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This is one of our many quizzes on Linear momentum. Click on this to access more such competitive quizzes on Dynamics.

Linear momentum Quiz 2

physics

 A body of mass $12\text{ }kg$ is having an initial momentum mentioned as $3\text{ }kgm{{s}^{-1}}$. Can we calculate the final velocity of the atom if no external force has been acted on the body? If yes, what will be the value? 

 The final and initial momentum will be equivalent if there is no external force acting on a system. Using this equates the momentum and finds the final velocity. 

  $\text{As there is no external force acting, the momentum will be conserved. That is,}\\{{P}_{i}}={{P}_{f}}\\\text{Expanding the final term,}\\{{P}_{i}}=m{{V}_{f}}\\\text{Substituting the values,}\\3=12\times {{V}_{f}} \\\therefore {{V}_{f}}=\dfrac{1}{4}\text{ }m{{s}^{-1}}$

 The total mass of a loaded gun has been mentioned as $1\text{ }kg$. The bullet inside the gun will be having a mass of $50\text{ }g$. What will be the recoil velocity of the gun if the forward velocity is $4\text{ }m{{s}^{-1}}$? Note that only one bullet has been loaded in the gun. 

 Find out the mass of the gun only first and then substitute this in the equation for the recoil velocity. Take care of the order of magnitude of the final answer.

  $\text{The mass of the bullet be,}\\m=50\text{ }g\\\text{Here the total mass of the gun is given.The mass of the gun only,}\\M=1\text{ }kg-50\text{ }g=0.95\text{ }kg\\\text{The recoil velocity,}\\u=\dfrac{-m}{M}v\\\text{Substituting the values,}\\u=\dfrac{-50\times {{10}^{-3}}}{0.95}\times 4 \\\therefore u=0.210\text{ }m{{s}^{-1}}$

 The linear momentum of a system of 4 particles has been mentioned as $5\text{ }kgm{{s}^{-1}}$,$1\text{ }kgm{{s}^{-1}}$,$3\text{ }kgm{{s}^{-1}}$ and $2\text{ }kgm{{s}^{-1}}$. The mass of the system has been given as $1\text{ }kg$. Determine the velocity of the centre of mass during its journey. 

 Find out the total momentum of the body which is the sum of individual momentum. Then divide it by the mass of the system. 

 $\text{The velocity of the centre of mass will be,}\\V=\dfrac{\sum{{{p}_{i}}}}{M}\\\text{The total linear momentum will be,}\\p={{p}_{1}}+{{p}_{2}}+{{p}_{3}}+{{p}_{4}}\\\text{Substituting the values,}\\V=\dfrac{1+2+3+5}{1} \\\therefore V=11\text{ }m{{s}^{-1}}$ 

 Ambika, a lady with a mass of $60\text{ k}g$, was driving her car through a smooth rod having a momentum of $5\text{ }kgm{{s}^{-1}}$. Her car was having a mass of $130\text{ k}g$. Find out the time taken for the travel if she covered a distance of $1\text{ }km$. 

 From the momentum equation, find the velocity of her travel. After that from this velocity, find out the time taken by her. 

 $\text{The momentum,}\\P=MV\\\text{Substituting the values,}\\5=190\times V \\\Rightarrow V=38\text{ }m{{s}^{-1}} \\\text{The velocity will be,}\\V=\dfrac{X}{T}\\\text{Substituting the values in it,}\\38=\dfrac{1000}{T} \\\therefore T=26.3\text{ }s$

 Let us assume that when a mass of $50\text{ }kg$ was travelling in a straight line, it collided with another mass of $150\text{ }kg$ elastically. The initial velocity of travel of $50\text{ }kg$ mass has been found to be $u=7{{t}^{2}}+5\text{ }m{{s}^{-1}}$ and after collision, it stopped the motion. Calculate the acceleration of the mass $150\text{ }kg$ if $t=3\text{ }s$? 

 In the case of an elastic collision, the linear momentum will be conserved. Acceleration can be found by taking the first derivative of the velocity. 

 $\text{The momentum will be conserved here and will be transferred to the next mass,}\\Mv=mu\\\text{Substituting the values,}\\150\times v=50\times \left( 7{{t}^{2}}+5 \right) \\\Rightarrow v=\dfrac{1}{3}\times \left( 7{{t}^{2}}+5 \right) \\\text{The acceleration of the 150 kg mass,}\\a=\dfrac{dv}{dt}\\\text{Substituting the values,}\\\Rightarrow a=\dfrac{1}{3}\times 14\times 3 \\\therefore a=14\text{ }s$