Question

Darth Maul fires a 20.0 g bullet into a 0.600 kg block that is
attached to a fixed horizontal spring. The spring constant for the
spring is 8.10 103 N/m. The block starts oscillating back and forth
with vibrations that have an amplitude of 21.5 cm. (That is, the
farthest away from equilibrium the block gets is this distance).
What was the speed of the bullet before impact if the bullet and
block move together after impact? Your answer is incorrect. .
**27.6 m/s the answer is not 27.6 m/s so please try to find
the right answer :)** "COLLISION!"

Answer #1

A = amplitude = 0.215 m

k =spring constant = 8100 N/m

m = mass of bullet = 0.02 kg

M = mass of block = 0.6 kg

v = speed of bullet

V = speed of bullet - block combination after collision

Using conservation of energy

Kinetic energy of combination after collision = Maximum spring potential energy

(0.5) (M + m) V^{2} = (0.5) k A^{2}

(0.6 + 0.02) V^{2} = (8100) (0.215)^{2}

V = 24.6 m/s

using conservation of momentum

mv = (M + m) V

(0.02) v = (0.02 + 0.6) (24.6)

v = 762.6 m/s

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