please help with full work shown! Car vs. Moose, Car vs. Deer. In some part of t

Question

please help with full work shown!

Car vs. Moose, Car vs. Deer. In some part of the country, collisions between vehicles and moose are a big problem. (Moose like to stand on roads in winter and lick up the salt put down by road crews to melt ice on the road.) Suppose a 1500 kg car slides into a stationary 500 kg moose on a very slippery road, and the moose is thrown through the windshield, making this a completely inelastic collision (and a very bad situation for the driver, who gets 500 kg of moose landing in his lap). What percentage of the original kinetic energy is lost in the collision to other forms of energy? Describe in a sentence or two where that kinetic energy actually goes. b. Deer are smaller than moose but more numerous, and they have a habit of "freezing" in place when they see car headlights approaching. What percentage of the original kinetic energy is lost if the car instead hits a stationary, 100 kg deer? In general, does the percentage of kinetic energy lost increase or decrease if the mass of the animal decreases? Based on this finding and your answer to part (a), which animal would you as the driver prefer (as if you had a choice) to hit with your car, and why?

Explanation / Answer

part a:

let initial velocity of car is v m/s

then initial momentum=mass*speed=1500*v kg.m/s

when the moose lands on the car , as it is an inelastic collision, energy is not conserved but total mometnum is conserved.

so if final speed of moose+car system is v1 m/s,

then (1500+500)*v1=1500*v

==>v1=0.75*v....(1)

initial kinetic energy=0.5*1500*v^2=750*v^2

final kinetic energy=0.5*(1500+500)*v1^2

=0.5*2000*(0.75*v)^2=562.5*v^2

hence kinetic energy lost=initial kinetic energy-final kinetic energy

=750*v^2-562.5*v^2

=187.5*v^2

so kinetic energy loss as a percentage of original kinetic energy=187.5*v^2/(750*v^2)

=25%

this energy is converted in thermal energy due to collision of two objects.

part b:

initial kinetic energy=750*v^2

final velocity=initial momentum/total final mass
=1500*v/(1500+100)=0.9375*v

then final kinetic energy=0.5*(1500+100)*(0.9375*v)^2

=703.125*v^2

then kinetic energy loss=46.875*v^2

so kinetic energy loss as a percentage of original kinetic energy=46.875/750=6.25%

part c:

comparing part a and part b, kinetic energy is loss is more in case of heavier moose

than the lighter deer.

so deer is preferrable as compared to moose as the damage will be lesser (relatively)

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