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PHYSICS, M2 EQ-Bank 4 MC

A 2 kg block is placed on a smooth inclined plane angled at 30\(^{\circ}\) to the horizontal. Assume friction is negligible.

What is the most likely motion of the block?

  1. The block will move down the incline with a constant acceleration.
  2. The block will move down the incline with increasing acceleration.
  3. The block will remain at rest.
  4. The block will move down the incline at a constant speed.
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\(A\)

Show Worked Solution
  • The block will experience a constant force down the slope which can be calculated using \(F = mg\ \sin \theta\).
  • As the block is subject to a constant force, it will also experience a constant acceleration down the slope (Newton’s 2nd Law).

\(\Rightarrow A\)

PHYSICS, M2 EQ-Bank 4

A 2 kg crate is being pushed up a rough incline that is angled at 30\(^{\circ}\) to the horizontal. The frictional force acting between the crate and the surface is 10 N.

  1. On the diagram provided, draw and label all the forces acting on the crate as it is pushed up the incline.   (3 marks)
       

  1. Calculate the magnitude of the applied force \(F\) required to accelerate the crate up the incline at 0.8 ms\(^{-2}\)   (2 marks)
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a.    
     

b.    \(F = 21.4\ \text{N}\)

Show Worked Solution

a.    
       

  • The frictional force acts down the slope in this scenario because frictional forces oppose the direction of motion.

b.    Weight force acting down slope \(= mg\sin \theta = 2 \times 9.8 \times \sin 30 = 9.8\ \text{N}\)

  • Total force down the slope \(=9.8 + 10 = 19.8\ \text{N}\)   (weight force down slope + frictional force)
\(F-19.8\) \(=ma_{\text{up slope}}\)  
\(F\) \(= 2 \times 0.8 +19.8\)  
  \(= 21.4\ \text{N}\)  

 
\(\therefore\) The force, \((F)\), acting up slope needs to be \(21.4\ \text{N}\).

PHYSICS, M2 EQ-Bank 3 MC

Two balls are dropped vertically from a cliff. Ball \(\text{X}\) is dropped first. One second later, Ball \(\text{Y}\), which has half the mass of Ball \(\text{X}\), is also dropped from the same height. Assume air resistance is negligible and both balls are dropped from rest.

Which of the following is correct?

  1. Ball \(\text{X}\) will hit the ground with twice the speed of Ball \(\text{Y}\).
  2. Ball \(\text{Y}\) will accelerate slower because it has less mass.
  3. The balls will strike the ground more than one second apart.
  4. The distance between the balls increases while they are falling.
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\(D\)

Show Worked Solution
  • Both balls will experience the same acceleration as they are both being acted on by gravity \((9.8\ \text{ms}^{-2})\).
  • The distance the balls fall during flight is given by  \(s= ut + \dfrac{1}{2}at^2 = \dfrac{1}{2}at^2\ \ (u = 0)\).

When  \(t=1:\)

Distance travelled by Ball \(\text{X} =\dfrac{1}{2} \times 9.8 \times 1= 4.9\ \text{m}\)

Distance travelled by Ball \(\text{Y = 0 m}\) (distance between balls = 4.9 m)
 

When  \(t=2:\)

Distance travelled by Ball \(\text{X} =\dfrac{1}{2} \times 9.8\times 4 = 19.6\ \text{m}\) 

Distance travelled by Ball \(\text{Y} =\dfrac{1}{2} \times 9.8 \times 1 = 4.9\ \text{m}\) (distance between the balls = 14.7 m)

  • Due to the factor of \(t^2\) in the distance equation, the distance between the balls will increase while they are falling.

\(\Rightarrow D\)

PHYSICS, M2 EQ-Bank 2

A block is being dragged along a flat surface with a horizontal force of 30 \(\text{N}\) being applied to it.

The mass of the block is 6 kilograms.

If the block is travelling at a constant velocity, determine the magnitude of the frictional forces acting on the block, giving reasons for your answer.   (2 marks)

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  • As the block is travelling at a constant velocity, the sum of all the forces acting on the block must add to 0. 
  • The frictional forces must be equal in magnitude and opposite to the applied force.
  • Therefore, the magnitude of the frictional forces is 30 \(\text{N}\).
Show Worked Solution
  • As the block is travelling at a constant velocity, the sum of all the forces acting on the block must add to 0. 
  • The frictional forces must be equal in magnitude and opposite to the applied force.
  • Therefore, the magnitude of the frictional forces is 30 \(\text{N}\).

