[Linear momentum] When to use which equation.

AI Thread Summary
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
Messages
22
Reaction score
0

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

 
Physics news on Phys.org
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.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top