# Tough Physics Question - Grade 12 - Kinematics/Forces/Energy all in one - Please show steps?

A bullet with a mass of 45 g is fired into a 8.3 kg block of wood resting on a floor against a spring (the coefficient of kinetic friction between the floor and the block is 0.24). This idea spring (k=76 N/m) has a maximum compression of 28 cm. What was the initial speed of the bullet?

### 2 Answers

- JimLv 77 years agoFavorite Answer
This question involves application of the two conservation principles:

Conservation of Momentum is applicable when bullet collides into wood block.

Conservation of Energy is applicable when block/bullet compress (ideal) spring.

Since the question asks for the initial speed of bullet let that unknown = v

and

Let the speed of block/bullet combination = V

The momentum of block/bullet combination = (0.045 + 8.3)V = 8.345V

The KE of block/bullet combination = 1/2(8.345)V² = 4.1725V²

SPE = 1/2kx² = (0.5)(76)(0.28)² = 2.9792 J

by conservation of energy => 2.9792 = 4.1725V²

V² = 0.714008388

V = 0.84499 m/s

The momentum of block/bullet combination = (8.345)(0.84499) = 7.05144 kgm/s

The momentum of bullet prior to hitting block = mv = 0.045v

by conservation of momentum => 0.045v = 7.05144

v = 7.05144/0.045 = 157 m/s ANS

- WhomeLv 77 years ago
To find the initial velocity of the bullet and block after impact we apply conservation of energy principles

The maximum spring potential plus the energy lost to friction will equal the initial kinetic energy

KE = PS + U

½mv² = ½kx² + Fd

as the spring is resting against the block before impact, then d = x

F will be the friction force or

F = μmg

½mv² = ½kx² + μmgx

v² = kx²/m + 2μgx

v = √(kx²/m + 2μgx)

v = √(76(0.28²) / (8.3 + 0.045) + 2(0.24)9.81(0.28))

v = 2.032 m/s

Now we use conservation of momentum to find the initial bullet velocity

mu1 + Mu2 = (m + M)v

as the block initial velocity, u2, is zero

u1 = (m + M)v / m

u1 = (0.045 + 8.3)2.032 / 0.045

u1 = 377 m/s