Sunday, July 5, 2015

Physics 9702 Doubts | Help Page 177

  • Physics 9702 Doubts | Help Page 177


Question 877: [Circular motion]
A large bowl is made from part of a hollow sphere.
A small spherical ball is placed inside the bowl and is given a horizontal speed. The ball follows a horizontal circular path of constant radius, as shown in Fig.1.

The forces acting on the ball are its weight W and the normal reaction force R of the bowl on the ball, as shown in Fig.2.

The normal reaction force R is at an angle θ to the horizontal.
(a)
(i) By resolving reaction force R into two perpendicular components, show that the resultant force F acting on the ball is given by the expression
W = F tan θ.

(ii) State significance of the force F for the motion of the ball in the bowl.

(b) The ball moves in circular path of radius 14 cm. For this radius, the angle θ is 28°. Calculate the speed of the ball.

Reference: Past Exam Paper – November 2014 Paper 41 & 42 Q2



Solution 877:
(a)
(i)
{The weight acts vertically down, so it does not have any horizontal component. The force component of the force F is then only due to the horizontal component of the reaction force R.}
F = R cosθ
{The path is circular with a constant radius. This means that the ball does not move up or down in the surface of the sphere. Thus, the resultant vertical component of the force on the ball is zero – that is, the vertical upward component of the reaction force R is equal to the downward weight.}
W = R sinθ
{Dividing the equations eliminate R and gives the required expression.}
dividing, W = F tanθ

(ii) The force F provides the centripetal force.

(b)
{As shown above, W = F tanθ. The weight W is equal to mg. The resultant force F provides the centripetal force, given by mv2 / r. Replacing these in the equation (W = F tanθ) allows us to solve for speed v.}
EITHER F = mv2 / r and W = mg
{W = F tanθ giving mg = (mv2 / r) tan θ}
OR v2 = rg / tan θ
v2 = (14×10–2 × 9.8) / tan 28° = 2.58
Speed v = 1.6 m s–1









Question 878:
A thermometer can be read to accuracy of ± 0.5 °C. This thermometer is used to measure a temperature rise from 40 °C to 100 °C.
What is percentage uncertainty in the measurement of the temperature rise?
A 0.5 %                       B 0.8 %                       C 1.3 %                       D 1.7 %

Reference: Past Exam Paper – November 2003 Paper 1 Q4 & June 2014 Paper 12 Q5



Solution 878:
Answer: D.
Temperature rise = 100 – 40 = 60°C

The ± 0.5°C uncertainty happens at the measurement of both the low and the high temperatures.

Percentage uncertainty in measurement = [ΔT / T] × 100% = [2(0.5) / 60] × 100% = 1.7%





4 comments:

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    1. For question 4, see solution 880 at
      http://physics-ref.blogspot.com/2015/07/physics-9702-doubts-help-page-178.html

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  2. This comment has been removed by the author.

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    1. For 4/M/J/02 Q.7(b)(c), see solution 882 at
      http://physics-ref.blogspot.com/2015/07/physics-9702-doubts-help-page-179.html

      Delete

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