Bullet embedding in wooden block: momentum and motion analysis

[TextClick]

Homework Statement

A bullet of mass 0.0020 kg traveling at 600.0 m/s embeds in a block of wood sitting on the edge of a cliff. The mass of the block is 5.0 kg and it lands 0.72 m from the base of the cliff. How tall is the cliff?

Relevant Equations

m1v1+m2v2=(m1+m2)vf
d=vit+1/2at^2
Fdeltat=mdeltav
F=ma

I first plugged my given values into m1v1+m2v2=(m1+m2)vf.

(0.002)(600)+(5)(0)=((0.0020)+(5))vf
vf=0.24 m/s

Next, I plugged my given values into F=ma.

((0.002)+(5))(9.8)
F=49.02 N

Next, I plugged my given values into Fdeltat=mdeltav.

deltat=mdeltav/F

((0.002)+(5))(0.24)/(49.02)
deltat=0.02448 s

Finally, I plugged my given values into y=vit+1/2at^2.

y=1/2(9.8)(0.02448)
y=0.12 m

 

[haruspex]

F=49.02 N

Next, I plugged my given values into Fdeltat=mdeltav.

deltat=mdeltav/F

The force you found is that due to gravity, so leads to a downward acceleration, and this occurs after the block is knocked off the top of the cliff.
The change in velocity you used there is horizontal, and occurs during the impact, before the block becomes airborne.
I see no basis for using them in the same equation.

Note you did not use the information about where the block lands. How can you use that to find the descent time?

 

[TextClick]

So I got the problem correct up to the point that I derived the initial velocity in the x direction, 0.24 m/s. After this point, I do not understand the methodology behind this problem.

The teacher told me to use the equation x=v1xt, plug in my given values, and solve for t to get 3 seconds.

The next step was to use the equation y=1/2gt, plug in my given values, and solve to get 14.7.

Can you please help me to understand this part? Thanks.

 

[haruspex]

The teacher told me to use the equation x=v1xt, plug in my given values, and solve for t to get 3 seconds.

You found the horizontal velocity after impact. Thereafter there is no acceleration horizontally (we are ignoring drag). How far does it travel horizontally after the collision? How long will that take?

 

[Saptarshi Sarkar]

I first plugged my given values into m1v1+m2v2=(m1+m2)vf.

(0.002)(600)+(5)(0)=((0.0020)+(5))vf
vf=0.24 m/s

How will the trajectory of the particle look like? How many independent velocity components will it have at any given time mid fall?