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Confused! kindly explain, - Laws of motion - JEE Main

A block of mass m=10 kg rests on horizontal table. The coefficient of friction between the block and the table is 0.05. When hit by a bullet of mass 50 g moving with speed \nu, that gets embedded in it, the block moves and comes to stop after moving a distance of 2 m on the table. If a freely falling object were to acquire speed \frac{v }{10} after being dropped from height H, then neglecting energy losses and taking g=10 ms-2, the value of H is close to :

  • Option 1)

    0.2 km

  • Option 2)

    0.3 km

  • Option 3)

    0.4 km

  • Option 4)

    0.5 km

 
Answers (1)
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As we learnt in

Kinetic or Dynamic Friction -

f_{K}\;\alpha\ R

f_{K}=\mu_{K} R

f_{K}= kinetic friction 

\mu_{K}= coefficient of kinetic friction

R = reaction

- wherein

f_{K}<F_{l}

\therefore\ \mu_{K}<\mu_{s}

\mu_{K}=depends on the nature of surface in contact.

 

 f=\mu(M+m)g\ \; \Rightarrow\ \; a=\frac{f}{M+m}=\frac{\mu (M+m)g}{(M+m)}=\mu g

a=0.05\tims 10=0.5 ms^{-2}

u=v_{0}=\frac{0.05v}{10.05}

v^{2}=u^{2}-2as\ \; \Rightarrow\ \; v^{2}+u^{2}=2as

v = 0       u2 = 2as

\left(\frac{0.05}{10.05} \right )^{2}=2\times0.5\times2\ \; \Rightarrow\ \; v=201\sqrt{2}

v^{2}=u^{2}+2gH\ \; \Rightarrow\ \; v^{2}=2gH=v=\sqrt{2gH}

\frac{v}{10}=\sqrt{2gH}\ \; \Rightarrow\ \; \frac{201\sqrt{2}}{10}=\sqrt{2\times 10\times H}

H = 40 m = H = 0.04 km 

Correct option is 3.

 


Option 1)

0.2 km

This is an incorrect option.

Option 2)

0.3 km

This is an incorrect option.

Option 3)

0.4 km

This is the correct option.

Option 4)

0.5 km

This is an incorrect option.

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