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

`280.87` ms⁻¹

Step-by-step 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