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A bullet fired into a wooden block loses half of its velocity after penetrating 40 cm. It comes to rest after penetrating a further distance of?
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Calculation of Distance Traveled by Bullet

Given:

  • Initial velocity of the bullet = \( v_0 \)
  • Velocity of the bullet after penetrating 40 cm = \( \frac{v_0}{2} \)
  • Distance penetrated before coming to rest = 40 cm


Calculating the Distance Traveled

Let the total distance traveled by the bullet before coming to rest be \( d \).

When the bullet penetrates 40 cm, its velocity becomes \( \frac{v_0}{2} \).

Using the formula for distance traveled with constant acceleration:

\[ v^2 = u^2 + 2as \]
Where:

  • \( v \) = final velocity
  • \( u \) = initial velocity
  • \( a \) = acceleration
  • \( s \) = distance traveled


Substitute the given values into the formula:

\[ 0 = (\frac{v_0}{2})^2 + 2a(40) \]
As the bullet comes to rest, its final velocity is 0. Solve for acceleration:

\[ a = -\frac{(v_0)^2}{160} \]
Now, when the bullet comes to rest, the distance traveled is given by:

\[ d = 40 + \frac{v_0^2}{160} \]
Final Answer

The bullet will come to rest after traveling a total distance of \( 40 + \frac{v_0^2}{160} \) cm.

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A bullet fired into a wooden block loses half of its velocity after penetrating 40 cm. It comes to rest after penetrating a further distance of?
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