Additional Answer Options: 2.1

2023 - May/June - Paper 1 - Question 2

2.1

Option 1

Downwards as positive:

v_{f}^2=v_{i}^2+2a\Delta y check 

(3,41)^2=(0)^2+(2)a(1,5) checkcheck

\mathrm{a}=3,88\mathrm{~m}\cdot\mathrm{s}^{-2}

Or 

Upwards as positive:

v_{f}^2=v_{i}^2+2a\Delta y check 

(-3,41)^2=(0)^2+(2)a(-1,5) checkcheck

a=-3,88

a=3.88\mathrm{~m}\cdot\mathrm{s}^{-2}

light bulb Tips: 

  • The initial velocity of block B is 0 m·s-1 because the system is initially stationary.
  • The velocities, acceleration, and displacement are all vector quantities; therefore, they need a direction. In this specific question, all of them happen to have the same direction; this is not always the case so be vigilant and read the question carefully.
  • If the initial and final velocities are swapped around, you can only get the mark for the formula.

There are more ways to answer this question. To view other options, click here.

3 marks

(3 x 1 mark)

2.2

 

Or

 

wcheck  Fg / Fw / weight/mg/gravitational force/ FEarth on block/73,5N
Tcheck  Tension/ FTension/FRope/FT/F

light bulb Tip: only two vertical forces are acting on block B; the upward tension in the rope and the weight of block B.

2 marks

(2 x 1 mark)

2.3

When a resultant/net force acts on an object, the object will accelerate in the direction of the force with an acceleration that is directly proportional to the force and inversely proportional to the mass of the object. checkcheck 

Or

The resultant/net force acting on an object is equal to the rate of change of momentum of the object in the direction of the resultant/net force. checkcheck 

2 marks

(1 x 2 marks)

2.4

 light bulb How to approach this question:

  • Always choose the direction of acceleration as the positive direction.
  • In this case, choose to make the downward direction, the positive direction. This would mean that the weight is positive and the tension in the rope is negative.

Calculation of tension (Block B)

Downwards as positive:

\left.\begin{array}{rl}\mathrm{F}_{\mathrm{net}} & =\mathrm{ma}\\ \mathrm{F}_{\mathrm{g}}+\mathrm{T} & =\mathrm{ma}\\ \mathrm{mg}-\mathrm{T} & =\mathrm{ma}\end{array}\right\}Anyone (1) check 

7,5(9,8)-T=7,5(3,88)(2 + 3)  

\mathrm{T}=44,40\mathrm{~N}

Or

Upwards as positive:

\left.\begin{array}{rl}\mathrm{F}_{\mathrm{net}} & =\mathrm{ma}\\ \mathrm{T}-\mathrm{F}_{\mathrm{g}} & =\mathrm{ma}\\ \mathrm{T}-\mathrm{mg} & =\mathrm{ma}\end{array}\right\}Anyone (1) check 

T-7,5(9,8)=7,5(-3,88) (2 + 3) check

\mathrm{T}=44,40\mathrm{~N}

 

Calculation of mass (Block A)

Upwards as positive:

  • Since block A accelerates upwards, it is recommended to make upwards positive.

\mathrm{F}_{\mathrm{net}}=\mathrm{ma}

\mathrm{T}-\mathrm{F}_{\mathrm{g}}=\mathrm{ma}

\mathrm{T}-\mathrm{mg}=\mathrm{ma}

  • The tension acting on block A has the same magnitude as the tension that the rope exerted on block B because of Newton's third law and the fact that the pulley is frictionless.
  • The acceleration of blocks A and B are the same because they are tied together.

44,40-m(9,8)=m(3,88) (4) check 

\mathrm{m}=3,25\mathrm{~kg} (5) check 

Or

Downwards as positive:

\mathrm{F}_{\mathrm{net}}=\mathrm{ma}

\mathrm{F}_{\mathrm{g}}-\mathrm{T}=\mathrm{ma}

\mathrm{mg}-\mathrm{T}=\mathrm{ma}

\mathrm{m}(9,8)-44,40=\mathrm{m}(-3,88) (4) check 

\mathrm{m}=3,25\mathrm{~kg}(5) check 

  • Make sure that the directions chosen as positive remain the same throughout your workings. If you change direction in the middle of a question, you can lose marks or get the answer incorrect.
  • For this question, choose the direction in which each object would accelerate positively, even if it is different for both objects. This rule only applies to objects tied together.

