Kinematics

JEE Main level, single correct, medium

45 of 308

Question 45 of 308

A bullet is fired into a fixed target looses one third of its velocity after travelling 4cm4cm. It penetrates further D×10−3mD\times {10}^{-3}m before coming to rest. The value of DD is :

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Answer: (A) 3232

When the bullet just enters the target, its equation of motion is given by
v2−u2=2aS...(1){v}^{2}-{u}^{2}=2aS...\left(1\right)
Equation (1) implies that
(2u3)2=u2+2(−a)(4×10−2)⇒4u29=u2−2a(4×10−2)⇒5u29=2a(4×10−2)…(2){\left(\dfrac{2u}{3}\right)}^{2}={u}^{2}+2\left(-a\right)\left(4\times {10}^{-2}\right) \Rightarrow \dfrac{4{u}^{2}}{9}={u}^{2}-2a\left(4\times {10}^{-2}\right) \Rightarrow \dfrac{5{u}^{2}}{9}=2a\left(4\times {10}^{-2}\right)\ldots \left(2\right)
When the bullet stops inside the target, its equation of motion can be written as
0=(2u3)2+2(−a)(x)...(3)0={\left(\dfrac{2u}{3}\right)}^{2}+2\left(-a\right)\left(x\right)...\left(3\right)
Equation (3) implies that
4u29=2ax...(4)\dfrac{4{u}^{2}}{9}=2ax...\left(4\right)
Dividing equation (2) by equation (4), we have
54=4×10−2x⇒x=165×10−2⇒x=3.2×10−2m=32×10−3m\dfrac{5}{4}=\dfrac{4\times {10}^{-2}}{x} \Rightarrow x=\dfrac{16}{5}\times {10}^{-2} \Rightarrow x=3.2\times {10}^{-2}m =32\times {10}^{-3}m
Hence, D=32D=32.

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