Saturday, December 27, 2014

Physics 9702 Doubts | Help Page 40

  • Physics 9702 Doubts | Help Page 40


Question 245: [Kinematics > Linear motion]
Sky-diver jumps from high-altitude balloon.
(a) Explain briefly why acceleration of the sky-diver
(i) decreases with time
(ii) is 9.8 ms-2 at start of the jump

(b) Variation with time t of the vertical speed v of sky-diver is shown.

Use Fig to determine magnitude of the acceleration of sky-diver at time t = 6.0 s.

(c) Sky-diver and his equipment have a total mass of 90 kg.
(i) Calculate, for sky-diver and his equipment,
1. total weight
2. accelerating force at time = 6.0s

(ii) Use answers in (i) to determine total resistive force acting on sky-diver at time t = 6.0 s.

Reference: Past Exam Paper – November 2009 Paper 21 Q2



Solution 245:
(a)
(i) (Air) resistance increases with speed. So, the resultant / accelerating force decreases.

(ii) EITHER Initially, (air) resistance is zero
OR Initially, the weight / gravitational force is the only force

(b)



(Consider the points (0.8, 10) and (10, 28). Gradient = (28 – 10) / (10 – 0.8) = 1.956 = 2.0 ms-2. The accepted range is 1.7 – 2.1 ms-2.)
Use of the gradient of a tangent (a time t = 6.0s)
(Gradient =) Acceleration = 1.9 (± 0.2) ms-2

(c)
(i)
1. Total weight (= mg) = 90 x 9.8 = 880N

2. Accelerating force (= ma) = 90 x 1.9 = 170N

(ii)
{(Resultant) Accelerating force = Weight – Resistive force}
Resistive force = 880 – 170 = 710N









Question 246: [Measurements > Errors]
In an experiment the external diameter D and internal diameter d of a metal tube were found to be (64 ± 2) mm and (47 ± 1) mm respectively. What is the maximum percentage error for the cross sectional area of the metal tube?

Reference:Pacific Physics A-Level” – Volume 1, by POH LIONG YONG, pg 23



Solution 246:
The cross-sectional area of the tube is the area of the metal along a cross-section, as shown shaded in blue below.




Area of shaded region = External Area – Internal Area
Area of shaded region, A = πD2/4 – πd2/4 = π (D2 – d2) / 4

The formula for the shaded area above consists of both a multiplication and a subtraction of the quantities involving uncertainties. That is, if calculated alone, both the external area and the internal area would have uncertainties of their own. Then, a subtraction is performed to find the area of the shaded region – this results in another uncertainty.

To simplify the analysis of errors, let’s write the formula as follows:
Since (D2 – d2) = (D + d) (D – d)
Area of shaded region = π (D + d) (D – d) / 4

In this way, we can find the uncertainties involved in each (D + d) and (D – d) and use the simple known formula for uncertainties of products.
Δ(D + d) = ΔD + Δd = 2 + 1 = 3 mm
Δ(D – d) = ΔD + Δd = 2 + 1 = 3 mm

Since, Area of shaded region = π (D + d) (D – d) / 4
ΔA / A = ± [{Δ(D + d) / (D + d)} + {Δ(D – d) / (D – d)}]

(Maximum) percentage error for cross-sectional area = (ΔA / A) x 100%
(ΔA / A) x 100% = ± [{3 / (64 + 47)} – {3 / (64 – 37)}] x 100%
(ΔA / A) x 100% = ± 20%








Question 247: [Kinematics > Linear motion]
Object is thrown with velocity 5.2 m s–1 vertically upwards on Moon. The acceleration due to gravity on Moon is 1.62 m s–2.
What is the time taken for object to return to its starting point?
A 2.5 s                         B 3.2 s                         C 4.5 s                         D 6.4 s

Reference: Past Exam Paper – June 2014 Paper 11 Q7



Solution 247:
Answer: D.
At the maximum height, velocity of the object is zero.

Taking the vertically upwards direction as the positive direction,
Acceleration, a = (v – u) / t
–1.62 = (0 – 5.2) / t
Time, t taken to reach maximum height = 5.2 / 1.62 = 3.2s

Now, the time taken for the object to move from the maximum position (under gravity) to the initial position is also the same t (= 3.2s) since in this case, the initial velocity is 0 and the acceleration is the same. The final velocity would be 5.2ms-1.
{the final velocity can be obtained from v2 = u2 + 2as. It can be seen that the velocity does not depend on time, but depends on the distance, on acceleration (these 2 are the same for the upward motion) and on the initial velocity}

So, the answer is obtained by doubling the value of t which gives 2(3.2) = 6.4s









Question 248: [Measurements > Errors]
Student measures time t for a ball to fall from rest through a vertical distance h. Knowing that the equation h = ½ gt2 applies, the student plots graph shown.

Which of the following is an explanation for intercept on the t axis?
A Air resistance has not been taken into account for larger values of h.
B There is a constant delay between starting timer and releasing the ball.
C There is an error in timer that consistently makes it run fast.
D The student should have plotted h against t2.

Reference: Past Exam Paper – June 2002 Paper 1 Q4



Solution 248:
Answer: B.
Choice A is irrelevant. 

If the timer runs faster, the time should be less and the intercept on the t axis should have a negative value. So, explanation C is incorrect.

Explanation D is also incorrect since a graph of h against t2 should have a t-intercept of zero. The equation is: h = ½ gt2. Comparing with y = mx + c, it can be concluded that the t-intercept of such a graph should be zero.

The only correct explanation is that there is a constant delay between starting the time and releasing the ball.










Question 249: [Vectors]
Cyclist is travelling due south with velocity u. Wind is blowing from north-east with velocity w.

Wind has a velocity v relative to the cyclist, where v = w – u.
Which vector diagram shows magnitude and direction of velocity v?


Reference: Past Exam Paper – June 2013 Paper 13 Q4



Solution 249:
Answer: A.
Velocity v is given by v = w – u. This can also be written as v = w + (– u)
So, the vector w is positive while the other vector should be – u, that is, the direction is opposite to that of u shown in the diagram.

Consider diagram A (the correct answer). To perform the vector additional graphically, we need to first draw a vector and then add the other vector to it. Consider the starting point to be the right end of the horizontal line.
The first vector to be drawn may EITHER be w, which is in the same direction as shown in the question OR it can be (– u) which would be upwards (opposite to u).

Let the first vector be w. Then, to perform the addition, we should draw the tail of the vector (– u) at the arrow head of w.

Finally, the resultant vector v is drawn from the tail of the first vector (w) to the head of the last vector (– u). This results in choice A.  









Question 250: [Measurements > Uncertainty]
Student wished to determine density ρ of lead. She measures mass and diameter of a small sphere of lead:
mass = (0.506 ± 0.005) g
diameter = (2.20 ± 0.02) mm.
What is the best estimate of percentage uncertainty in her value of ρ ?
A 1.9%                        B 2.0%                        C 2.8%                        D 3.7%

Reference: Past Exam Paper – November 2013 Paper 13 Q6



Solution 250:
Answer: D.
Density ρ = Mass m / Volume V
Volume, V of sphere = 4πr3 / 3 = 4π(d/2)3 / 3 = 4πd3 / (3x8) = πd3 / 6

Density ρ = m / (πd3 / 6) = 6m / πd3  
Δρ / ρ = (Δm / m) + 3(Δd / d)

Percentage uncertainty in ρ = (Δρ / ρ) x 100% = [(Δm / m) + 3(Δd / d)] x100%
Percentage uncertainty in ρ = [(0.005 / 0.506) + 3(0.02 / 2.20)] x100% = 3.7%





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