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This is one of our many quizzes on Newton's third law. Click on this to access more such competitive quizzes on Dynamics.

Newton's third law Quiz 2

physics

 Which of the following is the necessary requirement for Newton’s ${{3}^{rd}}$ law to be applicable? 

 Newton’s ${{3}^{rd}}$ law explains the nature of force. If and only if internal forces are balanced, then the action and reaction force will act on two different bodies which must be in contact with each other. 

  For Newton’s ${{3}^{rd}}$ law to be applicable, the two bodies should be necessarily in contact with each other so that the action and reaction forces could act on them accordingly. Also, the internal forces should be balanced so that these forces do not interfere between the action and reaction forces.

After analysing the below shown figure, which of the following conclusions brought up is false? Consider the string to be inextensible. <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/16e80f2a-1cea-4a4b-ac39-d1189312030a624838780233543790.png"/>

 Since the length of the string can not be changed, then if block $B$ goes upwards by distance $x$, then block $A$ will come down by $2x$ length. Equate and differentiate both sides w.r.t. time and relate them with options. 

  Here, it should be noted that the string is inextensible. Therefore, the length of the string can not change. Consider that the B moves by a certain distance ‘x’ during an interval of time. This will cause movement of pulley $K$ by that very distance. An extra length of ‘2x’ of string will come to the left of the pulley $L$. This must be coming from the right side of the pulleys. Hence, displacement of $A$ will be $2x$. On this basis, one can say that if the acceleration of block B is ‘a’ then the acceleration of $A$ is $2a$.

Two blocks $A$ and $B$ are joined together and moving as shown in the figure. if${{v}_{1}}$ and ${{v}_{2}}$ are the respective speeds of the block $A$ and $B$, then the relation between the two velocities is? <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/b72c6c39-42d0-475e-a0b3-2d861bbcb6c47860796611633168832.png"/>

 Write the constrained equation and differentiate it with respect to time to get the equations or formula for the velocities of the 2 blocks. Calculate the value of $\cos \theta $ from the given figure to get the required relation. 

<img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/d81fd766-0d3e-4cad-9934-4f124e43e8f63127549455086255548.png"/> $\text{Distances are taken from the center of the pulley as shown in the } \\ \text{figure} \\ \text{Constrained equation:} \\ \sqrt{{{x}_{1}}^{2}+{{h}_{1}}^{2}}+{{x}_{2}}=\text{ constant} \\ \text{Differentiating both sides w}\text{.r}\text{.t}\text{. time} \\ \dfrac{2{{x}_{1}}}{2\sqrt{{{x}^{2}}_{1}+{{h}^{2}}_{1}}}\dfrac{d{{x}_{1}}}{dt}+\dfrac{d{{x}_{2}}}{dt}=0 \\ \text{Since the ball moves so as to increase }{{x}_{1}}\text{ with time and block moves } \\ \text{so as to decrease }{{x}_{2}}\text{ with time, therefore,} \\ \dfrac{d{{x}_{1}}}{dt}=+{{v}_{1}}\text{ and }\dfrac{d{{x}_{2}}}{dt}=-{{v}_{2}} \\ \text{Also, }\dfrac{{{x}_{1}}}{\sqrt{{{x}^{2}}_{1}+{{h}_{1}}^{2}}}=\cos {{30}^{0}} \\ \text{Thus, }{{v}_{1}}\cos {{30}^{0}}-{{v}_{2}}=0 \\ \text{or }{{\mathbf{v}}_{2}}={{\mathbf{v}}_{1}}\cos {{30}^{0}}$

A massive rope of mass $2\,kg$ is used to pull a block of mass $20\,kg$ with a force of $25\,N$ as shown in the diagram. If the box starts moving in the direction of force applied, then what is the tension at the midpoint of the rope? <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/339ebaa1-3b8a-4146-ade7-84c2342f46522811083198768606245.png"/>

 Tension at a particular point on the string depends upon the weight it has to pull. Calculate tension at the midpoint using the effective mass it has to pull. 

 $\text{Since the rope also possess some mass here, therefore, at the mid point, total tension will be } \\ \text{because of the weight of the box and half the weight of the rope}\text{.} \\ \text{Let }a\text{ be the acceleration produced because of the applied force}\text{.} \\ \Rightarrow \text{ Tension at Midpoint = }\left( M+\dfrac{m}{2} \right)a\text{ where M is mass of box and m is mass of string}\text{.} \\ \text{Also,    } \\ \text{ }F=\left( M+m \right)a \\ \Rightarrow a=\dfrac{F}{M+m} \\ \Rightarrow \text{ Required tension = }\dfrac{\left( 2M+m \right)}{2\left( M+m \right)}F \\ \therefore \text{ Required tension} =\dfrac{\left( 2\times 20+2 \right)}{2\left( 20+2 \right)}=23.863\,N$

Consider 2 blocks $A$ and $B$ of masses $5\,kg\text{ and }10\,kg$ respectively, resting upon each other as shown. If a force of $50\,N$ is applied upon box $A$, then after what time will A will fall off from $B$? Coefficient of kinetic friction between $A$ and $B$ is $0.4$. <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/2289f48a-f33d-480f-86fa-c2deee18eb6d1971393369988073185.png"/>

 Calculate the individual accelerations of the blocks and hence find the acceleration $A$ relative to $B$.By using $S=\dfrac{1}{2}a{{t}^{2}}$,calculate the required time. 

 $\text{Calculating the individual accelerations of the 2 blocks} \\ \text{For block A,} \\ F-{{F}_{k}}=m{{a}_{A}}\text{      Where }{{\text{F}}_{k}}\text{ is the kinetic friction} \\ \Rightarrow {{a}_{A}}=\dfrac{\left( F-{{\mu }_{k}}mg \right)}{m} \\ \Rightarrow {{a}_{A}}=\dfrac{\left( 50-0.4\times 5\times 10 \right)}{5}=6m{{s}^{-2}} \\ \text{For block B,} \\ {{F}_{k}}=M{{a}_{B}} \\ \Rightarrow {{a}_{B}}=\dfrac{{{\mu }_{k}}Mg}{M} \\ \Rightarrow {{a}_{B}}=\dfrac{0.4\times 10\times 10}{10}=4m{{s}^{-2}} \\ \Rightarrow \text{ Acceleration of body A relative to B will be -} \\ \text{ }a={{a}_{A}}-{{a}_{b}}=6-4=2m{{s}^{-2}} \\ \text{Therefore, A will fall from B after time t given by - } \\ \sqrt{\dfrac{2L}{a}}..................\text{ where L is the length of the box B} \\ t=\sqrt{\dfrac{2\times 2}{2}}=1.414\,s$