Bomb explosion -- calculating the momentum of one of the fragments

Click For Summary
The discussion focuses on calculating the momentum of fragments from a bomb explosion, emphasizing that the total momentum must balance out regardless of the direction of the fragments. One participant initially expresses doubt about their approach but later acknowledges that the direction of movement does not affect the overall momentum balance. The importance of considering the resultant vector in opposite directions is highlighted. A request for verification of another post is made, indicating a desire for further clarification on related topics. The conversation ultimately reinforces the principle of momentum conservation in explosive events.
rudransh verma
Gold Member
Messages
1,067
Reaction score
96
Homework Statement
A bomb at rest explodes into 3 parts of the same mass. The momentum of the 2parts is -2pi^ and pj^. The momentum magnitude of the third part will be
Relevant Equations
##\vec R=\sqrt{{a\hat i}^2+{b\hat j}^2}##
I think if the two parts move in -x and +y direction, it must be balanced by the resultant of the two vectors but in opposite direction.
So ##p\sqrt5## will be the answer.
But I don’t think this is the right way to solve this.
 
Physics news on Phys.org
rudransh verma said:
I think if the two parts move in -x and +y direction, it must be balanced by the resultant of the two vectors but in opposite direction.
It doesn't matter what direction the two parts move. Their total momentum must be balanced by that of the third part.
rudransh verma said:
But I don’t think this is the right way to solve this.
Why not?
 
Doc Al said:
It doesn't matter what direction the two parts move. Their total momentum must be balanced by that of the third part.
Oh. So this is right. Thanks again.
Can you verify my other post?
 
rudransh verma said:
Can you verify my other post?
What post? Provide a link.
 
If have close pipe system with water inside pressurized at P1= 200 000Pa absolute, density 1000kg/m3, wider pipe diameter=2cm, contraction pipe diameter=1.49cm, that is contraction area ratio A1/A2=1.8 a) If water is stationary(pump OFF) and if I drill a hole anywhere at pipe, water will leak out, because pressure(200kPa) inside is higher than atmospheric pressure (101 325Pa). b)If I turn on pump and water start flowing with with v1=10m/s in A1 wider section, from Bernoulli equation I...

Similar threads

Replies
13
Views
2K
Replies
4
Views
2K
  • · Replies 9 ·
Replies
9
Views
5K
  • · Replies 40 ·
2
Replies
40
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 4 ·
Replies
4
Views
839
  • · Replies 38 ·
2
Replies
38
Views
5K
  • · Replies 41 ·
2
Replies
41
Views
12K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 3 ·
Replies
3
Views
7K