Free Study Material

NCERT Solutions

NCERT Books

Revision Notes

Practice Problems

This is one of our many quizzes on Motion in two dimension. Click on this to access more such competitive quizzes on Kinematics.

Motion in two dimension Quiz 2

physics

 If $\overset{\to }{\mathop{F}}\,=(2i+aj+7k)N$acts and displaces the body by $\overset{\to }{\mathop{s}}\,=(-2i+3j+2k)m$, then, the work done will be  $w=4\,J$. Find the acceleration of the body if the mass of the body is $2\text{ kg}$. 

 The work done is a scalar quantity and it is the product of vectors, that is, force and displacement.So, the work done will be the dot product of force and displacement. 

  $w=\overset{\to }{\mathop{F}}\,.\,\overset{\to }{\mathop{s}}\, \\ \Rightarrow 4=(2i+aj+7k).(-2i+3j+2k) \\ \Rightarrow 4=(2\times -2)+(a\times 3)+(7\times 2) \\ \Rightarrow 4=-4+3a+14 \\ \Rightarrow -6=3a \\ \therefore a=-2 \\ \text{The equation of force,} \\ \vec{F}=m\vec{a} \\ \text{Substituting the values,} \\ \Rightarrow 2\hat{i}+-2\hat{j}+7\hat{k}=2\left( {\vec{a}} \right) \\ \therefore \vec{a}=\hat{i}-\hat{j}+3.5\hat{k}$

The force experienced when the particle projects from point $P$ to point $Q$ for a time interval $t=2\text{ }s$ will be,<br><img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/6680f8eb-0302-4ac7-ab6a-f98e0a97330e7681250893419615287.png"/> 

 The momentum is the product of mass, the velocity and the sine of the angle of projection. The change in momentum equals the difference between the initial and final momentum.  

  $\text{The initial momentum, }{{p}_{i}}=mu\sin 30{}^\circ  \\ \text{The final momentum, }{{p}_{f}}=-mu\sin 30{}^\circ  \\ \text{The change in momentum, }p={{p}_{i}}-{{p}_{f}} \\ \Rightarrow p=mu\sin 30{}^\circ -(-mu\sin 30{}^\circ ) \\ \Rightarrow p=\dfrac{mu}{2}-\left( -\dfrac{mu}{2} \right) \\ \Rightarrow p=2\times \dfrac{mu}{2} \\ \therefore p=mu \\ \text{The equation of force experienced,} \\ F=\dfrac{\text{change in momentum}}{\text{change in time}} \\ \Rightarrow F=\dfrac{p}{t} \\ \text{Substituting the values,} \\ \text{F=}\dfrac{mu}{t} \\ \therefore F=\dfrac{mu}{2}$

 Consider two projectile motions with ${{u}_{1}}={{u}_{2}}$ and making ${{\theta }_{1}}=30{}^\circ $, ${{\theta }_{2}}=60{}^\circ $ with the horizontal. Find the ratio of $\dfrac{{{h}_{\max 1}}}{{{h}_{\max 2}}}$. 

 Consider the formula for computing the maximum height. As the maximum height is proportional to the sine of an angle, so, represent the heights in form of ratio in terms of angles and then find the ratio of heights by substituting the values of angles.  

 ${{h}_{\max }}=\dfrac{{{u}^{2}}{{\sin }^{2}}\theta }{2g} \\ \Rightarrow {{h}_{\max }}\propto {{\sin }^{2}}\theta  \\ \Rightarrow \dfrac{{{h}_{\max 1}}}{{{h}_{\max 2}}}=\dfrac{{{\sin }_{1}}^{2}\theta }{{{\sin }_{2}}^{2}\theta } \\ \Rightarrow \dfrac{{{h}_{\max 1}}}{{{h}_{\max 2}}}=\dfrac{{{\sin }^{2}}30{}^\circ }{{{\sin }^{2}}60{}^\circ } \\ \dfrac{{{h}_{\max 1}}}{{{h}_{\max 2}}}=\dfrac{1}{4}\times \dfrac{4}{3} \\ \therefore \dfrac{{{h}_{\max 1}}}{{{h}_{\max 2}}}=\dfrac{1}{3}$

 A passenger sitting on an exit seat jumps from an aeroplane moving with ${{u}_{H}}=630\,{km}/{hr}\;$ from $h=250\,m$. The horizontal range of the passenger will be,  

The speed unit must be converted to a SI unit. The product of speed and time is the horizontal range. By the acceleration due to gravity, time equals the root of twice the height. 

