Ampltitude problem:?

Question

A 20.0 g bullet strikes a 0.609 kg block attached to a fixed horizontal spring whose spring constant is 6.75 ✕ 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 two objects move together after impact?

Answer 1

Mass of bullet = m_b = 20 = 0.02 kg

Mass of block attached to spring = m_sb = 0.609 kg

Spring constant value k = 6.75 × 10³ N/m

Vibration amplitude s = 21.5 cm = 21.5 × 10^-2 m.

Total mass bullet and block m = 0.02 + 0.609 = 0.629 kg

The speed of the bullet before impact v_b = ?

The speed of the block and bullet after impact is v_bb= (m_b/m) v_b

= 0.02v_b/0.629 =  0.0317965 v_b.

 

The kinetic energy of the block and bullet is E_kinetic = ½m(v_bb)²

 = ½(0.629)( 0.0317965 v_b)²

= ½(0.629)( 0.0317965 v_b)²

= 0.317965× 10^-3  v_b²

 

The spring energy when compressed is E_spring = ½ks²

= ½(6.75 × 10³)( 21.5 × 10^-2)²

= 156.009375.

 

The two objects move together after impact. So the kinetic energy of the block and bullet is equals to the spring energy when compressed.

E_kinetic = E_spring

0.317965 v_b²  = 156.009375

v_b²  = 156.0093.75 / 0.317965

v_b²  = 490.6495212

v_b  = 22.15 m/s.

Velocity of bullet is 22.15 m/s.