Question

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.

Answer 1

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