Question

A bullet of mass 0.012kg and horizontal speed 70ms^-1 strikes a block of wood of mass 0.4kg and instantly comes to rest with respect to the block. The block is suspended from the ceiling by means of thin wires. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block.

Answer 1

If V be the velocity of the block after collision, the using law of conservation of momentum, we get

0.012 × 70 + 0 = (0.012 + 0.4)V

or V = \(\frac{0.012×70}{0.412}\)ms^-1 = 2.04ms^-1

If h be the height through which block rises, then

(M + m) gh = 1/2 (M + m)V²

or h = v²/2g or

h = \(\frac{2.04×2.04}{2×9.8}\)m = 0.212m = 21.2 cm

Amount of heat produced in the block = loss of K.E.

= \(\frac{1}{2}\) × 0.012 × 70 × 70 – \(\frac{1}{2}\) × 0.412 × 2.04 × 2.04

= 29.4J – 0.857J = 28.543J.