Answer:
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.