Question

A bullet with mass m1 = 3.00 g is fired into a wooden block of mass m2 = 1.00 kg, that hangs like a pendulum. The bullet is embedded in the block (complete inelastic collision). The block (with the bullet embedded in it) goes h = 30.0 cm high after collision. Calculate the speed of the bullet before it hit the block.

Answer 1

v = speed of bullet before collision

V = speed of block-bullet combination after collision

m₁ = mass of bullet = 3 g = 0.003 kg

m₂ = mass of block = 1 kg

using conservation of momentum

m₁v = (m₁ + m₂) V

v = (m₁ + m₂) V / m₁ eq-1

using conservation of energy

kinetic energy at bottom = Potential energy at Top

(0.5) (m₁ + m₂) V² = (m₁ + m₂) g h

V² = 2 gh

V² = 2 (9.8) (0.30)

V = 2.42 m/s

Using eq-1

v = (m₁ + m₂) V / m₁  

v = (0.003 + 1) (2.42)/0.003

v = 809.1 m/s