Question

A 0.0382-kg bullet is fired horizontally into a 3.78-kg wooden block attached to one end of a massless, horizontal spring (k = 833 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 0.190 m. What is the speed of the bullet?

Answer 1

Answer:

$280.87$ ms⁻¹

Explanation:

Consider the motion of the bullet-block combination after collision

$m$ = mass of the bullet = 0.0382 kg

$M$ = mass of wooden block = 3.78 kg

$V$ = velocity of the bullet-block combination after collision

$k$ = spring constant of the spring = 833 N m⁻¹

$A$ = Amplitude of oscillation = 0.190 m

Using conservation of energy

Kinetic energy of  bullet-block combination after collision = Spring potential energy gained due to compression of spring

$(0.5)(m + M)V^{2} = (0.5)kA^{2}$

$(0.0382 + 3.78)V^{2} = (833)(0.190)^{2}$

$V = 2.81$ ms⁻¹

$v_{o}$ = initial velocity of the bullet before striking the block

Using conservation of momentum for the collision between bullet and block

$m v_{o} = (m + M) V$

$(0.0382) v_{o} = (0.0382 + 3.78) (2.81)$

$v_{o} = 280.87$ ms⁻¹