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This is the second of our many quizzes on Add and subtract vectors. Click on this to access more such competitive quizzes on Mechanics.

Add and subtract vectors Quiz 2

physics

Two forces A and B $(\text{ at an angle with each other})$ are acting on a block such that resultant force displaces the block by $2\text{ }m$ and work done by the two forces is $4\sqrt{3}\text{ }J$. The resultant makes an angle of $30{}^\circ $ with the force B, is $\text{A }=2\text{N}$ then angle between A and B is: 

Parallelogram law of vector addition gives the resultant force. The resultant force displaces the block along its direction.

  <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/78c032bf-e118-4ee5-b762-58b394a3be813526207929497128054.png"/>$ \\ $ The resultant force will be the one performing the work. Therefore, $ \\ \dfrac{W}{s}=R \\\Rightarrow R=\dfrac{4\sqrt{3}\text{ }J}{2\text{ }m}=2\sqrt{3}\text{ }N \\ $ The perpendicular P can be written as $($considering the two right angled triangles$)$: $ \\P=R\sin 30{}^\circ\\ \Rightarrow P=A\sin \theta \\ \Rightarrow A\sin \theta =R\dfrac{1}{2} \\ \Rightarrow \sin \theta =\dfrac{\sqrt{3}}{2} \\ \therefore \theta =60{}^\circ$

A dart at a distance of $\text{800 m}$ from the target can be thrown with a velocity of $10\text{ }m{{s}^{-1}}$. If the wind speed is $5\text{ }m{{s}^{-1}}$, then at what angle should the dart be thrown? $(\text{ Neglect the effect of gravity})$ 

 Under the influence of the wind, the dart changes the direction as the velocity does not change and the resultant and of wind and dart velocity gives the required change to be made.

  <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/5f19b41d-1460-428f-a607-26d16b68fbf43974475998140849085.png"/>$ \\ $ If there was no wind, the dart would have hit the target, but wind makes it change the direction and it misses the target. $ \\ $ If we make a correction of this angle and throw the dart, the dart will hit the target again. $\\ $ The angle can be obtained as: $ \\ \sin \theta =\dfrac{\text{ wind}}{\text{dart}}=\dfrac{5}{10}=\dfrac{1}{2} \\ \therefore \theta =30{}^\circ\\ \text{ If we just turn the dart}\,30{}^\circ\\ $ away from the target $($opposing the wind flow$)$, then it lands on the target.

 A setting of a tug of war is made such that three pulling forces are $(\text{ centre being the origin}):$ $\\$ 1) $4\text{ N}$ along $+x$ direction, $\\$ 2) $\text{3 N}$ along the $+y$ direction, $\\$ 3) $\text{2 N}$ along the $-x$ direction, the resultant direction is given as:

All the given forces can be represented by vectors and resultant has to be found the same way i.e., vectorially.

 The difference of the two opposing forces is,$ \\2N\left( 4N-2N \right) \\ $ The two forces to consider now are: $ \\ 3\hat{j}\text{ }N\,\text{ and}\, 2\hat{i}\text{ }N\\ $ The sum of these two forces is just: $ \\ \overrightarrow{\text{R}}\text{=}2\hat{i}\text{+}3\hat{j}\\ $ Dividing this by magnitude, we find the unit vector or the direction of the force:$ \\ \dfrac{\overrightarrow{\text{R}}}{\left| R \right|}\text{=}\dfrac{2\hat{i}\text{+}3\hat{j}}{\sqrt{4+9}}\\ \therefore\dfrac{\overrightarrow{\text{R}}}{\left| R \right|}\text{= }\dfrac{2\hat{i}\text{+}3\hat{j}}{\sqrt{13}}\,N$

Two boys Phineas and Rodrick throw mud balls of the same mass at each other with constant velocities and they stick together. If the initial velocities of the two mud balls were<br>$\overrightarrow{{{\text{v}}_{1}}}=2\hat{i}+\hat{j}\text{ }m{{s}^{-1}}$ and $\overrightarrow{{{\text{v}}_{2}}}=-3\hat{i}+\hat{j}\text{ }m{{s}^{-1}}$. The final velocity has a magnitude of: 

The law of conservation of momentum helps us in determining the resultant velocities. 

 From momentum conservation, $ \\m\overrightarrow{{{v}_{1}}} + m\overrightarrow{{{v}_{2}}} = \vec{v}\left( m+m \right) \\ \Rightarrow \overrightarrow{{{v}_{1}}}+\overrightarrow{{{v}_{2}}}=2\vec{v} \\ \Rightarrow \left( 2\hat{i}+\hat{j} \right)+\left( -3\hat{i}+\hat{j} \right)=2\vec{v} \\ \Rightarrow -\dfrac{{\hat{i}}}{2}+\hat{j}=\vec{v} \\ $ The magnitude of this vector is: $ \\\left| {\vec{v}} \right|=\sqrt{{{\left( \dfrac{1}{2} \right)}^{2}}+{{1}^{2}}}\\\therefore\left| {\vec{v}} \right|=\dfrac{\sqrt{5}}{2}\text{ }m{{s}^{-1}}$

An electron moves with constant velocity along the $x$ axis, a constant electric field acts along the $y$ direction. Then, the velocity along $x$: 

Force has only been applied along the $y$ direction and not the $x$ direction, so the motion remains unaffected in the $x$ direction. 

 The electron motion after the application of the field can be described by two sets of equations: $ m{{a}_{y}}=eE $ Along the $y$ direction, and $m{{a}_{x}}=0$ Along the $x$ direction. This means that it suffers no acceleration along the $x$ direction. Therefore, the component of velocity along the $x$ direction remains the same. The resultant will definitely be different but along $x$, there will be no change.