A bullet of mass $8 g$ strikes a vertical wooden plank $5 c m$ thick with a velocity of $200 m / s$ in a horizontal direction and emerges out of at $150 m / s$ in same direction find retarding force bullet experience in wood for what additional thickness bullet will just emerge on other side

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Solution

Let $F$ be the retarding force
Initial Kinetic energy of bullet , $K_{1} = \frac{1}{2} m v_{1}^{2} = \frac{1}{2} \times 8 \times 10^{- 3} \times 200^{2} = 160 J$

Final Kinetic energy of bullet, $K_{2} = \frac{1}{2} m v_{2}^{2} = \frac{1}{2} \times 8 \times 10^{- 3} \times 150^{2} = 90 J$

Work done by retarding force , $W = - F (5 \times 10^{- 3})$

Using work- energy theorem we get ,
$- F (5 \times 10^{- 3}) = 90 - 160$

$F = \frac{70}{5 \times 10^{- 3}} = 14 \times 10^{3} N$

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A bullet of mass 8g strikes a vertical wooden plank 5cm thick with a velocity of 200m/s in a horizontal direction and emerges out of at 150m/s in same direction find retarding force bullet experience in wood for what additional thickness bullet will just emerge on other side

Let F be the retarding force Initial Kinetic energy of bullet , K1=12mv21=12×8×10−3×2002=160 JFinal Kinetic energy of bullet, K2=12mv22=12×8×10−3×1502=90 JWork done by retarding force , W=−F(5×10−3)Using work- energy theorem we get ,−F(5×10−3)=90−160F=705×10−3=14×103 N

A bullet of mass $8 g$ strikes a vertical wooden plank $5 c m$ thick with a velocity of $200 m / s$ in a horizontal direction and emerges out of at $150 m / s$ in same direction find retarding force bullet experience in wood for what additional thickness bullet will just emerge on other side