For a), I got that
0.1g/1000g=0.0001kg
0.0001kg*120km/hr=0.12kg*km/hr
But I don't know how to get b and c...Please help me with momentum problem?
a) You didn't make any reference to the mass of the car. The original momentum of the car is now that of the car plus the flysplat on the window.
Let the speed of the (car+ flysplat) be v, where v = 120km/h - 鈭?v. Then
1400*120 = (1400 + 0.0001)*v = (1400.0001)*(120 - 鈭?v),
while the car's new momentum is 1400*v.
Since the change in mass is so small, It seems reasonable to approximate 鈭?v, rather than trying obtain it to some ridiculous level of precision beyond many of our calculators! That %26quot;approximate%26quot; 鈭?v then will be given by
鈭?v / 120km/h = 0.0001/1400 = 7.143... x 10^(-8)
So, the FRACTIONAL LOSS of momentum by the car is
~ 7.143... x 10^(-8) of its original momentum, or
0.012kg*km/h in rather bizarre units.
[Ah, now I see what you did. In what you claimed was your part (a), you actually worked out (b), the change in momentum of the FLY (not the car) with the ASSUMPTION that the car + flysplat would STILL be moving at exactly 120km/h AFTER the splatting had occurred. (The momentum changes to car and fly separately will of course be equal and opposite.) In fact, your assumption has the same minute error in it that I was prepared to accept in my approach because bothering about the correction to IT seemed absolutely nit-picking. Then my answer to part (a) should have had PRECISELY the same magnitude as your answer to part (b). So why do they appear to differ by a factor of 10? --- Because YOU dropped a power of 10 when multiplying 0.0001 [= 10^(-4)] and 120 [= 1.2*10^2]. The answer should have been 1.2 * 10^(-2) = 0.012, not your result of 0.12.]
b) As already explained, this part IS (very closely) exactly what you worked out. But whether you're working out a) or b) most directly, the OTHER can be obtained simply from the principle that the two of them MUST be equal in magnitude, if opposite in sign!
c) If ONE FLY reduces the speed of the car by a fraction
7.143... x 10^(-8),
then to reduce it by a fraction 1/120 (since 1km/h is 1/120 * 120km/h) will require
1/120 * [7.143... x 10^(-8)]^(-1) flies = 1.4 x 10^7 / 120 flies
= 1.167... x 10^5 flies.
(A more accurate calculation shows that it should take
1.4 x 10^7 / 119 flies = 1.176... x 10^5 flies, so the difference is still very small between the approximate and more exact result. By the way, 'Helmut,' below, dropped the 119 divisor; that's why he simply got the result 1.4 x 10^7 !)
In either case, that's going to be one fairly filthy windshield/windscreen, by the time the car has slowed down 1km/h due to this effect!
Live long and prosperPlease help me with momentum problem?
You're most welcome. Thank you for your appreciation.
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Please help me with momentum problem?(1400 kg)(120 km/hr) + (0.0001kg)(0 km/hr) = (1400.0001)v
v = (1400)(120)/1400.0001) 鈮?119.99999 14285 72040
a) for the car, 鈭唒 鈮?- 0.01199 99991 42857
b) for the fly, 鈭唒 鈮?0.01199 99991 42857
c) (1400 kg)(120 km/hr) + n(0.0001kg)(0 km/hr) = (1400 + n(0.0001))(119)
n = 1400(120 - 119)/0.0001
n 鈮?14,000,000 flies
