masses are slightly different, we have one thousand kilograms is the incoming vehicle and also it results a automobile which is initially at rest of mass 2 thousand two hundred kilograms and also the coeffective of friction is instead zero allude 4 so for the fifth Edition the vehicle should have actually initially going fifteen meters per second.

I still do not understand also why the initial velocity of the moving vehicle is equal to the final velocity of both vehicles together? Could you explain this please?

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