Calculating Momentum of a Thrown Object at Maximum and Halfway Heights

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The discussion focuses on calculating the momentum of a 0.2 kg object thrown vertically upwards at an initial velocity of 20 m/s. At maximum height, the object's final velocity is zero, resulting in zero momentum. To find the momentum at halfway to maximum height, participants suggest using kinematic equations to first determine the height and then the velocity at that height. The correct approach involves calculating the maximum height and then applying the equations to find the velocity at half that height. The conversation emphasizes the importance of understanding kinematics and the consistent nature of mass in momentum calculations.
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Momentum problem,please check!

Homework Statement


A 0.2 kg object is thrown vertically upwards at an initial velocity of 20m/s.Find its momentum when :
a) it reaches its maximum height
b) its momentum at the halfway of its maximum height

Homework Equations


P=vm
V^2=Vo^2+2gh

The Attempt at a Solution


When it reaches its maximum height does it final velocity equal to zero?
If it is like this it will be the momentum also zero.
And for the halfway the answer should be 5sqrt(2),but I don't know how can I get it.
 
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You are correct on the 1st part.
For the 2nd, one method is to use the equation you have quoted to find h, the maximum height.
Then use the equation again to find v when h is half this value.
 


Thanks,but if it is like that The maximum height its 20m its half way 10m
V^2=400-2*10*10
V=20m/s not correct
 


Dear Zafer I believe that you are new to kinematics. If this so then welcome to mechanics. It will be better if you try to understand this question with association to real life.
When you throw the ball up it is pushed upwards with an initial velocity. Now since velocity is a vector we have to make a convention for its direction. Let's say that when it goes up it has +ive velocity. Now gravity pulls the ball down and when it is at its topmost position, it has zero velocity. Now ball changes its direction of motion and therefore the velocity now becomes negative.

divide problem in two parts-
in first part calculate the max height attained using u=20m/s v=0 and g=-10m/s2(as per the convention we adopted)
h=(v^2-u^2)/2g
=-20*20/-10=20
in second part
u=0 h=10 and g=10
v=sqrt(u^2+2gh)
=sqrt(0+2*10*10)
=10root2
m=0 at top
m=2root2 at middle- i think m=.5 kg then you will get the desired ans
cheers
Shryans
 


Mass can not be written as 0 ,because mass is always constant in every aspect.
 


m is for momentum...sorry it should have been p :)
 
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