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

Newton's laws (Forces) Quiz 2

physics

Identify which relation between the momentum and the acceleration of a body of fixed mass $m$ is true? 

The derivative of change in momentum will be the force. 

  $\begin{align} & F=ma \\ & F=\dfrac{dP}{dt} \\ & \Rightarrow ma=\dfrac{dP}{dt} \\ & \text{That is, }P\propto a \end{align}$

What will be the ratio between the mass of the body and the rate of change of momentum? 

The acceleration of the body can be found by taking the derivative of velocity.

  $\begin{align} & F=ma \\ & \Rightarrow \dfrac{dp}{dt}=m\dfrac{dv}{dt} \\ & \therefore \dfrac{m}{\dfrac{dp}{dt}}=\dfrac{dt}{dv} \end{align}$

 When a ball has been thrown to the air with a velocity $20\,m{{s}^{-1}}$, it comes back after $2\,s$.  Assume that there is a frictional force acting because of the air medium which has been found to be $10\,N$. Find the mass of the ball? 

The frictional force will be opposite to the direction of motion of the ball. Hence it will be subtracted from the force at which the ball is thrown upwards. 

 $\begin{align} & ma-{{f}_{f}}=mg \\ & \Rightarrow m\left( a-g \right)={{f}_{f}} \\ & \Rightarrow m=\dfrac{{{f}_{f}}}{a-g} \\ & \Rightarrow m=\dfrac{10}{\left( \dfrac{v}{t} \right)-g} \\ & \Rightarrow m=\dfrac{10}{\left( \dfrac{20}{2} \right)-9.8} \\ & \therefore m=50\,kg \end{align}$

 The sphere of mass $1\,kg$ having an initial momentum $2\,kgm{{s}^{-1}}$ is being moved in the x-axis. The momentum got changed into $4\,kgm{{s}^{-1}}$ by the action of an external force of magnitude $10\,N$.  What will be the displacement of the sphere?  

Momentum can be found by taking the product of mass and the velocity. 

 $\begin{align} & F=ma \\ & \Rightarrow a=\dfrac{F}{m} \\ & \Rightarrow a=\dfrac{10}{1}=10m{{s}^{-2}}\\ & {{p}_{2}}=mv \\ & \Rightarrow v=\dfrac{4}{1}=4m{{s}^{-1}} \\ & {{p}_{1}}=mu \\ & \Rightarrow u=\dfrac{2}{1}=2m{{s}^{-1}} \\ & \text{In accordance to the third equation of motion,} \\ & {{v}^{2}}={{u}^{2}}+2as \\ & \Rightarrow {{4}^{2}}={{2}^{2}}+2\times 10\times s \\ & \Rightarrow 12=20s \\ & \therefore s=0.6\,m \end{align}$

The pressure exerted by a metal ball having a mass $5\,kg$ on a wall has been given as $2.5\times {{10}^{5}}\,Pa$. The ball is found to be incident on the wall with the velocity $4\,m{{s}^{-1}}$ for a time interval of $8\,s$. What will be the area of the cross section of the wall where the metal ball will be exerting the pressure? 

Pressure can be defined as the force experienced per unit area.

 $\begin{align} & F=ma \\ & \Rightarrow F=m\dfrac{v}{t} \\ & \Rightarrow F=5\times \dfrac{4}{8}=2.5N \\ & \text{The pressure, }P=\dfrac{F}{A} \\ & \Rightarrow A=\dfrac{F}{P} \\ & \Rightarrow A=\dfrac{2.5}{2.5\times {{10}^{5}}} \\ & \therefore A={{10}^{-5}}\,{{m}^{2}} \end{align}$