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A canon shell moving along a straight line burst into two parts. Just after the burst one part moves with momentum 20 Ns making an angle 30° with the original line of motion. The minimum momentum of the other parts of shell just after the burst is :
  • a)
    0 Ns 
  • b)
    5 Ns
  • c)
    17.32 Ns
  • d)
    10 Ns
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
A canon shell moving along a straight line burst into two parts. Just ...
As shown in figure the component of momentum of one shell along initial direction and perpendicular to initial direction are  and P1y=10 Ns.
For momentum of the system to be zero in y-direction P2y must be 10 Ns. 2nd part of shell may or not have momentum in x-direction.
∴  P2min = 10 Ns.
The correct answer is: 10 Ns
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Most Upvoted Answer
A canon shell moving along a straight line burst into two parts. Just ...
P1 = 20  Ns
Let the minimum momentum of  other part be  P2
Applying conservation of momentum in y direction :
P1sin30= P2​sinθ
20×1/2 ​= P2​sinθ
⟹P2​sinθ = 10 N s
Now for P2 to be minimum, sinθ must be maximum   i.e  sinθ=1
⟹ P​= 10  N s
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