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
![\frac{1}{2} (m_1 + m_2) v^2 = \frac{1}{2} k A^2](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20%28m_1%20%2B%20m_2%29%20v%5E2%20%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20k%20A%5E2)
⇒ 0.62
= 7.5 ×
×
⇒ v = 23.65 m
The velocity of the bullet and block together is 23.65m
b.) Before impact,
![m_1v_1 + m_2v_2 = (m_1 + m_2)v](https://tex.z-dn.net/?f=m_1v_1%20%2B%20m_2v_2%20%20%3D%20%28m_1%20%2B%20m_2%29v)
⇒ (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.