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  • I have a doubt, kindly clarify. - Laws of motion - BITSAT

 A small ball of mass m starts at a point A . with speed vo and moves along a frictionless track AB as shown. The track BC has coefficient of friction \mu. The ball comes to stop at C after travelling a distance L which is :

  • Option 1)

    \frac{2h}{\mu }+\frac{v_{0}^{2}}{2\mu g}

  • Option 2)

    \frac{h}{\mu }+\frac{v_{0}^{2}}{2\mu g}

  • Option 3)

    \frac{h}{2\mu }+\frac{v_{0}^{2}}{\mu g}

  • Option 4)

    \frac{h}{2\mu }+\frac{v_{0}^{2}}{2\mu g}

 

Answers (1)

best_answer

As we learnt in

Impulse Momentum Theorem -

\vec{F}=\frac{d\vec{p}}{dt}

\int_{t_{1}}^{t_{2}}\vec{F}dt=\int_{p_{1}}^{p_{2}}\vec{dp}

- wherein

If \bigtriangleup t  is increased, average force is decreased

 

 Apply convervation of Energy

mgh+\frac{1}{2}mV_{o}^{2}=\frac{1}{2}mV^{2}

V^{2}=Vo^{2}+2gh

Let a be the acceleration produced because of friction

V^{2}=u^{2}+2as=>0=Vo^{2}+2gh-2al

L=\frac{Vo^{2}+2gh}{2a}

f=MN  =>  Mmg

a = Mg

L=\frac{Vo^{2+2gh}}{2Mg}=\frac{h}{M}+\frac{Vo^{2}}{2Mg}

 

 


Option 1)

\frac{2h}{\mu }+\frac{v_{0}^{2}}{2\mu g}

This is incorrect option

Option 2)

\frac{h}{\mu }+\frac{v_{0}^{2}}{2\mu g}

This is incorrect option

Option 3)

\frac{h}{2\mu }+\frac{v_{0}^{2}}{\mu g}

This is correct option

Option 4)

\frac{h}{2\mu }+\frac{v_{0}^{2}}{2\mu g}

This is incorrect option

Posted by

Aadil

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