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PHYSICS, M1 EQ-Bank 18

A physics student comes across a river which runs north to south and has a current of 3 ms\(^{-1}\) running south.

The student starts on the west side of the river at point A and paddles a kayak at 5 ms\(^{-1}\) directly across the river to finish at point B.

  1. Calculate the angle which he must position the boat to travel in a straight line across the river.   (2 mark)

  2. If the river is 100 metres wide, determine the time it takes for the student to cross the river.   (2 mark)

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a.    \(36.9^{\circ}\)

b.    \(\text{25 seconds}\)

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a.   
       

\(\sin \theta\) \(=\dfrac{3}{5}\)  
\(\theta\) \(=\sin^{-1}\Big(\dfrac{3}{5}\Big)=36.9^{\circ}\)  

 
The student must turn 36.9\(^{\circ}\) into the current as shown on the diagram.

 
b.    Using Pythagoras:

\(v=\sqrt{5^2-3^2}=4\ \text{ms}^{-1}\)

\(t=\dfrac{d}{s}=\dfrac{100}{4}=25\ \text{s}\)
 

\(\therefore\) It will take the student 25 seconds to travel from A to B.

PHYSICS, M1 EQ-Bank 15

Determine the magnitude and direction of the resultant velocity vector for an airplane that is simultaneously moving with a velocity of 350\(^{-1}\) directly towards the west and 275 ms\(^{-1}\) towards the northwest.   (3 marks)

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578 ms\(^{-1}\), N70.3°W

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Add vectors to show resultant vector:

Using the cosine rule to find the magnitude of \(v\):

\(v\) \(=\sqrt{350^2+275^2-2 \times 350 \times 275 \times \cos 135°}\)  
  \(=578.1376\dots\ \text{ms}^{-1}\)  

 

Using the sign rule to find \(\theta\):

\(\dfrac{\sin \theta}{275}\) \(=\dfrac{\sin 135°}{578.1376}\)  
\(\sin \theta\) \(=\dfrac{\sin135° \times 275}{578.1376}\)  
\(\theta\) \(=\sin^{-1}\Big(\dfrac{\sin135° \times 275}{578.1376}\Big)\)  
  \(=19.7^{\circ}\)  

 

\(\therefore\) Resultant velocity of the aeroplane is 578 ms\(^{-1}\), N70.3°W.

PHYSICS, M1 EQ-Bank 14

A car is travelling north and approaching an intersection at 50 kmh\(^{-1}\).

While maintaining a constant speed, the car turns left and continues east at 50 kmh\(^{-1}\). 

Using a vector diagram, calculate the change in velocity of the car.   (3 marks)

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\(\text{70.7 kmh}^{-1}, \text{S45°E} \)

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\(v= 50\ \text{kmh}^{-1}\ \text{east},\ \ u= 50\ \text{kmh}^{-1}\ \text{north.}\)

\(\Delta v= v-u = v+(-u),\ \ \text{where}\ -u= 50\ \text{kmh}^{-1}\ \text{south}\)
  

\(\Delta v=\sqrt{50^2 + 50^2}=\sqrt{5000}=70.7\ \text{kmh}^{-1}\)

\(\tan \theta \) \(=\dfrac{50}{50}\)  
\(\theta\) \(=\tan^{-1}\Big(\dfrac{50}{50}\Big)=45^{\circ}\)  

 

Change in velocity of the car = 70.7 \(\text{kmh}^{-1}\), S45\(^{\circ}\)E.

PHYSICS, M1 EQ-Bank 6

A boat is being rowed due north at a constant speed of 15 ms\(^{-1}\) when it encounters a current of 8 ms\(^{-1}\) going in the direction of S25°W.

Using vectors, determine the resultant velocity of the boat.   (3 marks)

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\(\overset{\rightarrow}R=8.455\ \text{ms}^{-1}\ \text{N}23.6^{\circ}\text{W}\)

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→ Using the cosine rule, the magnitude of \(\overset{\rightarrow}R\) is:

\(\overset{\rightarrow}R\) \(=\sqrt{15^2+8^2-2 \times 15 \times 8 \times \cos25°}\)  
  \(=8.455\ \text{ms}^{-1}\)  

  

→ Using the sine rule, the direction of \(\overset{\rightarrow}R\) is:

\(\dfrac{\sin\theta}{8}\) \(=\dfrac{\sin25°}{8.455}\)  
\(\sin\theta\) \(=\dfrac{8\sin25°}{8.455}\)  
\(\theta\) \(=\sin^{-1}\Big{(}\dfrac{8\sin25°}{8.455}\Big{)}\)  
  \(=23.6^{\circ}\)  

 
\(\therefore \overset{\rightarrow}R =8.455\ \text{ms}^{-1}\ \text{N}23.6^{\circ}\text{W}\)

PHYSICS, M1 2018 VCE 5*

Four students are pulling on ropes in a four-person tug of war. The relative sizes of the forces acting on the various ropes are \(F_{ W }=200 \text{ N} , F_{ X }=240 \text{ N} , F_{ Y }=180 \text{ N}\) and \(F_{ Z }=210 \text{ N}\). The situation is shown in the diagram below.
 

What is the resultant force vector \((F_{\vec{R}})\) acting at the centre of the tug-of-war ropes?   (3 marks)

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→ \(F_{\vec{R}}=36.1\ \text{N}, 33.7^{\circ}\) above the horizontal.

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→ By resolving the vertical vectors of \(F_{ W }=200 \text{ N}\) up and \(F_{ Y }=180 \text{ N}\) down, the net force in the vertical direction is \(F_v=20\ \text{N}\) up.

→ Similarly, the net force in the horizontal direction is \(F_h=30\ \text{N}\) to the right.

→ Thus the resultant vector \((F_{\vec{R}})\) can be calculated using the vector diagram below.
 

→ The magnitude of  \(F_{\vec{R}}=\sqrt{20^2+30^2}\)  and  \(\theta=\tan^{-1}\Big{(}\dfrac{20}{30}\Big{)}\)

→ Hence \(F_{\vec{R}}=36.1\ \text{N}, 33.7^{\circ}\) above the horizontal.