Category: International Baccalaureate Higher Level Examination (IBHLE)

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202301130941 Solution to 2014-IBHL-PHY-I-5

Two blocks of weight 5\,\mathrm{N} and 2\,\mathrm{N} are attached to two ropes, \textrm{X} and \textrm{Y}.

The blocks hang vertically. The mass of the ropes is negligible. What is the tension in \textrm{X} and the tension in \textrm{Y}?

Roughwork.

As always start with free-body diagrams. Take upward positive. First the lower block:

Then the higher block:

Or alternatively,

Ans. \begin{cases} T_\textrm{X}=7\,\mathrm{N} \\ T_\textrm{Y}=2\,\mathrm{N} \\ \end{cases}

This problem is not to be attempted.

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202301130918 Solution to 2011-IBHL-PHY-I-3

A skydiver of mass 80\,\mathrm{kg} falls vertically with a constant speed of 50\,\mathrm{m\,s^{-1}}. The upward force acting on the skydiver is approximately

A. 0\,\mathrm{N}.
B. 80\,\mathrm{N}.
C. 800\,\mathrm{N}.
D. 4000\,\mathrm{N}.

Roughwork.

As always start with a free-body diagram.

Take upward positive. Write the equation of motion by Newton’s 2nd law:

\begin{aligned} \mathbf{F}_{\textrm{net}} & = m\mathbf{a} \\ f-mg & = m(0) \\ f & = mg \\ & \approx (80)(10) \\ & = 800 \\ \end{aligned}

And the answer is C.

This problem is not to be attempted.

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202301120908 Solution to 2002-IBHL-PHY-I-9

A pendulum has a bob of mass m and swings in the arc shown below.

As the bob swings through the lowest point of its motion, the tension in the string will be

A. zero.
B. less than mg.
C. equal to mg.
D. greater than mg.

Roughwork.

Assumed the string is l long, taut and inextensible.

Derivation. Read More

    \begin{aligned} T & = \frac{1}{2}ml^2\dot{\theta}^2 \\ V & = mgl(1-\cos\theta ) \\ \mathcal{L} & = T-V \\ \frac{\partial \mathcal{L}}{\partial\theta} & = -mgl\sin\theta \\ \frac{\partial \mathcal{L}}{\partial\dot{\theta}} & = ml^2\dot{\theta} \\ 0 & =\frac{\mathrm{d}}{\mathrm{d}t}\bigg(\frac{\partial\mathcal{L}}{\partial \dot{\theta}}\bigg) - \frac{\partial\mathcal{L}}{\partial \theta} \\ \end{aligned}

    A simple pendulum, under gravity, has

    \textrm{EoM: }\qquad\displaystyle{\ddot{\theta}+\frac{g}{l}\cdot\sin\theta =0}

    the equation of motion, independent of mass m, neglecting friction.

    Lemma. (not in use) Read More

    \begin{aligned} \begin{bmatrix} \hat{\mathbf{e}}_r \\ \hat{\mathbf{e}}_\theta \end{bmatrix} & = \begin{bmatrix} \displaystyle{\frac{\partial x}{\partial r}} & \displaystyle{\frac{\partial y}{\partial r}} \\[2mm]\displaystyle{\frac{1}{r}\frac{\partial x}{\partial\theta}} & \displaystyle{\frac{1}{r}\frac{\partial y}{\partial\theta}} \end{bmatrix}\begin{bmatrix} \hat{\mathbf{e}}_x \\ \hat{\mathbf{e}}_y \end{bmatrix} \\ \begin{bmatrix} \hat{\mathbf{e}}_x \\ \hat{\mathbf{e}}_y \end{bmatrix} & = \begin{bmatrix} \displaystyle{\frac{1}{r}\frac{\partial y}{\partial \theta}} & \displaystyle{-\frac{\partial y}{\partial r}} \\[2mm]\displaystyle{-\frac{1}{r}\frac{\partial x}{\partial\theta}} & \displaystyle{\frac{\partial x}{\partial r}} \end{bmatrix} \begin{bmatrix} \hat{\mathbf{e}}_r \\ \hat{\mathbf{e}}_\theta \end{bmatrix} \\ \end{aligned}

    The net force for this non-uniform circular motion, being centripetal force producing radial plus tangential accelerations,

