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 cm. A bullet fired goes through the wood first and then penetrates 1 cm into iron. A similar bullet fired with the same velocity from opposite direction goes through iron first and then penetrates 2 cm into wood. If $a_{1} a n d a_{2}$ be the retardation offered to the bullet by wood and iron plates respectively, then

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$t_{1} + t_{2}$

$a_{2} = 2 a_{1}$

Data insufficient

$a_{1} = a_{2}$

answer is B.

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Detailed Solution

Let $a_{1} a n d a_{2}$ be the retardations offered to be bullet by wood and iron respectively.

For $A \to B \to C,$

$υ_{1}^{2} - u^{2} = 2 a_{1} (4)$

And $0^{2} - υ_{1}^{2} = 2 a_{1} (1)$

Adding, we get $- u^{2} = 2 (4 a_{1} + a_{2}) ................ (i)$

For $A' \to B' \to C',$

$υ_{2}^{2} - u^{2} = 2 a_{2} (2)$

and $0^{2} - υ_{2}^{2} = 2 a_{1} (2)$

Adding, we get $- u^{2} = 2 (2 a_{1} + 2 a_{2}) ............... (i i)$

Equating Eqs. (i) and (ii) and solving, we get

$4 a_{1} + a_{2} = 2 a_{1} + 2 a_{2}$

$\Rightarrow a_{2} = 2 a_{1}$

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