Question

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 thin wires. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block.

Answer 1

m₁ = 0.012 kg, u₁ = 70ms^-1

m₂ = 0.4 kg, u₂ = 0.

After the collision the bullet strikes in the block and both behave as a single body of mass m₁ + m₂ and moves with velocity V. 

By conservation of momentum (m₁ + m₂) v = m₁u₁ + m₂u₂

= 2.04 ms^-1

Let the block move to a height ‘h’

(m₁ + m₂)gh = \(\frac {1}{2}\) (m₁ + m₂) v²

h = \(\frac{v^2}{2g}\) = \(\frac {2.04 \times 2.04}{2 \times 9.8}\) = 0.212 m

To find the heat produced, calculate energy lost (w) 

w = initial KE of bullet – final KE of combination

= \(\frac {1}{2}\) m₁ u₁² \(\frac {1}{2}\)(m₁ + m₂) v²

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

w = 29.4 – 0.86 = 28.5 J

∴ Heat produced

H = \(\frac {w}{j}\) \(\frac {28.54}{4.2}\) 

= 68 cal.