Help with a problem about motion — bullet striking a block of wood

AI Thread Summary
The discussion centers on solving a physics problem involving a bullet striking a block of wood and determining the resulting motion. The initial calculations correctly derive the final velocity of the block after the impact as 0.24 m/s and the force due to gravity as 49.02 N. However, confusion arises regarding the methodology for finding descent time and horizontal travel after the block is knocked off a cliff. The teacher suggests using the equation x = v1xt to find time, leading to a result of 3 seconds, while also indicating to apply y = 1/2gt for vertical displacement. The conversation emphasizes the need to clarify the relationship between horizontal and vertical motion post-impact.
TextClick
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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
 
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TextClick said:
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?
 
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.
 
TextClick said:
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?
 
Last edited:
TextClick said:
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?
 

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