A target is made of two plates, one of wood and the other of iron. The thickness of the wooden plate is $4 cm$ and that of iron plate is $2 c m$ . A bullet fired goes through the wood first and then penetrates $1 c m$ into the iron. A similar bullet fired with the same velocity from the opposite direction goes through the iron first and then penetrates $2 cm$ into the wood. If $a_{1}$ and $a_{2}$ be the retardation offered to the bullet by the wood and iron plates respectively, then

a_{1} = 2 a_{2}

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a_{2} = 2 a_{1}

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a_{1} = a_{2}

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Data insufficient

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Solution

The correct option is B $a_{2} = 2 a_{1}$
Let $a_{1}$ and $a_{2}$ be the retardations offered to the bullet by wood and iron, respectively.

For $A \to B \to C$
$v_{1}^{2} - u^{2} = 2 a_{1} (4)$ and
$0^{2} - v_{1}^{2} = 2 a_{2} (1)$
Adding we get
$- u^{2} = (4 a_{1} + a_{2}) 2 \dots \dots (1)$

For $A^{^{'}} \to B^{^{'}} \to C^{^{'}}$
$v_{2}^{2} - u^{2} = 2 a_{2} (2)$ and
$0 - v_{2}^{2} = 2 a_{1} (2)$
Adding we get
$- u^{2} = (2 a_{1} + 2 a_{2}) 2 \dots \dots (2)$

From $(1)$ and $(2)$
$a_{2} = 2 a_{1}$

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