Abullet with mass 4.53 g is fired horizontally into a 1.953-kg block attached to a horizontal spring. the spring has a constant 5.73 102 n/m and reaches a maximum compression of 6.20 cm. find the initial speed of the bullet-block system.

The initial speed of the bullet-block system is 113 m/s

Explanation:

Hi there!

Let´s apply the theorem of conservation of energy. If we neglect friction, all the kinetic energy of the bullet-block system is used to compress the spring. In other words, the kinetic energy is converted into elastic potential energy. Then, the initial kinetic (KE) energy of the bullet-block system will be equal to the final elastic potential energy (EPE) of the spring:

KE = EPE

1/2 · m · v² = 1/2 · k · x²

Where:

m = mass of the bullet-block system.

v = initial speed of the bullet-block system.

k = spring constant.

x = compression of the spring

Then, solving for v:

v² = k · x² / m

v² = 5.73 × 10² N/m · (0.620 m)² / 1.957 kg

v = 113 m/s

The correct answer is option b and c.