Newtons Famous Equation that changed mechanics: R- (mv). The rate of change of m
ID: 2886662 • Letter: N
Question
Newtons Famous Equation that changed mechanics: R- (mv). The rate of change of momentum of body is equal to the resultant forces R on the body. This equation is on the front cover of K. Bullen's Theory of Mechanics" probably the most famous equation before EinsteinsE ook over the world. (a) A hail stone begins falling from a height of 5,000 metres. Assuming no air resistance (BIG assumption): i. Find the time it takes to reach the ground where we know: v()10 m/sec2 ii. Find the speed at which the hail stone would hit the ground (b) Of course there is air resistance, approximately: v(O 10- 3 V) m/sec' i. Find the approximate time it takes to reach the ground. ii. Find the approximate speed at which the hail stone would hit the groundExplanation / Answer
(a)
1. since s=ut +0.5gt2..........................(1)
where u is initial velocity which is zero in this case as its at rest. a is acceleration due to gravity which is 10m/s2 t is time taken and s is dispalcement (given as 5000m).
Substitue the value in equation (1)
and thus 5000= 0*t + 0.5*10*t2 => t=31.62 sec
2. here we neeed to find the final velocity with which it hit the ground i.e. v
so motion 1st equation: v= u+ at
we know u(=0) and t(=31.62) => v=0+10*31.62 => v=316.2 m/s
(b)
since air resistence is taken in account, acceleration will be altered and is air resistence is in opposite direction proportional to speed.
here final acceleration is already given, so not worry.
a= 10 - (1/3)v
since v = u + at => v = 0 + (10-(1/3)v)t => t = v/(10 - (v/3)) ........(2) and also s=ut +0.5gt2 => 5000 = 0.5* (10-(1/3)v) * t2 now put the value of t form eq.2
we get after re arranging....3v2 + 5000v -30*5000=0
solving this we get v= 29.47m/s and other negative one we will discard as it means retardation.
this is the answer for second part of this question which will use to get the time.
since t = v/(10 - (v/3)) => t = 166.81sec
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