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Question

A particle of mass 13 kg moves with 5 m/s in yz plane along the path 5y=12z+60. Here y and z are in metre. Magnitude of angular momentum of the particle about origin (in kg m2s1), when the particle is at z=10 m, is

[Your answer must be an integer]

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Solution

Given, 5y=12z+60 ... (1)
or 5dydt=12dzdt ... (2)
As v=5 m/s,
v2y+v2z=25
25=(dydt)2+(dzdt)2
25=14425(dzdt)2+(dzdt)2, from (2)
(dzdt)=2513
and so, dydt=2513×125=6013, from (2)
v=6013^j+2513^k
As z=10,
y=1805=36
i.e., P(0,36,10)
L=r×mv
=(6013^j+2513^k)×13(36^j+10^k)
=∣ ∣ ∣^i^j^k0602503610∣ ∣ ∣=^i(900600)=300^i
L=300

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