Bullet hits a block, penetration depth and velocity

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SUMMARY

The discussion centers on the physics problem of a bullet with mass m and velocity v penetrating a wooden block of mass M at rest on a frictionless surface. The common speed V of the bullet and block after the collision is derived using the conservation of momentum as V = mv/(m + M). The penetration depth l is calculated using the equation l = m(1 - m/(m + M))v²/(2F_{ret}), where F_{ret} is the constant retardation force opposing the motion. The equations provided are confirmed to be correct by participants in the discussion.

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AbigailM
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Homework Statement


A bullet with mass and speed v hits a wooden block of mass M that is situated at rest on a frictionless surface. It penetrates the block and gets trapped inside it as a result of a constant retardation force F_{ret} that opposes relative motion between the two objects. Find the common speed of the bullet and the block V, and the penetration length l in terms of m, M, v, and F_{ret}.

Homework Equations




mv=(m+M)V (Eq 1)

\frac{1}{2}mv^{2}=F_{ret}l+\frac{1}{2}(m+M)V^{2} (Eq 2)

The Attempt at a Solution


The common speed of m and M is V=\frac{mv}{m+M} (Eq 3) via conservation of momentum.

\frac{1}{2}mv^{2}=F_{ret}l+\frac{1}{2}\frac{m^{2}v^{2}}{(m+M)} (where I substituted Eq 3 into Eq 2)

Let's rearrange:
F_{ret}l=\frac{1}{2}m\left( 1-\frac{m}{m+M}\right) v^{2}

No let's solve for the penetration depth l:
l=\frac{m\left(1-\frac{m}{m+M}\right) v^{2}}{2F_{ret}}

Not sure if it is correct. Thanks for the help!
 
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AbigailM said:

Homework Statement


A bullet with mass and speed v hits a wooden block of mass M that is situated at rest on a frictionless surface. It penetrates the block and gets trapped inside it as a result of a constant retardation force F_{ret} that opposes relative motion between the two objects. Find the common speed of the bullet and the block V, and the penetration length l in terms of m, M, v, and F_{ret}.

Homework Equations




mv=(m+M)V (Eq 1)

\frac{1}{2}mv^{2}=F_{ret}l+\frac{1}{2}(m+M)V^{2} (Eq 2)

The Attempt at a Solution


The common speed of m and M is V=\frac{mv}{m+M} (Eq 3) via conservation of momentum.

\frac{1}{2}mv^{2}=F_{ret}l+\frac{1}{2}\frac{m^{2}v^{2}}{(m+M)} (where I substituted Eq 3 into Eq 2)

Let's rearrange:
F_{ret}l=\frac{1}{2}m\left( 1-\frac{m}{m+M}\right) v^{2}

No let's solve for the penetration depth l:
l=\frac{m\left(1-\frac{m}{m+M}\right) v^{2}}{2F_{ret}}

Not sure if it is correct. Thanks for the help!

Certainly your momentum and k.e. conservation equations are correct.
 

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