## Yahoo! Group Title 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) one year ago one year ago

1. Taufique

|dw:1358017831169:dw|

2. Diwakar

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

3. Taufique

|dw:1358018149270:dw|

4. Taufique

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

5. Yahoo!

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

6. Yahoo!

and..@Taufique i did nt get ur Approach

7. Taufique

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

8. Yahoo!

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

9. Yahoo!

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

10. Diwakar

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

11. Yahoo!

12. Taufique

@Yahoo! ..wait one minute

13. Taufique

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

14. Yahoo!

1120m

15. Taufique

sorry yahoo (xcom) without explosion =4m *R xcom after explosion =m*R/2 + k*3m ..then equate both then you will get right answer

16. Taufique

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

17. Taufique

@Yahoo! ..here k =1121.47 m answer

18. Yahoo!

No Problem bro

19. Yahoo!

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

20. Taufique

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

21. Yahoo!

Any Reason Behind it ?..

22. Taufique

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

23. Taufique

we also solve this problem using momentum conservation ...

24. Yahoo!

If u Dont Mind..Can u Show ?

25. Yahoo!

i appl that method..but didnt get the result

26. Taufique

ok..|dw:1358020235037:dw|

27. Taufique

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

28. Taufique

|dw:1358020473006:dw|

29. Taufique

|dw:1358020525435:dw|

30. Taufique

any query, yahoo!

31. Yahoo!

Got it..! THXXX

32. Taufique

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

33. Taufique

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 ???

34. Taufique

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

35. Yahoo!

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

36. Taufique

ok... try it.

37. Yahoo!

35 ??

38. Yahoo!

39. Taufique

y=-5cm

40. Yahoo!

hw....?

41. Taufique

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

42. Yahoo!

Huh..Silly....Careless

43. Yahoo!

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

44. Taufique

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?

45. Yahoo!

No..i Dont think

46. Yahoo!

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

47. Taufique

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

48. Yahoo!

If it was Along Y-Along

49. Yahoo!

*Axis

50. Taufique

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

51. Taufique

52. Yahoo!

Cleared...

53. Taufique

|dw:1358022098651:dw|

54. Taufique

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

55. Yahoo!

Got...It

56. Taufique

ok..

57. Yahoo!

Can u Take a luk at mmy Maths Question

58. Yahoo!