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

m_{1} = mass of bullet = 3 g = 0.003 kg

m_{2} = mass of block = 1 kg

using conservation of momentum

m_{1}v = (m_{1} + m_{2}) V

v = (m_{1} + m_{2}) V / m_{1} eq-1

using conservation of energy

kinetic energy at bottom = Potential energy at Top

(0.5) (m_{1} + m_{2}) V^{2} =
(m_{1} + m_{2}) g h

V^{2} = 2 gh

V^{2} = 2 (9.8) (0.30)

V = 2.42 m/s

Using eq-1

v = (m_{1} + m_{2}) V /
m_{1}

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

v = 809.1 m/s

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