In a 3D space without gravity, two particles are located at positions r1(vector) and r2(vector) respectively. They are moving with velocities v1(vector) and v2(vector) respectively.
@Mayank: what about the z component? @Shivam: I'm quite sure (v2-v1)X(r2-r1)=0 is correct. @Shourya: Though there's a significant flaw in your answer too.
If z component is taken into account,then would [(r2)x-(r1)x]/[(v1)x-(v2)x]=[(r2)y-(r1)y]/[(v1)y-(v2)y]=[(r2)z-(r1)z]/[(v1)z-(v2)z]>0 have been correct?
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ReplyDeleteX=7(24+13root(3))/23.
ReplyDeleteIs the answer approximately 14.05m?
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ReplyDeletetry not to use equations of motion for this question.
ReplyDelete14.15m
ReplyDeleteoh yes bhaiya u r right these type of problems r really easy.
ReplyDeleterelative velocity is the trick.
ReplyDeletesorry, a calculation error,
ReplyDeleteX=7(24+13√(3))/23 =14.15m ?
using relative velocity, it reduces to a simple trigonometry problem.
I think we have a correct answer... try this too:
ReplyDeleteIn a 3D space without gravity, two particles are located at positions r1(vector) and r2(vector) respectively. They are moving with velocities v1(vector) and v2(vector) respectively.
Write the condition for their collision.
not sure about this one, but is it
ReplyDelete(v2-v1)x(r2-r1)=0 ?
[(r2)x-(r1)x]/[(v1)x-(v2)x]=[(r2)y-(r1)y]/[(v1)y-(v2)y]>0 ?
ReplyDelete@shourya: Its correct.
ReplyDelete@Mayank: I think its wrong (what are x and y btw)
@All: Try the 2nd problem. Its better than the 1st one.
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ReplyDeletex and y represent the x-component and y-component respectively.
ReplyDeletei think it should be (r1-r2).(v1-v2)>0
ReplyDelete@Mayank: what about the z component?
ReplyDelete@Shivam: I'm quite sure (v2-v1)X(r2-r1)=0 is correct.
@Shourya: Though there's a significant flaw in your answer too.
Note that X denotes the vector cross product.
If z component is taken into account,then would
ReplyDelete[(r2)x-(r1)x]/[(v1)x-(v2)x]=[(r2)y-(r1)y]/[(v1)y-(v2)y]=[(r2)z-(r1)z]/[(v1)z-(v2)z]>0
have been correct?
i know bhaiya the particles might never collide!!!
ReplyDeletethats why i was uncertain the first time.
i think it should be (v2-v1)=k(r2-r1);k<0.
oh yes bahiya u r right their seperation and the velocity of approach have to be in same direction
ReplyDeleteI think Shourya's second post completes the answer, the velocity and the position must be antiparallel
ReplyDelete(v2-v1)x(r2-r1)=0
ReplyDelete(r1-r2).(v1-v2)>0
combining these two will make complete answer