A bullet of mass 0.012 kg and horizontal speed 70 ms-1 strikes a block

1.	A bullet of mass 0.012 kg and horizontal speed 70 ms-1 strikes a block of wood of mass 0.4 kg 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» By using law of conservation of momentum, m₁ u₁ + m₂ u₂ = (m₁ + m₂)v, we have v = {m₁ u₁ + m₂ u₂}/{m₁ + m₂} = {0.012 x 70 + 0.4 x 0}/{0.12 + 0.4} = 2.04 ms^-1 When bullet and block rise to the height h, using the law of conservation of total energy. 1/2 (m₁ + m₂)v² = (m₁ + m₂) gh i.e., h = v²/2g = (2.04)²/{2 x 9.8} = 0.212 m = 21.2 cm Also, heat produced = Loss in K.E. = 1/2 m₁u₁² - 1/2(m₁ + m₂)v² = 1/2(0.012)(70)² - 1/2 (0.012 + 0.4) x (2.04)² = 29.4 - 0.86 = 28.54 J

A bullet of mass 0.012 kg and horizontal speed 70 ms-1 strikes a block of wood of mass 0.4 kg 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»

By using law of conservation of momentum, m₁ u₁ + m₂ u₂ = (m₁ + m₂)v, we have

v = {m₁ u₁ + m₂ u₂}/{m₁ + m₂}

= {0.012 x 70 + 0.4 x 0}/{0.12 + 0.4} = 2.04 ms^-1

When bullet and block rise to the height h, using the law of conservation of total energy.

1/2 (m₁ + m₂)v² = (m₁ + m₂) gh

i.e., h = v²/2g = (2.04)²/{2 x 9.8}

= 0.212 m = 21.2 cm

Also, heat produced = Loss in K.E.

= 1/2 m₁u₁² - 1/2(m₁ + m₂)v²

= 1/2(0.012)(70)² - 1/2 (0.012 + 0.4) x (2.04)²

= 29.4 - 0.86 = 28.54 J