Question

A target is made of two plates, one of wood and the other of iron. The thickness of the wooden plate is $4\text{cm}$ and that of iron plates is $2\text{cm}$. A bullet fired goes through the wood first and then penetrates $1\text{cm}$ into the iron. A similar bullet fired with the same velocity from the opposite direction goes through the iron first and then penetrates $2\text{cm}$ 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. $a_1=2a_2$
• B. $a_2=2a_1$
• C. $a_1=a_2$
• D. $a_1=3a_2$

Answer 1

The correct option is B $a_2=2a_1$

Let $u$ be the velocity with which the bullet is fired
Case 1:
For $A\to B\to C$
Using third equation of motion, we get
$v_1^2-u^2=-2a_1\left(4\right)$ (for wood)
And $0^2-v_1^2=-2a_2\left(1\right)$ (for iron)
On adding, we get
$u^2=(4a_1+a_2)2$ ... (1)
Case 2:
For $A^{\prime }\to B^{\prime }\to C^{\prime }$
$v_2^2-u^2=-2a_2\left(2\right)$ (for iron)
And $0-v_2^2=-2a_1\left(2\right)$ (for wood)
On Adding, we get $u^2=2(2a_1+2a_2)$ ... (2)
From equation (1) and (2), we get
$a_2=2a_1$
Hence, the correct answer is option (b).