5 marks

(5 x 1 mark)

Marking criteria:

  • (1) Any correct

formula check 

  • (2 + 3) Correct

substitution to

calculate tension

checkcheck 

  • (4) Correct

substitution to

calculate the mass

of block A check 

  • (5) Correct final

answer: 3,25 kg check 

2.5

light bulb How to approach this question:

  • Block A experiences a tensile force until it reaches a height of 1,5m above the ground; at this point, block B will hit the ground with a velocity of 3,41 m·s-1 (both blocks A and B will be travelling at the same velocity because they are tied together).
  • Block A will continue to move upwards as a projectile because it will accelerate downwards at 9,8 m·s-2 since the only force acting on it is the force of gravity.
  • It is recommended that you keep using the upward direction as the positive direction.
  • At the highest point in the motion, the velocity of block A will be 0 m·s-1.

Option 1

Upwards as positive:

v_{f}^2=v_{i}^2+2a\Delta y (1) check 

\left(0^2\right)=(3,41)^2+(2)(-9,8)\Delta y(2 + 3) checkcheck 

  • Remember, the question asked for the maximum height above the ground.
  • This displacement calculated () is only the displacement of the block above the point of release, not the maximum height reached so you need to add the height above the ground with the displacement calculated above.

\text{ Maximum height }=0,59+1,5 (4) check 

=2,09\mathrm{~m}(5)  

Or

Downwards as positive:

v_{f}^2=v_{i}^2+2a\Delta y (1) check 

\left(0^2\right)=(-3.41)^2+(2)(9.8)\Delta y (2 + 3) check 

\Delta y=-0,59

\Delta y=0,59m

\text{ Maximum height }=0,59+1,5 (4) check 

=2,09\mathrm{~m} (5) check 

There are more ways to answer this question. To view other options, click here.

5 marks

(5 x 1 mark)

Marking criteria:

  • (1) Any correct

formula check 

  • (2) Correct

substitution of vi

and vcheck 

  • (3) Correct

substitution of

9,8 m·s-2 check 

  • (4) Adding 1,5m

to calculate Δcheck 

  • (5) Correct final

answer: 2,09m check 


Or

Option 2

Downwards as positive:

Part 1

\Delta\mathrm{y}=\left(\frac{\mathrm{v}_{\mathrm{i}}+\mathrm{v}_{\mathrm{f}}}{2}\right)\Delta\mathrm{t}

1,5=\left(\frac{0+3,41}{2}\right)\Delta t

\Delta t=0,88\mathrm{~s}

Part 2

\mathrm{v}_{\mathrm{f}}=\mathrm{v}_{\mathrm{i}}+\mathrm{a}\Delta\mathrm{t}

-3,41=(0)+a(0,88)

a=-3,88

\mathrm{a}=3,88\mathrm{~m}\cdot\mathrm{s}^{-2}

Or

\Delta\mathrm{y}=\mathrm{v}_{\mathrm{i}}\Delta\mathrm{t}+1/2\mathrm{a}\Delta\mathrm{t}^2

-1,5v=(0)(0,88)+1/2a(0,88)^2

a=-3,88

a=3,88\mathrm{~m}\cdot\mathrm{s}^{-2}


Or

Upwards as positive:

Part 1

\Delta\mathrm{y}=\left(\frac{\mathrm{v}_{\mathrm{i}}+\mathrm{v}_{\mathrm{f}}}{2}\right)\Delta\mathrm{t}

-1,5=\left(\frac{0-3,41}{2}\right)\Delta t

\Delta\mathrm{t}=0,88\mathrm{~s}

Part 2

\mathrm{v}_{\mathrm{f}}=\mathrm{v}_{\mathrm{i}}+\mathrm{a}\Delta\mathrm{t}

3,41=(0)+a(0,88)

\mathrm{a}=3,88\mathrm{~m}\cdot\mathrm{s}^{-2}

Or

\Delta\mathrm{y}=\mathrm{v}_{\mathrm{i}}\Delta\mathrm{t}+1/2\mathrm{a}\Delta\mathrm{t}^2

1,5=(0)(0,88)+1/2a(0,88)^2

\mathrm{a}=3,88\mathrm{~m}\cdot\mathrm{s}^{-2}

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