# Physics: Post your doubts here!

Discussion in 'International A And AS Level' started by XPFMember, Sep 29, 2011.

1. ### Clark20

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I understand why option A and D are wrong. However, I dont get how option C is justified.

The er says under conditions where mass keeps reducing, the bodys speed can change and it can accelerate. My question is, when a body is moving with constant velocity (zero resultant force) and even if its mass starts reducing, resultant force still is 0 and hence acceleration also is zero even if mass is lost. Speed shouldnt change.

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3. ### Physicist

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go to
http://physics-ref.blogspot.com/2019/02/all-external-forces-on-body-cancel-out.html

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4. ### Clark20

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I did. Still having this confusion.

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5. ### Hamnah Zahoor

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The main thing you have to focus on here is that not the ' Speed ' remains unchanged but the ' Velocity ' remains unchanged. For the acceleration to be zero the change in velocity should be zero not speed.

6. ### Clark20

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The question implies there is 0 net force (by saying all external forces on a body cancel out which means it must be moving with constant velocity and 0 net force). When the net force is 0, by N2L there shouldnt be any change in speed as fnet=ma. Fnet=0 and 0=ma for even a reducing mass gives 0 acceleration just due to 0 net force. With 0 net force f=ma shouldnt work as an explanation for change in speed even if mass is changing.

But, if there is net force, then fnet=ma can explain how change in mass causes the body to accelerate/change its speed

My question is, with 0 net force how can change in mass cause a body to change its speed?

Last edited: Feb 24, 2019
7. ### Physicist

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you should check it again.
http://physics-ref.blogspot.com/2019/02/all-external-forces-on-body-cancel-out.html
it has been updated and not the same as previously

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8. ### Physicist

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it can be true but is NOT always.

that is why the questions says 'MUST'

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9. ### Clark20

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Thank you. I understand why C is wrong.

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10. ### Alis alis

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how to do this one?

11. ### Ebrahim12

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A2 op-amp question.
This is from the cie application booklet:

Can somebody explain how D2 blocks the back emf from affecting the op-amp exactly? Thanks.

12. ### Clark20

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Need help understanding this question. D is the answer.

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13. ### Emmris

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I assume that u already know that gravity is always acting on the object through the motion ... therefore at its Max height u will have g acting downwards .... now let's study the horizontal component of the velocity, u will find out that the horizontal component throughout the journey stays constant even at its Max height which implies that air resistance is acting to balance the force , hence the acceleration of the horizontal component is zero n as u know air resistance act on opposite direction which in this case it's to the right as the object is moving to the left ...

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14. ### Emmris

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Actually my logic was wrong .... it's only g that acts on the object throughout the journey .... so ur answer maybe wrong ....

15. ### Hamnah Zahoor

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I hope the examination report help understand the question:
Question 12
When a body is moving through air to the left, there will be a drag force acting on it to the right, as well as its weight. The answer here is D, but both A and C were more popular responses.

16. ### Emmris

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https://www.google.com/search?q=for...AQ&biw=360&bih=511&dpr=2#imgrc=1Mi7aLR1qrdwYM

It says dat only g acts on the object ....

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17. ### Hamnah Zahoor

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The site you mentioned is for projectile motion in which we assume that air resistance is negligible or equal to zero while doing our calculations.
However, in the question mentioned above it does not states that the the air resistance was zero or it is projectile motion.

18. ### Clark20

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Thank you.

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19. ### Clark20

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In this question, is the journey of the object between X and Y considered as projectile motion? As it follows a parabolic path.

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20. ### Hamnah Zahoor

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Yes it is considered a projectile for the motion is without air resistance and it's a parabolic path.

21. ### MShaheerUddin

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Can anyone draw b(i) and b(ii) for me?

Thanks

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