    \mathbf{a} = -l(\dot{\theta}^2\cos\theta +\ddot{\theta}\sin\theta)\,\hat{\mathbf{i}} + l(\ddot{\theta}\cos\theta -\dot{\theta}^2\sin\theta )\,\hat{\mathbf{j}}

    is the resultant of tension

    \begin{aligned} T & = |\mathbf{T}| \\ \mathbf{T} & =T\sin\theta\,\hat{\mathbf{i}}+T\cos\theta\,\hat{\mathbf{j}}} \end{aligned}

    and weight \mathbf{W}=m\mathbf{g}=-mg\,\hat{\mathbf{j}}. Write, with a free-body diagram in mind,

    \begin{aligned} \because\enspace & \mathbf{F}_{\textrm{net}} =\mathbf{T}+\mathbf{W}=m\mathbf{a} \\ \therefore\enspace & \begin{cases} T\sin\theta = -ml(\dot{\theta}^2\cos\theta+\ddot{\theta}\sin\theta ) \\ T\cos\theta - mg = ml(\ddot{\theta}\cos\theta-\dot{\theta}^2\sin\theta ) \\ \end{cases} \\ \end{aligned}

    When the bob is at the lowest point, i.e., \theta =0,

    \begin{aligned} T & = mg + ml\ddot{\theta} \\ & = mg - mg\sin\theta \qquad (\textrm{by EoM})\\ & \approx mg-mg\theta \qquad (\because\enspace \sin\theta\approx\theta\textrm{ for small }\theta)\\ & \lneq mg \\ \end{aligned}

    The official answer is D but mine B.  Could that be E: more or less mg? If I am wrong, I must have omitted some simple facts of noble truth.

    This problem is not to be attempted.

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202301111028 Solution to 2003-IBHL-I-4

Two forces of magnitudes 7\,\mathrm{N} and 5\,\mathrm{N} act at a point. Which one of the following is not a possible value for the magnitude of the resultant force?

A. 1\,\mathrm{N}
B. 3\,\mathrm{N}
C. 5\,\mathrm{N}
D. 7\,\mathrm{N}

Roughwork.

Suppose

\begin{aligned} \mathbf{F}_1 & = 7\sin\alpha\,\hat{\mathbf{i}} + 7\cos\alpha\,\hat{\mathbf{j}} \\ \mathbf{F}_2 & = 5\sin\beta\,\hat{\mathbf{i}} + 5\cos\beta\,\hat{\mathbf{j}} \\ \end{aligned}

where \alpha ,\beta\in [0,2\pi ).

\begin{aligned} \mathbf{F}_3 & = \mathbf{F}_1+\mathbf{F}_2 \\ & = (7\sin\alpha + 5\sin\beta )\,\hat{\mathbf{i}} + (7\cos\alpha +5\cos\beta )\,\hat{\mathbf{j}} \\ F_3 & = |\mathbf{F}_3| \\ & = \sqrt{(7\sin\alpha + 5\sin\beta )^2 + (7\cos\alpha +5\cos\beta )^2}\\ & = \sqrt{74+70(\sin\alpha\sin\beta +\cos\alpha\cos\beta )}\\ & = \sqrt{74+70\cos (\alpha-\beta)} \\ \end{aligned}

\begin{aligned} \because \quad & \begin{cases} \max (F_3) = \sqrt{74+70(1)} = 12 \\ \min (F_3) = \sqrt{74+70(-1)} = 2 \\ \end{cases} \\ \therefore\quad & F_3 \in [2,12] \\ \end{aligned}

One can actually do without vector analysis for the sake of a mental exercise.

This problem is not to be attempted.

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202301101150 Solution to 1999-IBHL-PHY-I-7

Two 10\,\mathrm{kg} blocks on a smooth horizontal surface are tied together. They are accelerated by a horizontal force of 30\,\mathrm{N} which acts as shown below:

If frictional effects are negligible, what is the tension in the connecting rope?

Roughwork.

As always start with free-body diagrams. Take rightward positive. Considering first the bodies as a whole:

so as to write the equation of motion by Newton 2nd law:

\begin{aligned} F_{\textrm{net}} & =ma \\ 30 & = 20a\\ a & = 1.5\,\mathrm{m\,s^{-2}} \\ \end{aligned}

where the tension regarded as internal force does not count. Considering then as individuals,

so as to write:

\begin{aligned} F_{\textrm{net}} & =ma \\ 30 - T & = (10)(1.5)\\ T & = 15\,\mathrm{N} \\ \end{aligned}

One may also do with the left block, but an exercise left the reader.