Projectile problem (This is a quickie)

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The discussion focuses on solving a projectile motion problem involving momentum and kinetic energy. The user attempts to calculate the mass of a projectile given its momentum of 200 kg·m/s and kinetic energy of 1000 J. Confusion arises regarding the units and calculations, particularly when determining velocity and mass. Participants clarify that velocity should be expressed in m/s, not Joules, and emphasize the importance of understanding the relationships between force, work, and energy. The conversation highlights the necessity of mastering fundamental physics equations for accurate problem-solving.
gcombina
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Hello, I need help on my basic math

This is the problem
A projectile is launches with 200kg.m/s of momentum and 1000 J of kinetic energy. What is the mass of the projectile.
(a) 20 kg

This is my math:

P = mv so
m = P/v
m = (200 kg.m/s)/v

KE = 1000 J
1000 J = 1/2 mv ^2
1000 J = 1/2 (mv) v
1000 J = 1/2 (200) v
v = 10 J

IF m = 200/v
m= (200 kg.m/s)/ (10 J)
m= (200 kg.m/s) / (10 NM)
m = (200 kg.m/s) / (10 ma.m)
m = (200 kg.m/s) / (10 kg.m/s.m)
m = (20/m) WHY IS THAT I HAVE THIS ANSWER?? 20/M? Instead of 20kg?
 
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Both sides of every equation has units, including this one:
1000 J = 1/2 (200) v

You wrote v = 10 J
but v does not have units of Joules.
 
gcombina said:
1000 J = 1/2 mv ^2
1000 J = 1/2 (mv) v
1000 J = 1/2 (200) v
v = 10 J
You forgot that the 200 has units of kg*m/s so when you divide Joules by kg*m/s you get m/s
So the answer is v=10 m/s

gcombina said:
m= (200 kg.m/s)/ (10 J)
m= (200 kg.m/s) / (10 NM)
m = (200 kg.m/s) / (10 ma.m)
m = (200 kg.m/s) / (10 kg.m/s.m)
m = (20/m) WHY IS THAT I HAVE THIS ANSWER?? 20/M? Instead of 20kg?

Now try it with 10 m/s instead of 10 J and see if it works out.
 
when I divide J/(kg*m/s), I get m/s? how so??

isn't J = NM?

so it would be [(nm) / (kg*m/s )] ?
 
gcombina said:
when I divide J/(kg*m/s), I get m/s? how so??

isn't J = NM?

so it would be [(nm) / (kg*m/s )] ?

Yes, and a Newton is kg*m/s^{2} so a Joule is kg*m^{2}/s^{2}
So divide kg*m^{2}/s^{2} (Joules) by kg*m/s and what do you get?
 
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gcombina said:
when I divide J/(kg*m/s), I get m/s? how so??

isn't J = NM?

so it would be [(nm) / (kg*m/s )] ?
This is where mastery and easy recall of equations is important.

W=Fs
But F=ma

So W=mas
This tels you that the Joule is a kg m m / sec2
 
@nascent Oxygen

I am sorry , I do not understand ...what is fs? what is mas?
 
@nathaneal,
thanks!
 
gcombina said:
@nascent Oxygen

I am sorry , I do not understand ...what is fs? what is mas?
These are equations you must commit to memory. They are your tools of trade.

W=F.s where F is force & s is displacement (both F and s are vectors).

F=m.a where F is force & a is acceleration (both a anf F are vectors).
 

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