Elastic collision in one dimension

AI Thread Summary
The discussion revolves around calculating the speeds of a pendulum and a block after an elastic collision. The pendulum, weighing 0.5 kg and swinging from a height, has an initial velocity of 3.7 m/s calculated using gravitational potential energy. Participants explore the conservation of momentum and kinetic energy equations but face challenges in solving them without using a specific formula for one-dimensional elastic collisions, which requires a lengthy derivation. A proposed approach involves rewriting the conservation equations to express one variable in terms of another to facilitate solving for the final velocities. The conversation highlights the complexities of applying conservation laws in elastic collisions while adhering to academic requirements.
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Homework Statement


A Pendulum weighs .5kg and has a string length of 70cm swings from a horizontal position downwards to hit a block that weighs 2.5 kg and is on a frictionless plain. Calculate the speed of both the ball and the block after the elastic collision.


Homework Equations


mgh=.5mV^2
conservation of momentum
m1V= m1V1+m2V2
conservation of kinetic energy
.5m1V^2=.5m1V1^2+.5m2V2^2


The Attempt at a Solution



I used the first equation to solve for the initial velocity of the pendulum, which is 3.7m/s.
From here comes the elastic collision part.
I try to use both conseravation equations and substitute for one of the variables by solving for either V1 or V2 in either equation and thus substituting. However, when i try to solve it that way it ends up canceling out.
I know there is an equation for one dimension elastic collisions, but my professor says we can't use it unless we write the derivation of it, and that is a tedious thing to do and is rather lengthy.

Please help, ask questions if things are unclear
 
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lone21 said:

Homework Statement


A Pendulum weighs .5kg and has a string length of 70cm swings from a horizontal position downwards to hit a block that weighs 2.5 kg and is on a frictionless plain. Calculate the speed of both the ball and the block after the elastic collision.

Homework Equations


mgh=.5mV^2
conservation of momentum
m1V= m1V1+m2V2...(1)
conservation of kinetic energy
.5m1V^2=.5m1V1^2+.5m2V2^2...(2)
You can rewrite eq.(1) as
m1v - m1v1= m2v2...(3)
Similarly you can rewrtie eq.(2) as
m1v^2 - m1v^21= m2v^22..(4)
From 3 and 4 you can get
(v - v1)/v^2 - v1^2= v[/SUB]2[/SUB]/v[/SUB]2[/SUB]^2
After simplification you get
(v + v1) = v2
Substitute the value of v2 in equation (1) and solve for v1.
 
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...

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