PHYSICS, M2 2018 VCE 8

Two blocks, \(\text{A}\) of mass 4.0 kg and \(\text{B}\) of mass 1.0 kg, are being pushed to the right on a smooth, frictionless surface by a 40 N force, as shown in the diagram.
 

  1. Calculate the magnitude of the force on block \(\text{B}\) by block \(\text{A}\) (\(\left.F_{\text {on B by A}}\right)\). Show your working.   (2 marks)

  2. State the magnitude and the direction of the force on block \(\text{A}\) by block \(\text{B}\) (\(\left.F_{\text {on A by B}}\right)\).   (2 marks)

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a.    \(8\ \text{N}\)

b.    \(8\ \text{N}\) to the left.

Show Worked Solution

a.   Using \(F=ma\), calculate the acceleration of the entire system:

\(a=\dfrac{F}{m}=\dfrac{40}{5}=8\ \text{ms}^{-2}\)

\(F_{\text {on B by A }}=m \times a=1 \times 8=8\ \text{N}\)

♦♦ Mean mark (a) 36%.

 
b.   Newton’s third law of motion:

  • \(F_{\text {on A by B }}\) will be equal in magnitude and opposite in direction.
  • \(F_{\text {on A by B }}= 8\ \text{N}\) to the left.

PHYSICS, M2 2020 VCE 9-10 MC

Two blocks of mass 5 kg and 10 kg are placed in contact on a frictionless horizontal surface, as shown in the diagram below. A constant horizontal force, \(F\), is applied to the 5 kg block.
 
 


 

Question 9

Which one of the following statements is correct?

  1. The net force on each block is the same.
  2. The acceleration experienced by the 5 kg block is twice the acceleration experienced by the 10 kg block.
  3. The magnitude of the net force on the 5 kg block is half the magnitude of the net force on the 10 kg block.
  4. The magnitude of the net force on the 5 kg block is twice the magnitude of the net force on the 10 kg block.

 
Question 10

If the force \(F\) has a magnitude of 250 N, what is the work done by the force in moving the blocks in a straight line for a distance of 20 m?

  1. \(5 \text{ kJ}\)
  2. \(25 \text{ kJ}\)
  3. \(50 \text{ kJ}\)
  4. \(500 \text{ kJ}\)
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\(\text{Question 9:}\ C\)

\(\text{Question 10:}\ A\)

Show Worked Solution

\(\text{Question 9}\)

Using Newton’s second Law:  \(F=ma\ \ \Rightarrow\ \ a=\dfrac{F}{m}\).

  • The blocks will experience the same acceleration.
  • Both blocks will have the same force to mass ratio. Since the 5 kg block is half the mass of the 10 kg block, it will experience half the magnitude of the net force as the 10 kg block.

\(\Rightarrow C\)

♦ Mean mark 49%.

 
\(\text{Question 10}\)

\(W\) \(=F_{\parallel}s\)  
  \(=250 \times 20\)  
  \(=5000\ \text{J}\)  
  \(=5\ \text{kJ}\)  

 
\(\Rightarrow A\)

PHYSICS, M5 2018 HSC 21

  1. Compare the force of gravity exerted on the moon by Earth with the force of gravity exerted on Earth by the moon.    (2 marks)
  1. The acceleration due to gravity on the moon is `1.6 \ text{m s}^(-2)` and on Earth it is `9.8 \ text{m s}^(-2)`. Quantitatively compare the mass and weight of a 70 kg person on the moon and on Earth.   (2 marks)
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a.   Using Newton’s Third Law:

  • The force of gravity of the Earth on the moon is equal in magnitude and opposite in direction to the force of gravity exerted on Earth by the moon.
     

b.   Comparison of mass:

  • The mass of the person on both Earth and the moon is 70 kg.

Comparison of weight:

  • The weight of the person on Earth is given by  `W_e=mg=70 xx9.8=686\ text{N.}`
  • The weight of the person on the moon is given by  `W_m=mg=70xx1.6=112\ text{N.}`
  • The persons weight on Earth is greater than it is on the moon.
Show Worked Solution

a.   Using Newton’s Third Law:

  • The force of gravity of the Earth on the moon is equal in magnitude and opposite in direction to the force of gravity exerted on Earth by the moon.

♦♦ Mean mark (a) 33%.

b.   Comparison of mass:

  • The mass of the person on both Earth and the moon is 70 kg.

Comparison of weight:

  • The weight of the person on Earth is given by  `W_e=mg=70 xx9.8=686\ text{N.}`
  • The weight of the person on the moon is given by  `W_m=mg=70xx1.6=112\ text{N.}`
  • The persons weight on Earth is greater than it is on the moon.