Get Answers to all your Questions

header-bg qa

The velocity of the bullet becomes one third after it penetrates 4 \mathrm{~cm} in a wooden block. Assuming that bullet is facing a constant resistance during its motion in the block. The bullet stops completely after travelling at (4+x) \mathrm{cm} inside the block. The value of \mathrm{x} is :

Option: 1

2.0


Option: 2

1.0


Option: 3

0.5


Option: 4

1.5


Answers (1)

best_answer

\mathrm{V_{1}=\frac{u}{3}\left ( Given \right )}

\mathrm{V_{2}=0}

* For constant retardation by the equations of motion

\mathrm{V_1^2=u^2+2 a s }

\mathrm{\frac{u^2}{9}-u^2=2 a\left(4 \times 10^{-2}\right) }

\mathrm{-\frac{4 u^2}{9 \times 4 \times 10^{-2}}=a }

\mathrm{a=\frac{-100 u^2}{9} \rightarrow(1) }

\mathrm{v_2^2=u^2+2 a s }

\mathrm{0=u^2+2\left(-\frac{100 u^2}{9} \times(4+x) \times 10^{-2}\right) }

\mathrm{u^2=\frac{2 u^2(4+x)}{9} }

\mathrm{x=0.5 \mathrm{~cm}}

Hence 3 is correct option

Posted by

chirag

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE