Question

A block of mass $2\,\,Kg$ is moving with a velocity of $2i - j + 3k\,\,m{s^{ - 1}}$. Find the magnitude and direction of momentum of the block with the $x$−axis.

Answer 1

Hint
A quantity that has both the magnitude and the direction is known as the vector quantity. It is generally represented by an arrow where the direction is the same as that of the quantity and whose length is directly proportional to the quantity's magnitude.
The formula for momentum of the block along the $x$-axis is given as;
$\Rightarrow \overrightarrow P = m \times \overrightarrow v$
Where, $\overrightarrow P$ denotes the momentum of the block along the $x$-axis, $\overrightarrow v$ denotes the velocity by which the block moves, $m$ denotes the mass of the moving block.

Complete step by step answer
The data given the above problem are;
Mass of the block mass is, $m = 2\,\,Kg$.
Velocity of the moving block is, $\overrightarrow v = 2i - j + 3k\,\,m{s^{ - 1}}$
momentum of the block along the $x$-axis is given as;
$\overrightarrow P = m \times \overrightarrow v$
Substitute the values of the mass of the block and the velocity of the block in the above equation;
$\Rightarrow \overrightarrow P = 2 \times \left( {2i - j + 3k} \right)$
$\Rightarrow \overrightarrow P = 4i - 2j + 6k\,\,m{s^{ - 1}}$
magnitude of momentum of the block along the $x$−axis:
$\Rightarrow \left| P \right| = \sqrt {{{\left( 4 \right)}^2} - {{\left( 2 \right)}^2} + {{\left( 6 \right)}^2}}$
$\Rightarrow \left| P \right| = \sqrt {16 - 4 + 36}$
$\Rightarrow \left| P \right| = 2\sqrt {14} \,\,m{s^{ - 1}}$
That is the magnitude of momentum of the block along the x−axis is $\left| P \right| = 2\sqrt {14} \,\,m{s^{ - 1}}$.
The direction of momentum of the block along the $x$-axis:
$\Rightarrow \tan \theta = \dfrac{4}{{\sqrt {{{\left( 4 \right)}^2} - {{\left( 2 \right)}^2} + {{\left( 6 \right)}^2}} }}$
$\Rightarrow \tan \theta = \dfrac{4}{{\sqrt {16 - 4 - 36} }}$
$\Rightarrow \tan \theta = \sqrt {\dfrac{2}{7}}$
$\Rightarrow \theta = {\tan ^{ - 1}}\sqrt {\dfrac{2}{7}}$
That is the magnitude of momentum of the block along the x−axis is $\theta = {\tan ^{ - 1}}\sqrt {\dfrac{2}{7}}$.
Therefore, the magnitude of momentum of the block along the x−axis is $\left| P \right| = 2\sqrt {14} \,\,m{s^{ - 1}}$and the magnitude of momentum of the block along the x−axis is $\Rightarrow \theta = {\tan ^{ - 1}}\sqrt {\dfrac{2}{7}}$.

Note
While a vector has magnitude and direction, it does not have a particular position. Momentum definitely has a direction, that is it is directly proportional to the velocity. Velocity is a measure of a body's speed with its direction as well.