DLS
At the same instant that a 0.50-kg ball is dropped from 25 m above Earth, a second ball, with a
mass of 0.25 kg, is thrown straight upward from Earth’s surface with an initial speed of 15 m/s.
They move along nearby lines and pass without colliding. At the end of 2.0 s the velocity of
the center of mass of the two-ball system is:
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mayankdevnani
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options??
DLS
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A. 11 m/s, down
B. 11 m/s, up
C. 15 m/s, down
D. 15 m/s, up
E. 20 m/s, down
DLS
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V(a)=20m/s
V(b)=-5m/s
mayankdevnani
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Since this is multiple choice, lets use 10 m/s^2 as the acceleration due to gravity.
DLS
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so
0.50*20+0.25*(-5)/0.75
mayankdevnani
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The trick is to look at the momentum of the system after 2 seconds. So for the heavy ball, after two seconds, it has a velocity of -20 m/s.
DLS
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10+1.25/0.75 (wil we have a minus sign or are we considering just the mag?
mayankdevnani
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Thus it has a momentum of -10 kg*m/s.
DLS
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u took downwards negative?
mayankdevnani
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yaa, i took downwards negative.
The light ball will have a velocity of -5 m/s (15 m/s going up minus 20 m/s due to gravity gives 5 m/s downward). So it has a momentum of -1.25 kg*m/s.
mayankdevnani
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In both cases, the negative sign indicates the momentum is downward, and the sum is -11.25 kg*m/s.
mayankdevnani
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Find the velocity of the system by dividing by the total mass (0.75 kg)??
then solve it and you get your answer.
DLS
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i took downwards positive and upwards negative,im getting something different
mayankdevnani
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no, always took downwards negative and upwards positive
DLS
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v=0+10*2
v=20m/s downwards
and
v=15-20
v=-5m/s upwards
mayankdevnani
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no!!!!
mayankdevnani
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Find the velocity of the system by dividing by the total mass (0.75 kg)
and you get 15 m/s down.
so your answer is option C) 15 m/s up.
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wait u messed up!!
mayankdevnani
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no,, 15 m/s down.
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im taking downwards negative and upwards positive so
for the 0.25 kg ball,
v=15+20=35?
we are taking upwards positive right
mayankdevnani
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right
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0.25*35=8.75
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-20+8.75
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oh :o
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-11.25/0.75=-15m/s
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i see! :D
mayankdevnani
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okk