Question:
A vector $Q$ has a magnitude of 8 is added to the vector $p$ which ties along the X-axis. The resultant of these two vectors is a third vector $R$ which lies along the Y-axis and has a magnitude twice that of $P$ The magnitude of $P$ is
A vector $Q$ has a magnitude of 8 is added to the vector $p$ which ties along the X-axis. The resultant of these two vectors is a third vector $R$ which lies along the Y-axis and has a magnitude twice that of $P$ The magnitude of $P$ is
Updated On: Apr 4, 2024
- $6/ \sqrt{5}$
- $8/ \sqrt{5}$
- $12/ \sqrt{5}$
- $16/ \sqrt{5}$
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The Correct Option is B
Solution and Explanation
Given Q = 8 units
$ R = 2 P$
Since. R is along Y-axis and P is along x-axis.
Therefore. P and R are perpendicular vectors.
Hence, $ Q^2 = R^2 + P^2$
Putting the given values in E (i), we get
$ (8)^2 = (2P)^2 + P^2 = 4P^2 + P^2 = 5 P^2 \, \, \, or \, \, \, 5 P^2 = 64$
$ P^2 = \frac{64}{5}$
$\therefore P = \frac{8}{\sqrt5}$
$ R = 2 P$
Since. R is along Y-axis and P is along x-axis.
Therefore. P and R are perpendicular vectors.
Hence, $ Q^2 = R^2 + P^2$
Putting the given values in E (i), we get
$ (8)^2 = (2P)^2 + P^2 = 4P^2 + P^2 = 5 P^2 \, \, \, or \, \, \, 5 P^2 = 64$
$ P^2 = \frac{64}{5}$
$\therefore P = \frac{8}{\sqrt5}$
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Concepts Used:
Motion in a Plane
It is a vector quantity. A vector quantity is a quantity having both magnitude and direction. Speed is a scalar quantity and it is a quantity having a magnitude only. Motion in a plane is also known as motion in two dimensions.
Equations of Plane Motion
The equations of motion in a straight line are:
v=u+at
s=ut+½ at2
v2-u2=2as
Where,
- v = final velocity of the particle
- u = initial velocity of the particle
- s = displacement of the particle
- a = acceleration of the particle
- t = the time interval in which the particle is in consideration