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edison (3966)

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 A 2000 kg car whose brakes have failed coasts at 30 mph toward a 1000 kg car whose brakes are locked in "park" mode. If the coefficient of sliding friction between the 1000 kg car and the road is 0.7, determine how long after the first collision the second collision occurs. (Assume elastic collisions.) Find the distance between collisions. Will there be a third collision? Prove or disprove.





 


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Aatish (2261)

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I might look foolish while saying this........


Time after first collision to second collision according to my calcs...is........


0.198 seconds.........is it correct sir????.....even i myself cannt believe.......someone else also pls try.......


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akshay.khare91 (167)

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I cannot understand why the second collision will take place
since the collision is elastic....

<TABLE CELLSPACING="1" CELLPADDING="1" BORDER="0">
<TR><TD>


<DIV ALIGN="right">Glitter Graphics</DIV></TD></TR></TABLE>
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Aatish (2261)

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See buddy.....when firstly the car (break failed) hits the other car (at rest) then due to elastic collision....


the other car.......goes away with some velocity.....but as there is friction between car and surface.......


it starts retarding after sometime.......BUT THE CAR (break failed) HAS NOT STOPPED>>>.........


it comes again and again hits the same car.,......


 


WORK IN PROGRESS.........for remaining question.......


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There's Light at the end of every Tunnel, so KEEP MOVING....
Best of luck to all my mates....
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rohitkuruvila (123)

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is the answer to the first part 60/7 seconds

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Aatish (2261)

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how u proceeded rohit.....can u pls telll.........

have converted the velocity from milesperhour to meterpersec////

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rohitkuruvila (123)

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sir,i need the exact value of 1 mile in km/hr in the last step.

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Aatish (2261)

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1 mile = 1609.344 m

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rohitkuruvila (123)

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is the answer 3.83 sec

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Aatish (2261)

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Now I m getting........1.98 seconds.......

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