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1. Two balls with exactly the same size and shape are laughed with same initial velocity from the surface of a perfectly flat plane. When air resistance is considered, the ball with greater mass will have a:
A. longer flight time and a greater maximum height
B. longer flight time and a lower maximum height
C.shorter flight time and a greater maximum height
D. shorter flight time and a lower maximum height.
The answer is A.
The following is how I reasoned:
1.Air resistance affect up trip differently from down trip.
2. For the up trip, Mg+f=Ma1
ma+f=ma2
(f stands for air resistance, which is same for both objects. M is the bigger mass and m is smaller mass., a1 stands for the acceleration of object with greater mass while a2 stands for the acceleration of object with smaller mass.)
a1= g+f/M,
a2=g+f/m
M>m
So, we have a1 is less than a2
Vt=Vo-at, Vt=0,
Vo=at
Vo=a1t1
Vo=a2t2
The two balls have the same initial velocity, since a1<a2, t1>t2.
So, for the up trip, the object with greater mass has longer time
For the height, Vt^2=V0^2-2ah, Vt=o,
V0^2=2ah,
The two balls have the same initial velocity, since a1<a2, h1>h2,
So, the bigger object has a greater maximum height.
3. For the down trip, it is free falling, we all know the time for the bigger object has less time.
Mg-f=Ma1
mg-f=ma2,
a1=g-f/M
a2=g-f/m
M>m
a1>a2,
So, Vt=at if the two balls have the same final velocity, then t1< t2.
4. Combing the results of 2 and 3, we can say that the flight time of the bigger object is greater then smaller one only when air resistance affect up trip much more then down trip.
5. Thanks for your patience to read this long post. Please let me know what you think
thanks a lot.
A. longer flight time and a greater maximum height
B. longer flight time and a lower maximum height
C.shorter flight time and a greater maximum height
D. shorter flight time and a lower maximum height.
The answer is A.
The following is how I reasoned:
1.Air resistance affect up trip differently from down trip.
2. For the up trip, Mg+f=Ma1
ma+f=ma2
(f stands for air resistance, which is same for both objects. M is the bigger mass and m is smaller mass., a1 stands for the acceleration of object with greater mass while a2 stands for the acceleration of object with smaller mass.)
a1= g+f/M,
a2=g+f/m
M>m
So, we have a1 is less than a2
Vt=Vo-at, Vt=0,
Vo=at
Vo=a1t1
Vo=a2t2
The two balls have the same initial velocity, since a1<a2, t1>t2.
So, for the up trip, the object with greater mass has longer time
For the height, Vt^2=V0^2-2ah, Vt=o,
V0^2=2ah,
The two balls have the same initial velocity, since a1<a2, h1>h2,
So, the bigger object has a greater maximum height.
3. For the down trip, it is free falling, we all know the time for the bigger object has less time.
Mg-f=Ma1
mg-f=ma2,
a1=g-f/M
a2=g-f/m
M>m
a1>a2,
So, Vt=at if the two balls have the same final velocity, then t1< t2.
4. Combing the results of 2 and 3, we can say that the flight time of the bigger object is greater then smaller one only when air resistance affect up trip much more then down trip.
5. Thanks for your patience to read this long post. Please let me know what you think
thanks a lot.