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A bullet of mass 0.02 kg is moving with a speed of 10 ms-1 . It can penetrate 10 cm of a wooden block, and comes to rest. If the thickness of the target would be 6 cm only, find the K.E. of the bullet when it comes out. |
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Answer» For x = 10 cm = 0.1 m, Fx = \(\frac{1}{2}\)\(mv^2_1\) = 1J ∴ F = 10N For x = 6 cm = 0.06 m, Fx = \(\frac{1}{2} mv^2_1 - \frac{1}{2}{mv^2_2}\) or Fx = \(\frac{1}{2}\) \(mv^2_1\) - Final K.E. Final K.E. = \(\frac{1}{2}\) \(mv^2_1\) - Fx = 1 - 10 x 0.06 =1 - 0.06 = 0.4 J |
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