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Physics: Post your doubts here!

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akshay 999

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can someone help me out wiz ziz question plz.

An early experimenter, working in other than Si units, obtained the following ten values for the magnitudes of the changes on small oil drops:
9.82, 19.64, 39.28, 39.28, 34.37, 19.64, 19.64, 29.46, 19.64, 39.28.
What value do these results suggest for the magnitude of the charge of the electron as measured in these units?
A 2.45
B 4.91
C 9.82
D 19.64
 
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akshay 999 said:
can someone help me out wiz ziz question plz.

An early experimenter, working in other than Si units, obtained the following ten values for the magnitudes of the changes on small oil drops:
9.82, 19.64, 39.28, 39.28, 34.37, 19.64, 19.64, 29.46, 19.64, 39.28.
What value do these results suggest for the magnitude of the charge of the electron as measured in these units?
A 2.45
B 4.91
C 9.82
D 19.64
Click to expand...
I dont get the question. :p
Moreover this is the first time I have came across this type. Which paper it is!
 
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Czechoslovakia said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_qp_11.pdf
question 14 - pleaseee! Thought blocker
Click to expand...
14)
sagar had explained v.well..
Suppose we take the upward direction to be positive, and the downward direction to be negative, we can again write the momentum equations and use them to find the final velocity of the system - the forces between the clay and the lead pellet are huge compared to other external forces, so even though there are external forces acting on the system during the collision, we can assume the momentum stays approximately constant.

So, the initial momentum is the momentum of the bullet alone, which is equal to mv = (5.0/1000) * 200 = 1.0 kg ms^-1 . The clay block is stationary, so it does not contribute any momentum to the system initially.
When the bullet collides with the clay, the bullet gets stuck in the clay and they both move off with the same velocity, which we'll call v(f). The mass of the lead pellet+ the clay block = (95/1000 + 5/1000) = (100/1000) = 0.1 kg.
Their final velocity = v(f)
Therefore, the final momentum of the system is 0.1 * v(f)

Since this is equal to the initial momentum, we can write (0.1 kg) * v(f) = (1.0 kg ms^-1) and so v(f) = 10 ms^-1.

This is the speed with which they rise after the collision. While the system moves upwards, gravity accelerates them downwards at a constant rate of -9.81 ms^-2. When they reach the highest point above the original position, their velocity is 0. Therefore, we can use the formula v^2 = u^2 + 2as to give us "s", which is the maximum height. So,

(0)^2 = (10)^2 +2(-9.81)s
19.62 * s = 100
s = 5.09 m = 5.1 meters = A.
 
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