Question

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 the 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 the opposite direction goes through iron first and then penetrates $2 \, cm$ into the wood. If $a_{1}$ and $a_{2 \, }$ be the retardation offered to the bullet by wood and iron plates respectively, then

• A. $a_{1}=2a_{2}$
• B. $a_{2}=2a_{1}$
• C. $a_{1}=a_{2}$
• D. Data insufficient

Answer 1

Answer: (B)

Let $a_{1}$ and $a_{2}$ be the retardations offered to be bullet by wood and iron respectively.
For $A \rightarrow B \rightarrow C,$
$v_{1}^{2}-u^{2}=2a_{1}\left(4\right),$ and $0^{2}-v_{1}^{2}=2a_{2}\left(\right.1\left.\right)$
Adding, we get
$-u^{2}=2\left(\right.4a_{1}+a_{2}\left.\right)$ ...(i)
For $A^{′} \rightarrow B^{′} \rightarrow C^{′},$
$v_{2}^{2}-u^{2}=2a_{2}\left(2\right)$
and $0^{2}-v_{2}^{2}=2a_{1}\left(2\right)$
Adding, we get
$-u^{2}=2\left(\right.2a_{1}+2a_{2}\left.\right)$ ....(ii)
Equating Eqs. (i) and (ii) and solving, we get
$4a_{1}+a_{2}=2a_{1}+2a_{2}$
$\Rightarrow \, a_{2}=2a_{1}$