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A projectile is fired upwards at a speed of 100 m/s at an angle of 37 with the horizontal.it explodes into two parts at the highest point in the mass ratio 1:3 and the lighter one comes to rest. The distance from the point of projection where the heavier mass lands is (g=10m/s^2)
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Taufique
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|dw:1358017831169:dw|
Diwakar
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At highest point v=100cos37=80m/s
If total mass of body is m, then masses of two pieces will be m/4 and 3m/4.
The lighter one comes to a rest .
By momentum conservation, velocity of 3m/4=4*80/3=320/3 m/s
This new projectile will now cover a horizontal distance from its point of formation i.e. the highest point in the trajectory of previous projectile ,s=v*sqrt(2h/g) where v=320/3 as found before. and h is the point of maximum height found by applying kinematics to the projectile of mass m.
The total distance from point of projection =s+ half of the range of previous projectile.
The range can be calculated just like maximum height i.e. apply kinematics to it.
I find it convinient to remember the formulas for range and max height
h=(v sintheta)^2/2g
Range,r=v^2 sin(2*theta)/g
here
theta=37
v=100
Taufique
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|dw:1358018149270:dw|
Taufique
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R is the range ..(X com) is the centre of mass of the system..
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@Diwakar i also used Same Kind of ur Approach But Didnt Get the Result..
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and..@Taufique i did nt get ur Approach
Taufique
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there is no external force so centre of mass will be same at intial and final so ,i have applied this concept..
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Centre of mass Will Be near to Heavier Body....
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and using the Formula for Range i calculated Pos ( CM) = 960 m..ryt ?
Diwakar
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The c.m. will finally be at R. It will follow a parabolic trajectory
Yahoo!
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The answer shuld be1120m
Taufique
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@Yahoo! ..wait one minute
Taufique
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@Yahoo! ..what is the answer in your text book for this question
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1120m
Taufique
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sorry yahoo (xcom) without explosion =4m *R
xcom after explosion =m*R/2 + k*3m ..then equate both then you will get right answer
Taufique
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@ yahoo! ..i am sorry for this mistake ...
Taufique
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@Yahoo! ..here k =1121.47 m answer
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No Problem bro
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4mR = m*R/2 + k*3m
hw R/2 ????????
Taufique
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here explosion is self explosion i.e no external force is here so we apply this method
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Any Reason Behind it ?..
Taufique
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because small part is in rest and it will drop vertically downward at R/2 distance from origin
Taufique
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we also solve this problem using momentum conservation ...
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If u Dont Mind..Can u Show ?
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i appl that method..but didnt get the result
Taufique
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ok..|dw:1358020235037:dw|
Taufique
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|dw:1358020365409:dw|
|dw:1358020404165:dw|
and find k..
Taufique
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|dw:1358020473006:dw|
Taufique
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|dw:1358020525435:dw|
Taufique
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any query, yahoo!
Yahoo!
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Got it..! THXXX
Taufique
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now i am giving you a question which is similar to it ..you are interested to solve this
Taufique
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An isolated particle of mass m is moving in horizontal plane (X-Y),along x-axis,at a certain height above the ground.it suddenly exlodes into two fragment of masses m/4 and 3m/4.An instant later the smaller fragment is at y=15 cm .what will be the position of larger fragment at this instant ???
Taufique
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@Yahoo! ..solve this ..it will give you more concept on this type question
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Thxx...Brother.....i will try to Solve it
Taufique
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ok... try it.
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35 ??
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@Taufique wat is the answer
Taufique
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y=-5cm
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hw....?
Taufique
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HINT: the body is moving on the x axis that is there is no motion along y axis this means the momentum will be zero along y-axis at any time
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Huh..Silly....Careless
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Didnt See it was on x-axis.......@Taufique if it was y axis..then the answer would be 35 ryt ?
Taufique
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now find momentum along y-axis for this situation.. after the body will explode ..the momentum will be zero along Y-axis ..am i right?
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No..i Dont think
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Since it is moving..along Y-axis...it will have an Initial P
Taufique
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no in the question ,it has been given that the body is moving along horizontal direction i.e X-axis
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If it was Along Y-Along
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*Axis
Taufique
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then the momentum will be zero along x -axis at any time..because there is no component of velocity along Xaxis
Taufique
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where is your confusion???
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Cleared...
Taufique
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|dw:1358022098651:dw|
Taufique
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y=k (let ) for 3m/4 mass
|dw:1358022439653:dw|
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Got...It
Taufique
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ok..
Yahoo!
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Can u Take a luk at mmy Maths Question