Bug vs. Bus physics problem?

Question

Bug vs. Bus physics problem?

Answer 1

The straightforward answer, as far as I can see, is c) both for all of them (with a caveat that that's not totally accurate for the momentum).



1: Force. Newton's third law implies that during an impact such as this an equal force acts on the bus and the bug. Therefore the "greater force" acts on both objects.



2: Impulse. Assuming the bus and the bug are traveling a lot slower than the speed of light, the difference in proper time between their inertial reference frames is going to be less than negligible. Since impulse is simply force exerted times time of impact, and we've already established that the force is the same, then this is c again, both.



3: Momentum. Momentum is always conserved. Always. If total momentum before equals total momentum afterwards, then the decrease in momentum for the fly must equal the increase in momentum of the bus (the momentum of the bus increases because although it might slow down, its mass has increased as it has bits of bug on it, p=mv).

Answer 2

Bug Bus

Answer 3

the situation calls so you might equate the forces at a component interior the trail of the motor vehicle, on the around arc, against the attitude of the arc ? => ?(t) i.e. is a function of time. If the arc has a radius r then arc length is only s = r? (the place s(t) is likewise a function of time). We define ?= 0 the place motor vehicle is up at h = 0.3m and ?= 90deg while motor vehicle is on the backside. The vertical stress is only the burden of the motor vehicle W= mg (g = 9.8m/s² of course) The component to this stress alongside the process action is the tangential component = Wcos?, there's a common reaction from the song it is Wsin?, there is hence a frictional stress opposing the action F = ? Wsin? the place ? is a parameter for the coefficient of resistance. From the stress diagaram we only upload up all of the forces on the physique: Unbalanced stress: relaxing = W + F a vector eqn. Fun_tangential = Wcos? - ?Wsin? this stress ought to equate to mass x tangential acceleration, for useful around action it is (?²?/?t²)r or written ?''r for brevity, so; m?''r = mg(cos? - ?sin?) => ?''r = g(cos? - ?sin?) _ 2d Order non-linear diff _ Eqn a million we additionally equate the radial component to Wsin? to the centripital acceleration (??/?t)²r = ?²r [?']²r = gsin? it is a 1st Order diff eqn. _ Eqn 2 you will might desire to do extra analyze to remedy those, remember ? => ?(t). even even with the incontrovertible fact that your boundary circumstances state that at t=0, h(0) = 36cm and r = 0.3m const. from geometry on your image; height above backside of song is h =r(a million - sin?) 0.36 = 0.3[a million - sin?(0)] or sin ?(0) = - 0.12, ?(0) = -6.9deg or -0.12rad. additionally at t=0 starts from relax skill that ?'(0) = 0. As t -> T the time while particle at last is composed of relax h(T) = 0, ?(T) = 90deg, ?'(T) = 0, ?''(T) = 0. [option: you will possibly have used PE at good (t=0) = mgh(0) = KE(t) + artwork(t) on arc yet you will nevertheless finally end up with differential eqn's in ?(t) to remedy! PE = mgr(a million - sin?) KE = a million/2mv² = a million/2m(?' r)² artwork = Fs = ?mgsin?.r? gr(a million - sin?) = a million/2(?' r)² + ?gsin?.r? ___Eqn 3]

Answer 4

Isn't it the bug for all because the bus gets the bug moving in the opposite direction. The bus barely reduces speed and the impact on the bus is unfelt. The greater force acts on the bug because the bus has a greater mass and is moving much faster than the bug. The bug has a greater change in momentum because it is stopped (squashed by the impact) and then forced the opposite way. The momentum of the bug therefore suddenly becomes negative. Impulse is the change in momentum of an object therefore the answer is the bug as the second and third are the same question just phrased differently.