Monday, March 25, 2019

One end of a wire is connected to a fixed point. A load is attached to the other end so that the wire hangs vertically.


Question 17
One end of a wire is connected to a fixed point. A load is attached to the other end so that the wire hangs vertically.

The diameter d of the wire and the load F are measured as
d = 0.40 ± 0.02 mm,
F = 25.0 ± 0.5 N.

(a) For the measurement of the diameter of the wire, state
(i) the name of a suitable measuring instrument, [1]

(ii) how random errors may be reduced when using the instrument in (i). [2]


(b) The stress σ in the wire is calculated by using the expression
σ = 4F / πd2 .

(i) Show that the value of σ is 1.99 × 108 N m-2. [1]

(ii) Determine the percentage uncertainty in σ. [2]

(iii) Use the information in (b)(i) and your answer in (b)(ii) to determine the value of σ, with its absolute uncertainty, to an appropriate number of significant figures. [2]
 [Total: 8]






Reference: Past Exam Paper – November 2017 Paper 22 Q1






Solution:
(a)
(i) micrometer (screw gauge)/digital calipers

(ii) Random errors can be reduced by taking several readings along the wire and then averaging them.


(b)
(i)
σ = 4 × 25 / [π × (0.40×10-3)2] = 1.99 × 108 N m-2
or
{σ = 4F / πd2 = 4F / π(2r)2 = F / πr2 }
σ = 25 / [π  × (0.20 × 10-3)2] = 1.99 × 108 N m-2

(ii)
EITHER
{%F = (ΔF/F) × 100%             and %d = (Δd/d) × 100%
%d = (0.02/0.40) × 100% = 5%
%F = (0.5/25) × 100% = 2%}
%F = 2% and %d = 5%
{%σ = %F + 2(%d)}
%σ = 2% + (2 × 5%)
%σ = 12%                                          


OR
{Δσ / σ = ΔF/F + 2(Δd/d)}
ΔF / F = 0.5 / 25          and Δd / d = 0.02 / 0
{%σ  = (Δσ / σ) × 100%}
%σ  = [0.02 + (2×0.05)] ×100 %
%σ = 12%                              


(iii)
{ Δσ / σ = 12 % = 12 / 100
Δσ = 12 % × σ = (12 / 100) × σ }
absolute uncertainty Δσ = (12 / 100) × 1.99×108 = 2.4×107
{uncertainty should be given to only 1 s.f.
absolute uncertainty Δσ = 2×107 = 0.2×108}
σ = 2.0 × 108 ± 0.2 × 108 N m-2              or (2.0 ± 0.2) × 108 N m-2

Sunday, March 24, 2019

Two blocks, A and B, are on a horizontal frictionless surface. The blocks are joined together by a spring, as shown in Fig. 2.1.


Question 14
(a) State the principle of conservation of momentum. [2]


(b) Two blocks, A and B, are on a horizontal frictionless surface. The blocks are joined together by a spring, as shown in Fig. 2.1.


Fig. 2.1

Block A has mass 4.0 kg and block B has mass 6.0 kg.

The variation of the tension F with the extension x of the spring is shown in Fig. 2.2.


Fig. 2.2

The two blocks are held apart so that the spring has an extension of 8.0 cm.
(i) Show that the elastic potential energy of the spring at an extension of 8.0 cm is 0.48 J. [2]

(ii) The blocks are released from rest at the same instant. When the extension of the spring becomes zero, block A has speed vA and block B has speed vB.

For the instant when the extension of the spring becomes zero,
1. use conservation of momentum to show that
kinetic energy of block A / kinetic energy of block B = 1.5
[3]

2. use the information in (b)(i) and (b)(ii)1 to determine the kinetic energy of block A. It
may be assumed that the spring has negligible kinetic energy and that air resistance
is negligible. [2]

(iii) The blocks are released at time t = 0.
On Fig. 2.3, sketch a graph to show how the momentum of block A varies with time t until the extension of the spring becomes zero.
Numerical values of momentum and time are not required.


Fig. 2.3
[2]
[Total: 11]





Reference: Past Exam Paper – March 2017 Paper 22 Q2





Solution:
(a) The principle of conservation of momentum states the sum of momentum of the bodies before an interaction is equal to the sum of momentum of the bodies after the interaction for an isolated (closed) system (where no (resultant) external force acts).


(b)
(i)
{The elastic potential energy can be obtained by the area under the force-extension graph.}

EPE = area under graph         or ½ Fx           or ½ kx2 and F = kx

{Using EPE = ½ Fx}
energy = ½ × 12.0 × 8.0 × 10-2 = 0.48 J

or
{Using EPE = ½ kx2               where k = F / x = 12.0 / 0.08 = 150 N m-1}
energy = ½ × 150 × (8.0 × 10-2)2 = 0.48 J

(ii)
1.
{Momentum is always conserved.
Sum of momentum before interaction = 0 as both blocks were stationary.
Sum of momentum after interaction = 4.0 vA - 6.0 vB

Sum of momentum before = Sum of momentum after
4.0 vA - 6.0 vB = 0}

4.0 vA = 6.0 vB
EK = ½ mv2      

{EK of block A = ½ × 4.0 × vA2

Since 4.0 vA = 6.0 vB,
vB = 2 vA / 3
EK of block B = ½ × 6.0 × (2vA/3)2

Ratio = EK of block A / EK of block B}
Ratio = (½ × 4.0 × vA2) / (½ × 6.0 × (2vA/3)2)
Ratio = 1.5     

2.
{Energy is also conserved.
Initial energy (EPE) = 0.48 J
Total energy after interaction = EK of A + EK of B}
0.48 = EK of A + EK of B
{EK of A / EK of B = 1.5
So, EK of B = EK of A / 1.5}

0.48 = EK of A + (EK of A / 1.5)
0.48 = 5/3 × EK of A               
EK of A = 0.29 (0.288) J

(iii)
curve starts from origin and has decreasing gradient
final gradient of graph line is zero                              




{The block is initially at rest. Its speed is zero. So, the momentum is also zero initially.

Momentum depends on velocity (p = mv).
The force in the stretched spring is the tension. As the spring is released, this force causes the block to acceleration. So, its velocity increases with time and so does its momentum.

But the force on the spring depends on the extension (F = kx). As the spring is released, the extension decreases, and so, the force and acceleration produced also decreases with time. That is, the increase in velocity (and momentum) decreases with time. So, the gradient of the graph decreases. (The gradient of a momentum-time graph actually gives the force. F = Δp / Δt.)

When the extension of the spring becomes zero the force on block A will become zero so that the gradient of the graph becomes zero.}
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