anonymous
  • anonymous
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)
Physics
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anonymous
  • anonymous
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)
Physics
katieb
  • katieb
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anonymous
  • anonymous
|dw:1358017831169:dw|
anonymous
  • anonymous
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
anonymous
  • anonymous
|dw:1358018149270:dw|

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anonymous
  • anonymous
R is the range ..(X com) is the centre of mass of the system..
anonymous
  • anonymous
@Diwakar i also used Same Kind of ur Approach But Didnt Get the Result..
anonymous
  • anonymous
and..@Taufique i did nt get ur Approach
anonymous
  • anonymous
there is no external force so centre of mass will be same at intial and final so ,i have applied this concept..
anonymous
  • anonymous
Centre of mass Will Be near to Heavier Body....
anonymous
  • anonymous
and using the Formula for Range i calculated Pos ( CM) = 960 m..ryt ?
anonymous
  • anonymous
The c.m. will finally be at R. It will follow a parabolic trajectory
anonymous
  • anonymous
The answer shuld be1120m
anonymous
  • anonymous
@Yahoo! ..wait one minute
anonymous
  • anonymous
@Yahoo! ..what is the answer in your text book for this question
anonymous
  • anonymous
1120m
anonymous
  • anonymous
sorry yahoo (xcom) without explosion =4m *R xcom after explosion =m*R/2 + k*3m ..then equate both then you will get right answer
anonymous
  • anonymous
@ yahoo! ..i am sorry for this mistake ...
anonymous
  • anonymous
@Yahoo! ..here k =1121.47 m answer
anonymous
  • anonymous
No Problem bro
anonymous
  • anonymous
4mR = m*R/2 + k*3m hw R/2 ????????
anonymous
  • anonymous
here explosion is self explosion i.e no external force is here so we apply this method
anonymous
  • anonymous
Any Reason Behind it ?..
anonymous
  • anonymous
because small part is in rest and it will drop vertically downward at R/2 distance from origin
anonymous
  • anonymous
we also solve this problem using momentum conservation ...
anonymous
  • anonymous
If u Dont Mind..Can u Show ?
anonymous
  • anonymous
i appl that method..but didnt get the result
anonymous
  • anonymous
ok..|dw:1358020235037:dw|
anonymous
  • anonymous
|dw:1358020365409:dw| |dw:1358020404165:dw| and find k..
anonymous
  • anonymous
|dw:1358020473006:dw|
anonymous
  • anonymous
|dw:1358020525435:dw|
anonymous
  • anonymous
any query, yahoo!
anonymous
  • anonymous
Got it..! THXXX
anonymous
  • anonymous
now i am giving you a question which is similar to it ..you are interested to solve this
anonymous
  • anonymous
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 ???
anonymous
  • anonymous
@Yahoo! ..solve this ..it will give you more concept on this type question
anonymous
  • anonymous
Thxx...Brother.....i will try to Solve it
anonymous
  • anonymous
ok... try it.
anonymous
  • anonymous
35 ??
anonymous
  • anonymous
@Taufique wat is the answer
anonymous
  • anonymous
y=-5cm
anonymous
  • anonymous
hw....?
anonymous
  • anonymous
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
anonymous
  • anonymous
Huh..Silly....Careless
anonymous
  • anonymous
Didnt See it was on x-axis.......@Taufique if it was y axis..then the answer would be 35 ryt ?
anonymous
  • anonymous
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?
anonymous
  • anonymous
No..i Dont think
anonymous
  • anonymous
Since it is moving..along Y-axis...it will have an Initial P
anonymous
  • anonymous
no in the question ,it has been given that the body is moving along horizontal direction i.e X-axis
anonymous
  • anonymous
If it was Along Y-Along
anonymous
  • anonymous
*Axis
anonymous
  • anonymous
then the momentum will be zero along x -axis at any time..because there is no component of velocity along Xaxis
anonymous
  • anonymous
where is your confusion???
anonymous
  • anonymous
Cleared...
anonymous
  • anonymous
|dw:1358022098651:dw|
anonymous
  • anonymous
y=k (let ) for 3m/4 mass |dw:1358022439653:dw|
anonymous
  • anonymous
Got...It
anonymous
  • anonymous
ok..
anonymous
  • anonymous
Can u Take a luk at mmy Maths Question

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