 $\text{The horizontal distance formula} \\ x=ut \\ \text{The speed of the aeroplane equals the speed with which the passenger jumps}\text{.} \\ \Rightarrow R={{u}_{H}}\times \sqrt{\dfrac{2h}{g}} \\ \Rightarrow R=630\times \dfrac{1000}{60\times 60}\times \sqrt{\dfrac{2\times 250}{9.8}} \\ \therefore R=1250\,m$

 The position vectors of particles $\overset{\to }{\mathop{{{a}_{1}}}}\,=(-i-2j)\,m$and $\overset{\to }{\mathop{{{a}_{2}}}}\,=(2i+2j)\,m$moving with velocities $\overset{\to }{\mathop{{{v}_{1}}}}\,=(3i+4j)\,{m}/{s}\;$and $\overset{\to }{\mathop{{{v}_{2}}}}\,=(xi+3j)\,{m}/{s}\;$ collide after $5\,s$. Find out the resultant momentum if the total mass is $6\text{ }kg$. 

 The displacement vector is the difference between the position vectors. The resultant  velocity vector in this question is also the difference between the velocity vectors. The velocity is displacement by time. Using these calculations, compute the value of the variable. 

 $\text{The displacement vector, }\overset{\to }{\mathop{a}}\,=\overset{\to }{\mathop{{{a}_{2}}}}\,-\overset{\to }{\mathop{{{a}_{1}}}}\, \\ \Rightarrow \text{ }\overset{\to }{\mathop{a}}\,=(2i+2j)-(-i-2j) \\ \Rightarrow \text{ }\overset{\to }{\mathop{a}}\,=3i+4j \\ \therefore \left| \overset{\to }{\mathop{a}}\, \right|=\sqrt{{{3}^{2}}+{{4}^{2}}}=5\,m \\ \text{The resultant velocity, }\overset{\to }{\mathop{v}}\,=\overset{\to }{\mathop{{{v}_{2}}}}\,-\overset{\to }{\mathop{{{v}_{1}}}}\, \\ \Rightarrow \text{ }\overset{\to }{\mathop{v}}\,=(xi+3j)-(3i+4j) \\ \Rightarrow \text{ }\overset{\to }{\mathop{v}}\,=(x-3)i-j \\ \Rightarrow \left| \overset{\to }{\mathop{v}}\, \right|=\sqrt{{{(x-3)}^{2}}+{{(-1)}^{2}}} \\ \therefore \left| \overset{\to }{\mathop{v}}\, \right|=\sqrt{{{(x-3)}^{2}}+1}\,m{{s}^{-1}} \\ \text{The velocity, }\overset{\to }{\mathop{v}}\,=\dfrac{\overset{\to }{\mathop{a}}\,}{t} \\ \Rightarrow \sqrt{{{(x-3)}^{2}}+1}\,=\dfrac{5}{5} \\ \Rightarrow {{(x-3)}^{2}}+1=1 \\ \Rightarrow {{(x-3)}^{2}}=0 \\ \therefore x=3 \\ \text{Substituting this in the equation of resultant velocity}, \\ \vec{v}=\sqrt{{{(x-3)}^{2}}+1}\text{ }\,m{{s}^{-1}} \\ \Rightarrow \vec{v}=\sqrt{{{\left( 3-3 \right)}^{2}}+1}\text{ }m{{s}^{-1}} \\ \therefore \vec{v}=1\text{ }m{{s}^{-1}} \\ \text{The equation of resultant momentum,} \\ \vec{p}=m\vec{v} \\ \text{Substituting the values,} \\ \vec{p}=6\times 1 \\ \therefore \vec{p}=6\text{ }kgm{{s}^{-1}}$