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“A particleis travelling is a straightline with acceleration a passes A whilst movingwith velocityu , when it reaches a point B it has velocityv and at this point its acceleration changes to -a show that when it again passes by A its
Speed is squre root of 2v^2-u^2
Speed is squre root of 2v^2-u^2
6 years ago
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2 Answers
First part path AB: using suvat eqns get v^2= u^2 + 2as where s is the fixed displacement AB (remember all these quantities are vectors so +ve is direction from AB).
so solving for s= (v^2 - u^2)/2a
[each part of motion has its own set of suvat values]
Second part, particle must travel from pt B back again to A (eventually) so its displacement here will be -s (note the negative value)
using v`^2= u^2 + 2as again, but this time substituting v for u, -a for a and -s for s, then letting v` be the final velocity as it returns to pt.A:
v`^2= v^2 +2(-a)(-s) = v^2 +2as = v^2 + 2a[(v^2 - u^2)/2a]
= v^2 + [v^2 - u^2] = 2v^2 - u^2
taking sqr. get speed |v`|...
speed = SQR[2v^2 - u^2] as rqd.
note that the displacement s has same magnitude both parts of the problem, we are using displacement not distance.
so solving for s= (v^2 - u^2)/2a
[each part of motion has its own set of suvat values]
Second part, particle must travel from pt B back again to A (eventually) so its displacement here will be -s (note the negative value)
using v`^2= u^2 + 2as again, but this time substituting v for u, -a for a and -s for s, then letting v` be the final velocity as it returns to pt.A:
v`^2= v^2 +2(-a)(-s) = v^2 +2as = v^2 + 2a[(v^2 - u^2)/2a]
= v^2 + [v^2 - u^2] = 2v^2 - u^2
taking sqr. get speed |v`|...
speed = SQR[2v^2 - u^2] as rqd.
note that the displacement s has same magnitude both parts of the problem, we are using displacement not distance.
This is a SUVAT problem.
V^2 = U^2 + 2as [1], where s = dsitance A to B.
Now on the way back, B to A, its initial speed U = V.
i.e. After B, The paticle will slow to zero, stop and return to point B with
speed V now in the direction towards A by symmetry.
Lets call its new final velocity when it returns to A, V1.
V1^2 = V^2 + 2as [2],
But [1] ==> 2as = V^2-U^2,
Sub this in [2],
V1^2 = V^2 + (V^2-U^2),
V1^2 = 2V^2 -U^2 QED
V^2 = U^2 + 2as [1], where s = dsitance A to B.
Now on the way back, B to A, its initial speed U = V.
i.e. After B, The paticle will slow to zero, stop and return to point B with
speed V now in the direction towards A by symmetry.
Lets call its new final velocity when it returns to A, V1.
V1^2 = V^2 + 2as [2],
But [1] ==> 2as = V^2-U^2,
Sub this in [2],
V1^2 = V^2 + (V^2-U^2),
V1^2 = 2V^2 -U^2 QED
Answered by [Deleted Member]
