Question

A 12 g bullet is fired into a 9.0 kg wood block that is at rest on a wood table. The block, with the bullet embedded, slides 5.0 cm across the table. The coefficient of kinetic friction for wood sliding on wood is 0.20.
What was the speed of the bullet?

Answer 1

 The bullet has a certain pre-impact momentum, p(bul), that is given as the product of the bullet's mass, m(bul), and it's pre-impact velocity, v(bul) 

p(bul) = m(bul) x v(bul) 

Since the block is at rest prior to being struck, its original momentum is 0 kg*m/s. The total momentum of the system pre-impact is therefore equal to the bullet's original momentum. Find that and you can easily find the bullet's original velocity. 

We know that when the bullet strikes the block the block absorbs the momentum and the bullet-block system continues traveling in the same direction. The force of friction decreases the momentum over time: 

Δp = Ff x t 

Let's start by finding Ff. 

Ff = μ x Fn 

Fn is the normal force, or the force exerted by the tabletop perpendicularly against the block. For horizontal surfaces, the normal force is the same as the block's weight. Since we have no reason to assume that the tabletop isn't horizontal, Fn = Fw. The weight is the product of the block's mass and gravity: 

Fn = Fw = m x g 

So... 

Ff = μ x m x g 

Ff = (0.20) x (9.012 kg) x (9.81 m/s²) = 17.7 N 

Now we have to find the time over which the block stops. When you're dealing with accelerations that either start or end at rest, you can use the following equation: 

Δx = 1/2at² 

Where Δx is the displacement of the block while it was accelerating (speeding up or slowing down). In this case the block's displacement was 5.0 cm, or 0.050 m. We don't know the block's acceleration yet, but we can find it using Newton's second law: 

a = Ff / m = (17.7 N) / (9.012 kg) = 1.96 m/s² 

Side note: You have to be careful here. In reality the acceleration should be negative since it opposes the direction of the block's initial motion, but we're ignoring that for the time being. Don't let it bite you on the butt in other problems, though! 

Now that we know the acceleration we'll get the time: 

0.050 m = 1/2 (1.96 m/s²) t² 

0.050 m = (0.982 m/s²) t² 

t² = 0.0509 s² 

t = 0.226 s 

FINALLY...we can plug the time into the formula Δp = Ff x t and figure out the system's change in momentum. 

Δp = (17.7 N)(0.226 s) = 3.99 N*s 

The bullet's original momentum was 3.99 N*s. Now we can find its original velocity: 

p(bul) = m(bul) x v(bul) 

3.99 N*s = (0.012 kg) v(bul) 

v(bul) = 333 m/s 

Properly rounded to two sig-figs, thats 330 m/s, or even better, 3.3x10² m/s.