Question

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?

Answer 1

Answer:

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

Explanation:

Mass of bullet, $m_1$ = 0.02 kg

Mass of the block , $m_2$ = 0.6 kg

The spring stiffness constant, k = 7.5 × $10^3$ 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

           $\frac{1}{2} (m_1 + m_2) v^2 = \frac{1}{2} k A^2$

⇒      0.62$v^2$ = 7.5 × $10^3$ × $(0.215)^2$

⇒      v = 23.65 m

The velocity of the bullet and block together is 23.65m

b.) Before impact,

             $m_1v_1 + m_2v_2 = (m_1 + m_2)v$

⇒      (0.02$\times v_1$)  +  ( 0.6 × 0)  = 0.62 × 23.65

⇒        $v_1$   =   $\frac{0.62 \times 23.65}{0.02}$ = 733.15 m /s

where $v_1$ is the speed of the bullet before impact.

and $v_2$ 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.