An object is thrown straight upward with an initial velocity of 1217 pixels per second. The vertical velocity of the object after 2 seconds is 0 due to gravity and the object is 1188 pixels above its starting point. The acceleration on the object due to gravity is 100 pixels per second squared. What is the mass of the object?
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1"pixels per second squared" is not a unit of force. – preferred_anon Aug 20 '13 at 23:05
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@DanielLittlewood: Would it make sense to replace the word "pixels" with "meters"? – Joncom Aug 20 '13 at 23:07
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No; in either case it would be a unit of acceleration. Do you perhaps mean the gravitational acceleration (usually called $g$)? – preferred_anon Aug 20 '13 at 23:08
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My bad, yes, I mean acceleration. – Joncom Aug 20 '13 at 23:28
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Just like with real gravity, the acceleration and motion of an object is independent of its mass whether measured in meters or pixels. So you cannot determine its mass other than to say that it is not zero. In order to determine mass you need force.
Tpofofn
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This is only an approximation (discovered by Galileo, and valid only when the object doesn't move too far and doesn't weigh too much), but it's the best you can do with the given information. Newtonian gravitationis a little more complex. – dfeuer Aug 21 '13 at 03:25
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@dfeuer, I disagree. It is not an approximation, but an exact fact. Galileo stated the general observation, to be further confirmed by Newton and Einstein. The fundamental principal is that gravitational mass and inertial mass are exactly the same. The principal holds in for all masses (from a BB to a star) and for all distances traveled (from a dropped ball to an orbiting planet). – Tpofofn Aug 22 '13 at 21:04
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@Tpofofn, here's a hint: Newton's second and third laws come into play as soon as you look beyond the instantaneous. – dfeuer Aug 23 '13 at 00:19
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@Tpofofn: You noted, correctly, that any two objects, $m_1$ and $m_2$, will accelerate the same when placed in the same gravitational field, produced by object $M$. But object $M$ will accelerate because it is in the field produced by $m_1$ or $m_2$, and its motion in response to that acceleration changes the field it generates. – dfeuer Aug 23 '13 at 00:29
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Yes, but this is now Newton's law of gravitation. I think that the OP assumed that M >> m where M is akin to Earth's mass and m is associated with a projectile. This is indicated by the fact that the OP specified constant gravitational acceleration. I don't think he/she is modeling celestial motion. – Tpofofn Aug 23 '13 at 02:57