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laila [671]
3 years ago
6

Which is a possible measurement of the momentum of a moving object?

Physics
2 answers:
Delvig [45]3 years ago
8 0

Momentum is a vector, although we don't hammer on that.  In order to completely describe a momentum, you need a magnitude AND a direction ... just like force and velocity.

So choice #2 is the magnitude of a momentum, without its direction.

<em>Choice #3</em> is  the full package, with both the magnitude and the direction.

Choice #1 has units of energy, and choice #4 has units of acceleration, so neither of those can be it.

VMariaS [17]3 years ago
5 0

choice 1 is a  possible measurement of the momentum of a moving object?


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a 300kg motorboat is turned off as it approaches a dock and coasts towards it at .5 m/s. Isaac, whose mass is 62 kg jumps off th
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-- Before he jumps, the mass of (Isaac + boat) = (300 + 62) = 362 kg,
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After the jump:

-- Just as Isaac is coiling his muscles and psyching himself up for the jump,
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    His momentum relative to the dock is (62 x 3.5) = 217 kg-m/s toward it.

But there was only 181 kg-m/s total momentum before the jump, and Isaac
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=======================================

Another way to do it . . . maybe easier . . . in the frame of the boat.

In the frame of the boat, before the jump, Isaac is not moving, so
nobody and nothing has any momentum.  The total momentum of
the boat-centered frame is zero, which needs to be conserved.

Isaac jumps out at 3 m/s, giving himself (62 x 3) = 186 kg-m/s of
momentum in the direction <u>toward</u> the dock.

Since 186 kg-m/s in that direction suddenly appeared out of nowhere,
there must be 186 kg-m/s in the other direction too, in order to keep
the total momentum zero.

In the frame of measurements from the boat, the boat itself must start
moving in the direction opposite Isaac's jump, at just the right speed 
so that its momentum in that direction is 186 kg-m/s.
The mass of the boat is 300 kg so
                                                         (300 x speed) = 186

Divide each side by 300:  speed = 186/300 = <em>0.62 m/s</em>    <u>away</u> from the jump.

Is this the same answer as I got when I was in the frame of the dock ?
I'm glad you asked. It sure doesn't look like it.

The boat is moving 0.62 m/s away from the jump-off point, and away from
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To somebody standing on the dock, the whole boat, with its intrepid passenger
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  yay !

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