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)

- anonymous

- chestercat

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- anonymous

|dw:1358017831169:dw|

- 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

|dw:1358018149270:dw|

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## More answers

- anonymous

R is the range ..(X com) is the centre of mass of the system..

- anonymous

@Diwakar i also used Same Kind of ur Approach But Didnt Get the Result..

- anonymous

and..@Taufique i did nt get ur Approach

- anonymous

there is no external force so centre of mass will be same at intial and final so ,i have applied this concept..

- anonymous

Centre of mass Will Be near to Heavier Body....

- anonymous

and using the Formula for Range i calculated Pos ( CM) = 960 m..ryt ?

- anonymous

The c.m. will finally be at R. It will follow a parabolic trajectory

- anonymous

The answer shuld be1120m

- anonymous

@Yahoo! ..wait one minute

- anonymous

@Yahoo! ..what is the answer in your text book for this question

- anonymous

1120m

- 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

@ yahoo! ..i am sorry for this mistake ...

- anonymous

@Yahoo! ..here k =1121.47 m answer

- anonymous

No Problem bro

- anonymous

4mR = m*R/2 + k*3m
hw R/2 ????????

- anonymous

here explosion is self explosion i.e no external force is here so we apply this method

- anonymous

Any Reason Behind it ?..

- anonymous

because small part is in rest and it will drop vertically downward at R/2 distance from origin

- anonymous

we also solve this problem using momentum conservation ...

- anonymous

If u Dont Mind..Can u Show ?

- anonymous

i appl that method..but didnt get the result

- anonymous

ok..|dw:1358020235037:dw|

- anonymous

|dw:1358020365409:dw|
|dw:1358020404165:dw|
and find k..

- anonymous

|dw:1358020473006:dw|

- anonymous

|dw:1358020525435:dw|

- anonymous

any query, yahoo!

- anonymous

Got it..! THXXX

- anonymous

now i am giving you a question which is similar to it ..you are interested to solve this

- 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

@Yahoo! ..solve this ..it will give you more concept on this type question

- anonymous

Thxx...Brother.....i will try to Solve it

- anonymous

ok... try it.

- anonymous

35 ??

- anonymous

@Taufique wat is the answer

- anonymous

y=-5cm

- anonymous

hw....?

- 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

Huh..Silly....Careless

- anonymous

Didnt See it was on x-axis.......@Taufique if it was y axis..then the answer would be 35 ryt ?

- 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

No..i Dont think

- anonymous

Since it is moving..along Y-axis...it will have an Initial P

- anonymous

no in the question ,it has been given that the body is moving along horizontal direction i.e X-axis

- anonymous

If it was Along Y-Along

- anonymous

*Axis

- anonymous

then the momentum will be zero along x -axis at any time..because there is no component of velocity along Xaxis

- anonymous

where is your confusion???

- anonymous

Cleared...

- anonymous

|dw:1358022098651:dw|

- anonymous

y=k (let ) for 3m/4 mass
|dw:1358022439653:dw|

- anonymous

Got...It

- anonymous

ok..

- anonymous

Can u Take a luk at mmy Maths Question

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