A 18 g bullet traveling at 525 m/s strikes an 0.8 kg block of wood that is balanced on a table edge?

Question

0.8 m above ground (Figure 8-66). If the bullet buries itself in the block, find the distance D at which the block hits the floor.

Answer 1

You must treat this like an inelastic collision, as energy is being lost when the bullet is embedded in the block. So you use this conservation of momentum equation:

m1v1 + m2v2 = (m1 + m2)vf

Pretend the numbers are subscripts, and vf = final velocity of the entire moving system, both m1 (the bullet) and m2 (the block).

Convert the weight of the bullet to kilograms to avoid any errors, giving you .018kg. Then plug in the numbers you have to get:

(.018)(525) + (.8)(0) = (.8 + .018)vf

Solve for vf, and you should get 11.55m/s.

Now you have a projectile motion problem. In all projective motion problems, the horizontal velocity always stays the same. This block/bullet system thus begins moving with a horizontal velocity of 11.55m/s and continues until it hits the ground. What you need is how much time it takes for the system to hit the ground, then. You can find that with

y = vo*t + 1/2gt^2

vo, the initial velocity, of the block in the y direction, is 0. So you just have y = 1/2gt^2. You know the y distance to the ground is .8m, so plug in what you have to get:

.8 = .5(9.8)t^2

Solve for t= .4041s

Now D = Vox*t, the initial x velocity multiplied by the time. This is (11.55)(.4041) =

4.67m

Answer 2

The gravity vector is orthogonal to the horizontally travelling bullet's vector. This makes falling time dependent only on the value of g and the distance to the floor.
It also makes horizontal momentum independent of the vertical force due to gravity. This lets you find the horizontal velocity of the shot block from conservation of momentum.
You now have a time and a velocity.

Answer 3

The initial momentum Mb of the bullet = .018*525 kg-m/s
The initial momentum of the block = 0
Total momentum M = Mb + 0
After collision, total momentum still = M due to conservation. The horizontal velocity Vh of the bullet and block = M/(.8 + .018) m/s
The time T to fall 0.8 m = sqrt(2*.8/g) sec
The horizontal distance D = Vh*T m

Answer 4

Neglecting friction: 4.7 meters; The combined object will fall .8 meters in .404 seconds. The combined mass will have a velocity of 11.552 m/s resulting in displacement of 4.7 meters.