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Impulse Quiz 2

physics

Refer the diagram and calculate the impulse of the body for a time interval  from  $t=2\sec $ to $t=6\sec $. Mass of the body is $m$ $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/4682e881-cd3d-40e0-8044-748499ba38df839602475231305678.png"/> 

Remember the definition of Impulse in terms of momentum. Finding velocities for the required instances is the best way to approach problems like that. 

  Now By the definition of Impulse we know that, I impulse is the change in momentum. It can also be defined as the product of mass and change in velocity. So let us fid first the change in velocity; $ \\\Delta v={{v}_{6}}-{{v}_{2}} \\\Rightarrow \Delta v=8-8 \\\Rightarrow \Delta v=0m{{s}^{-1}} \\ $ Now if we calculate the impulse we get; $ \\J=m\Delta v \\\Rightarrow J=m\times 0 \\\therefore J=0$ 

$30$ bullets are fired in $5\sec $ from a gun. If the mass of one single bullet is $50\,g$ what should be the initial velocity of each bullet if $260\,N$ of rebounding force is acting on the gun? 

The Impulse for one bullet is given so find out the total impulse of all the bullets being fired.  Rebounding force is the force that the bullet exerts of the gun when it is shot.  

  Momentum for each bullet is 50 v. Thus, momentum for 30 bullets will be given by; $ \\{{P}_{t}}=50\times 30v=J\\ $ But we also know that $ \\ J=F\times t \\\Rightarrow 1500v=260\times 5 \\\therefore v=\dfrac{13}{15}m{{s}^{-1}}$ 

 A body $A$ of mass $m$ moving with velocity $u$ collides with another body $B$ of mass $2m$ . After collision both bodies move with equal velocities. Now body B collides with a third body $C$ initially at rest. After collision bodies $B$ and $C$ move with equal velocities. Find the impulse delivered on $A$, $B$ and $C$. If $C$ comes to rest after sometime. 

The momentum is conserved so conservation of momentum should be the best way to approach the problem. 

 Let us conserve the momentum for bodies A and B. So we have; $ \\m{{v}_{A}}+2m{{v}_{B}}=mu \\\Rightarrow {{v}_{B}}=\dfrac{u}{3} \\ $ Now impulse on A is $ \,mu-\dfrac{mu}{3}=\dfrac{2mu}{3}\\{{J}_{A}}=\dfrac{2mu}{3}\\ $ According to the second condition we have; $ \\2m{{v}_{B}}+3m{{v}_{C}}=\dfrac{2mu}{3} \\\Rightarrow 5m{{v}_{C}}=\dfrac{2mu}{3} \\\Rightarrow {{v}_{C}}=\dfrac{2u}{15} \\ $ Thus, Impulse on B is given by; $ \\{{J}_{B}}=\dfrac{2mu}{3}-\dfrac{4mu}{15} \\\Rightarrow {{J}_{B}}=\dfrac{6mu}{15}=\dfrac{2mu}{5} \\ $ And impulse on C is given by; $ \\{{J}_{C}}=mv=\dfrac{3m\times 2u}{15} \\\therefore {{J}_{C}}=\dfrac{2mu}{5}$

 A mass of $M$ is resting on the horizontal plane. This mass is connected to a string that passes over a pulley and is connected to another mass of $m$ in the air. To what height should the mass $m$ be lifted to and then allowed to fall such that the mass $M$ is lifted for a time $t$. 

The other definition of impulse is the product of force and the time interval for which the force is acting.

 The mass m undergoes a free fall motion so its velocity when the string is taut again can be given as; $ \\{{v}_{m}}=\sqrt{2gh}\\ $ Now for M if J is the impulse due to the tension of the string. We have; $ \\J=MV\\ $ But for m we have to write; $ \\m{{v}_{m}}=mV-J \\\Rightarrow m{{v}_{m}}=\left( m-M \right)V \\\Rightarrow V=\dfrac{m{{v}_{m}}}{\left( m-M \right)} \\\Rightarrow V=\dfrac{m\sqrt{2gh}}{\left( m-M \right)} \\ $ Now according to the question, we know that the time for which the lifting force is acting is t. Thus, for the mass M to be lifted $ \\Mg=\dfrac{J}{t} \\\Rightarrow Mg=\dfrac{Mm\sqrt{2gh}}{t\left( m-M \right)} \\\Rightarrow g=\dfrac{m\sqrt{2gh}}{t\left( m-M \right)} \\\Rightarrow 2gh=\dfrac{{{g}^{2}}{{t}^{2}}{{\left( m-M \right)}^{2}}}{{{m}^{2}}} \\\therefore h=\dfrac{g{{t}^{2}}{{\left( m-M \right)}^{2}}}{2{{m}^{2}}}$ 

A wall of strength $Y$ is at rest. A crane is to be used to break the wall. Now the bob of the crane weighs $m$ , if the area of the face the bob will strike is $A$ and strain produced is $e$ , what is the minimum initial velocity of the bob for which the wall will break? $(\text{Assume the time taken by the bob to reach the wall is t})$ 

The wall will break if we provide the minimum force. This minimum force actually corresponds to the strength of the material. The velocity corresponding to this force will be the minimum velocity for which the wall will break. 

 We know that Impulse can also be written as a product of force and time. $ \\J=Ft\\\Rightarrow F=\dfrac{P}{t}\\\Rightarrow P=m\Delta v=m\left( {{v}_{f}}-{{v}_{i}} \right)\\ $ Assuming that after striking the bob comes at rest, we get $ {{v}_{f}}=0\\J=-m{{v}_{i}} \\\Rightarrow \left| J \right|=m{{v}_{i}} \\ $ Now strength Y can be written as; $ \\Y=\dfrac{Stress}{Strain}\\\text{stress}=\dfrac{F}{A}\,\text{and strain}=e\\ $ Thus for finding the minimum velocity we write; $ \\Y\le \left| \dfrac{F}{Ae} \right| \\\Rightarrow Y\le \left| \dfrac{P}{tAe} \right| \\\Rightarrow Y\le \dfrac{m{{v}_{i}}}{Ate} \\\therefore {{v}_{i}}\le \dfrac{AteY}{m}$ 