A bullet with initial velocity 150 m/s penetrates a rubber block of thickness 10 cm retarding at 30 $m / s^{2}$ and then enters a wooden block that stops the bullet after a distance of 6 cm. What will be the retardation when the bullet enters the rubber block?

$- 1000 m / s^{2}$

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$- 100 m / s^{2}$

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$- 10 m / s^{2}$

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$- 1 m / s^{2}$

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Solution

The correct option is B
$- 100 m / s^{2}$

Let us consider rubber block first.

Bullet starts moving just before the block.

Given, $i n i t i a l v e l o c i t y u_{1}^{2} = 150 m / s, a_{1} = - 30 m / s^{2}$ and $S_{1} = 10 c m = \frac{10}{100} m = 0.1 m$

Using $v_{1}^{2} - u_{1}^{2} = 2 a_{1} S_{1}$ ,

$\Rightarrow v_{1}^{2} = u_{1}^{2} + 2 a_{1} S_{1}$

$\Rightarrow v_{1}^{2} = 150 + 2 \times (- 30) \times 0.1$

$\Rightarrow v_{1}^{2} = 144 m / s \Rightarrow v_{1} = 12 m / s$

Velocity of leaving the rubber block and velocity of entering the wooden block are the same.

So, $v_{1}$ = $u_{2}$ = 12 m/s and given distance travelled by the bullet in rubber block, $S_{2}$ = 6 cm $= \frac{6}{100} m = 0.06 m$

We can use the equation $v_{2}^{2} - u_{2}^{2} = 2 a_{2} S_{2}$ to find the retardation

$\Rightarrow 0 - 12 = 2 \times a_{2} \times 0.06$

$\Rightarrow 2 \times a_{2} \times 0.06 = - 12$

$\Rightarrow a_{2} = - \frac{12}{2 \times 0.06}$

$\Rightarrow a_{2} = - 100 m / s^{2}$

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A bullet with initial velocity 150 m/s penetrates a rubber block of thickness 10 cm retarding at 30 m/s2 and then enters a wooden block that stops the bullet after a distance of 6 cm. What will be the retardation when the bullet enters the rubber block?

A bullet with initial velocity 150 m/s penetrates a rubber block of thickness 10 cm retarding at 30 $m / s^{2}$ and then enters a wooden block that stops the bullet after a distance of 6 cm. What will be the retardation when the bullet enters the rubber block?