Question

A 0.0233-kg bullet is fired horizontally into a 2.41-kg wooden block attached to one end of a massless, horizontal spring (k = 845 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 0.196 m. What is the speed of the bullet?

Answer 1

Answer:

$v= 381.82\ m/s$

Explanation:

Given,

mass of the bullet, m = 0.0233 Kg

Mass of the block, M = 2.41 Kg

horizontal spring constant, k = 845 N/m

Amplitude of oscillation, A = 0.196 m

Using conservation of energy when the bullet is embedded

PE = KE

$\dfrac{1}{2}kA^2 = \dfrac{1}{2} Mv^2$

$v =A \sqrt{\dfrac{k}{M}}$

$v =0.196\times \sqrt{\dfrac{845}{2.41+0.0233}}$

$v= 3.65\ m/s$

Now using conservation of  momentum to calculate the initial velocity of bullet

$m V = M v$

$0.0233\times v=(2.41+ 0.0233)\times 3.65$

$v= 381.82\ m/s$