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Physics / Chapter 3: Vector / Section 3-1

Vectors and Scalars — Question 2

Section 3-1 · Physics pages 314–315

Question 2 Moving an arrow does not necessarily change its vector Four displacement arrows lie in the same horizontal plane with a common north reference and a common drawing scale. Their exact specifications are: A and B are each 5 m at 30◦north of east, with B drawn at another location. C is 5 m at 30◦south of west; D is 3 m at 30◦north of east. Here displacements are compared as free vectors: only magnitude and direction enter equality. Start and finish locations are separate event data. A: 5 m B: 5 m C: 5 m D: 3 m east −→; north ↑ Common drawing scale: A, B and D lie 30 degrees north of east; C lies 30 degrees south of west. (a) Identify every equal pair of vectors and every pair with equal magnitudes. Which arrows have the same direction? Which have opposite directions? (b) Could sliding C across the page, without rotating or resizing it, make it equal to A? Explain. (c) A student says A and B cannot be equal because their tails differ. State the error and what genuinely differs between the two drawings. (d) Do equal displacement vectors establish identical journeys or identical final positions? Construct two journeys with equal displacement and different endpoints, and explain why traveled distances can also differ.
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Question 2 - Solution Moving an arrow does not necessarily change its vector A: 5 m B: 5 m C: 5 m D: 3 m east −→; north ↑ (a) Test both magnitude and direction.⃗ A =⃗B = −⃗C = 5 3⃗D , A = B = C = 5 m, D = 3 m Only A and B are equal. Equal-magnitude pairs are A–B, A–C and B–C. A, B and D point together; C points opposite to all three. (b) Translation cannot reverse an arrow. (⃗rf +⃗a) −(⃗ri +⃗a) =⃗rf −⃗ri =⃗C = −⃗A Translating both endpoints leaves their difference unchanged. C needs a 180◦rotation to match A; sliding alone cannot do it. (c) Separate the vector from its representative placement.⃗ rf,B −⃗ri,B =⃗rf,A −⃗ri,A,⃗ ri,B̸ =⃗ri,A Drawn tails and tips identify different events, whereas the directed separation is the same. Equality of free vectors does not require their representatives to share a tail. (d) A common change does not specify common endpoints or paths.⃗ ri,B =⃗ri,A + 10ˆj m ⇒⃗rf,B =⃗rf,A + 10ˆj m; sdetour = 5 m + ℓ Choose the second start 10 m north of the first, then make the same 5 m displacement. Endpoints differ by 10 m. An added closed detour of length ℓ> 0 increases distance without changing displacement.

Question 2

Moving an arrow does not necessarily change its vector

Four displacement arrows lie in the same horizontal plane with a common north reference and a common drawing scale. Their exact specifications are: A and B are each 55 m at 30∘30^\circ north of east, with B drawn at another location. C is 55 m at 30∘30^\circ south of west; D is 33 m at 30∘30^\circ north of east. Here displacements are compared as free vectors: only magnitude and direction enter equality. Start and finish locations are separate event data.

Diagram shown in the original SVG worksheet.

  1. Identify every equal pair of vectors and every pair with equal magnitudes. Which arrows have the same direction? Which have opposite directions?

  2. Could sliding C across the page, without rotating or resizing it, make it equal to A? Explain.

  3. A student says A and B cannot be equal because their tails differ. State the error and what genuinely differs between the two drawings.

  4. Do equal displacement vectors establish identical journeys or identical final positions? Construct two journeys with equal displacement and different endpoints, and explain why traveled distances can also differ.

Question 2 - Solution

Moving an arrow does not necessarily change its vector

Diagram shown in the original SVG worksheet.

(a) Test both magnitude and direction.

A→=B→=−C→=53D→,A=B=C=5m,D=3m\boxed{\vec A=\vec B=-\vec C=\tfrac53\vec D},\qquad A=B=C=5\,\mathrm m,\ D=3\,\mathrm m Only A and B are equal. Equal-magnitude pairs are A–B, A–C and B–C. A, B and D point together; C points opposite to all three.

(b) Translation cannot reverse an arrow.

(r→f+a→)−(r→i+a→)=r→f−r→i=C→=−A→(\vec r_f+\vec a)-(\vec r_i+\vec a)=\vec r_f-\vec r_i=\vec C=-\vec A Translating both endpoints leaves their difference unchanged. C needs a 180∘180^\circ rotation to match A; sliding alone cannot do it.

(c) Separate the vector from its representative placement.

r→f,B−r→i,B=r→f,A−r→i,A,r→i,B≠r→i,A\vec r_{f,B}-\vec r_{i,B}=\vec r_{f,A}-\vec r_{i,A},\qquad\vec r_{i,B}\ne\vec r_{i,A} Drawn tails and tips identify different events, whereas the directed separation is the same. Equality of free vectors does not require their representatives to share a tail.

(d) A common change does not specify common endpoints or paths.

r→i,B=r→i,A+10ĵm⇒r→f,B=r→f,A+10ĵm;sdetour=5m+ℓ\vec r_{i,B}=\vec r_{i,A}+10\hat j\,\mathrm m\Rightarrow\vec r_{f,B}=\vec r_{f,A}+10\hat j\,\mathrm m;\quad s_{\mathrm{detour}}=5\,\mathrm m+\ell Choose the second start 10 m north of the first, then make the same 5 m displacement. Endpoints differ by 10 m. An added closed detour of length ℓ>0\ell>0 increases distance without changing displacement.

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