[Linear momentum] When to use which equation.

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In inelastic collisions, such as a bullet embedding into a block, the correct equation to use is (mbullet)(vinitial) = (mbullet + mblock)(vfinal). This equation accounts for the conservation of momentum, where the combined mass moves with a common final velocity after the collision. If the block is initially at rest, the equation simplifies to (m1)(vi)bullet = (m1+m2)(vf)system. The final velocities of both objects are the same because they stick together, forming a single system post-collision. Understanding these relationships is crucial for correctly applying momentum conservation principles in physics problems.
hiuting
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Homework Statement


I'm confused about when to use which formula.
I thought for inelastic collisions,
for example, when a bullet is shot into a block,
I should use
(mbullet)(vinitial) = (mbullet + mblock)(vfinal)

but it turns out that I must use
b207134634.jpg


why do i have to use this one?

Homework Equations





The Attempt at a Solution

 
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eqn w/ velocities you use
 
They should really be the same equation. (m1)(vi)+(m2)(vi)=(m1)(vf)+(m2)(vf) will reduce to the top equation you wrote if the block "m2" is initally at rest (vi=0) and upon impact they stick and form a system. Thus they have a combined mass (m1+m2) and the same final velocity (vf). So it reduces to:

(m1)(vi)bullet = (m1+m2)(vf)system
 
kjohnson said:
They should really be the same equation. (m1)(vi)+(m2)(vi)=(m1)(vf)+(m2)(vf) will reduce to the top equation you wrote if the block "m2" is initally at rest (vi=0) and upon impact they stick and form a system. Thus they have a combined mass (m1+m2) and the same final velocity (vf). So it reduces to:

(m1)(vi)bullet = (m1+m2)(vf)system

then v1 final and v2 final are the same??
 
Yes, this is true because they stick together to form a system. They must have the same velocity.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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