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A particle moves with constant momentum = (10 kg. m/s) The particle has an angular momentum about the origin of = (20 kg. m2/s) when t = 0 s. (a) The magnitude of the angular momentum of this particle is (A) decreasing. (B) constant. (C) increasing. (D) possibly nut not necessarily constant. (b) The trajectory of this particle (A) definitely passes through the origin (B) might pass through the origin. (C) will not pass through the origin, but it is uncertain how close it will pass to the origin. (D) will not pass through the origin, but one can calculate exactly how close it will pass to the origin.

A particle moves with constant momentum  = (10 kg. m/s) The particle has an angular momentum about the origin of  = (20 kg. m2/s)when t = 0 s.
(a) The magnitude of the angular momentum of this particle is
(A) decreasing.                
(B) constant.                    
(C) increasing.               
(D) possibly nut not necessarily constant.
(b) The trajectory of this particle
(A) definitely passes through the origin
(B) might pass through the origin.
(C) will not pass through the origin, but it is uncertain how close it will pass to the origin.
(D) will not pass through the origin, but one can calculate
exactly how close it will pass to the origin.

Grade:upto college level

2 Answers

Deepak Patra
askIITians Faculty 471 Points
8 years ago
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zhi zheng ong
26 Points
one year ago
since momentum is constant, its velocity must also be constant also\
 
\begin{aligned} L &= r \times p \\ \frac{dL}{dt} &= \left( \frac{dr}{dt} \times p \right ) + \left(r \times\frac{dp}{dt} \right ) \\ &= (v \times mv) + (r \times \frac{dv}{dt}) \\ &= 0 \end{aligned}
 
on the 2nd last line, any vector crossed with itself will be zero
also, dv/dt is zero as well, because it is constant
 
now since dL/dt is zero, and it is initially non-zero (20kgm^2/s) then that means angular momentum is always constant.

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