A 20.0 g bullet strikes a 0.600 kg block attached to a fixed horizontal spring whose spring stiffness constant is 7.50 103 N/m and sets it into vibration with an amplitude of 21.5 cm. What was the speed of the bullet before impact if the bullet and block move together after impact?

The speed of the bullet before impact was 733.15 m/s.

Explanation:

Mass of bullet, = 0.02 kg

Mass of the block , = 0.6 kg

The spring stiffness constant, k = 7.5 × N/m

Amplitude of the spring, A  = 0.215 m

a.) Therefore equating the energy of the bullet and mass with he energy of the spring we get

⇒      0.62 = 7.5 × ×

⇒      v = 23.65 m

The velocity of the bullet and block together is 23.65m

b.) Before impact,

⇒      (0.02)  +  ( 0.6 × 0)  = 0.62 × 23.65

⇒           =   = 733.15 m /s

where is the speed of the bullet before impact.

and is the speed of the block before impact.

and v is the speed of the block and bullet together after impact.

Therefore the speed of the bullet before impact was 733.15 m/s.

they should be similar or the same

an electric circuit