A bullet fired into a fixed target loses half its velocity after penetratin 3 cm How much further it will penetrate before coming to rest assuming that it aces constant resistance to motion ?

  • Option 1)

    1.5 cm

  • Option 2)

    1.0 cm

  • Option 3)

    3.0 cm

  • Option 4)

    2.0 cm

 

Answers (1)

As we learnt in

Law of Consevation of Momentum -

 \vec{F}=\frac{\vec{dp}}{dt}

\vec{F}=0            then \vec{p}=constant

\vec{p}=\vec{p}_{1}+\vec{p}_{2}+\cdots =const

- wherein

\ast Independent of frame of reference

 

 

 

For first part of penetration, by equation of motion,

\left ( \frac{u}{2} \right )^{2}= \left ( u^{2} \right )-2f\left ( 3 \right )

or\: \: \: \: \: \: \: 3u^{2}= 24f\cdots \cdots \cdots \left ( i \right )

For latter part of penetration,

0= \left ( \frac{u}{2} \right )^{2}-2fx

or \: \: \: \: \: u^{2}= 8fx\cdots \cdots \cdots \left ( ii \right )

From (i) and (ii)

3\times \left ( 8\: fx \right )= 24f

or \: \: \: \: x= 1 cm

Correct option is 2.

 


Option 1)

1.5 cm

This is an incorrect option.

Option 2)

1.0 cm

This is the correct option.

Option 3)

3.0 cm

This is an incorrect option.

Option 4)

2.0 cm

This is an incorrect option.

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