Kinematics

JEE Main level, single correct, medium

50 of 308

Question 50 of 308

The velocity of the bullet becomes one third after it penetrates 4cm4cm 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)cm\left(4+x\right)cm inside the block. The value of xx is

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Answer: (C) 0.50.5

According to the given information for the first 4cm4cm, u=v0u={v}_{0} and v=v03v=\dfrac{{v}_{0}}{3}. Applying equation of motion for constant acceleration,
v2=u2+2as{v}^{2}={u}^{2}+2as
⇒(v03)2=v2+2a(4)⇒a=−8v029(8)=−v029\Rightarrow {\left(\dfrac{{v}_{0}}{3}\right)}^{2}={v}^{2}+2a\left(4\right) \Rightarrow a=-\dfrac{8{{v}_{0}}^{2}}{9\left(8\right)}=-\dfrac{{{v}_{0}}^{2}}{9}
When the body stops, v=0v=0
⇒0=v02+2a(4+x)\Rightarrow 0={{v}_{0}}^{2}+2a\left(4+x\right)
⇒v02=2(v029)(4+x)\Rightarrow {{v}_{0}}^{2}=2\left(\dfrac{{{v}_{0}}^{2}}{9}\right)\left(4+x\right)
⇒4.5=4+x\Rightarrow 4.5=4+x
⇒x=0.5cm\Rightarrow x=0.5cm

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