A 0.0346-kg bullet is fired horizontally into a 3.67-kg wooden block attached to

Question

A 0.0346-kg bullet is fired horizontally into a 3.67-kg wooden block attached to one end of a massless, horizontal spring (k = 860 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 0.170 m. What is the speed of the bullet?

Explanation / Answer

Here

Momentum remains Constant

Therefore

Momentum Before Collision = Momentum After Collision

Therefore

0.0346*V = (3.67+0.0346)*u

Aso As we know in Oscilatory motion

At the Extreme Point

All the Kinetic Energy will Change into Elastic potenteial Energy of the Spring

Theefore

0.5*k*A^2 = 0.5*M*u^2

Therefore

u = A*sqrt(k/M)

= 0.170*sqrt(860/(0.0346+3.67))

= 2.59 m/sec

Therfore From Equation(1) , by putting the Value of u , we get

0.0346*V = (3.67+0.0346)*2.59

V = 277.31 m/sec

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