A 0.0144-kg bullet is fired horizontally into a 3.03-kg wooden block attached to one end of a massless, horizontal spring (k = 800 N/m). The other end of the spring is fixed in place, and the spring is unstrained initially. The block rests on a horizontal, frictionless surface. The bullet strikes the block perpendicularly and quickly comes to a halt within it. As a result of this completely inelastic collision, the spring is compressed along its axis and causes the block/bullet to oscillate with an amplitude of 0.208 m. What is the speed of the bullet?

Explanation:

Given that,

Mass of bullet

m = 0.0144kg

Mass of wooden block

M = 3.03kg

Spring constant

k = 800 N/m

Let the initial velocity of the bullet be u'.

The initial velocity of the block is zero since the block is not moving

u = 0

Amplitude is 0.208m

x = 0.208m

inelastic collision occurs between the bullet and the block, this shows that after collision both the bullet and the block moves together with common velocity, let the common velocity be V

Using conservation of momentum

Initial momentum = final momentum

mu' = (m + M) V

u' = (m+M)V / m

Using conservation of energy

Energy from the spring is transfers to K.E in the block

Ux = K.E

½kx² = ½(m+M)V²

½ × 800 × 0.208² = ½(0.0144 + 3.03) V²

17.3056 = 1.5222V²

V² = 17.3056 / 1.5222

V² = 11.37

V = √11.37

V = 3.37 m/s

Final speed of the bullet and the block is 3.37 m/s

Then,

u' = (m+M)V / m

u' = (0.0144 + 3.03) × 3.37 / 0.0144

u' = 3.0444 × 3.37 / 0.0144

u' = 712.47 m/s

u' ≈ 712.5 m/s

The initial speed of the bullet is 712.5 m/s.

beta is electron from nucleus

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reserve is the answer