a 20 g bullet strikes a 0.7 kg block attached to a fixed horizontal spring whose spring costant is 5600 N/m and sets it into vibration with an amplitude of 25 cm. What was the speed of the bullet before impact if the two objects move together after impact?

Answer 1

Mass before impact, m = 20 g = 0.02 kg
Combined mass after impact, M = 0.7+0.02=0.72 kg
Velocity of bullet before impact = v m s^-1
Combined velocity after impact, V = mv/M

Kinetic energy before impact, e1 = (1/2)mv²
Kinetic energy after impact, e2 = (1/2)MV²

Potential energy of spring when compressed
= (1/2)Kx²
= (1/2)5600*0.25²
= 175 J

Equate kinetic and potential energies
175 J = (1/2)MV²
V = 22.05 m s^-1

v = MV/m
= 22.05*0.72/0.02
= 794 m s^-1

Check my work.

Answer 2

your approach makes a lot of sense.
thanks for your help!