Question
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?
Answers
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<sup>-1</sup>
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<sup>-1</sup>
v = MV/m
= 22.05*0.72/0.02
= 794 m s<sup>-1</sup>
Check my work.
Combined mass after impact, M = 0.7+0.02=0.72 kg
Velocity of bullet before impact = v m s<sup>-1</sup>
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<sup>-1</sup>
v = MV/m
= 22.05*0.72/0.02
= 794 m s<sup>-1</sup>
Check my work.
your approach makes a lot of sense.
thanks for your help!
thanks for your help!
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