Additional Answer Options : 3.3

2018 - November - Paper 1 - Question 3

3.1

Downwards.

The only force acting on the object is the gravitational force/weight which acts downwards.

The only force acting is gravitational/weight.

Gravitational force/weight acts downwards.

The ball is in free fall.

(Gravitational) acceleration is downwards.

Tip: when an object is in free fall, regardless of whether it is moving upward or downward in motion, the only force acting on the object is the force of gravity which acts downwards.

2 marks

(2 x 1 mark)

3.2

Option 1

Upwards as positive

$v_{f}=v_{i}+a\Delta t$

$0=7,5+(-9,8)\Delta t$

$\Delta\mathrm{t}=0,77\mathrm{~s}$

Downwards as positive

$v_{f}=v_{i}+a\Delta t$

$0=-7,5+(-9,8)\Delta t$

$\Delta\mathrm{t}=0,77\mathrm{~s}$

Tips:

The velocity of the ball at the highest point in its motion, when it is thrown upwards is always zero.
Remember, that all velocities as well as accelerations are vector quantities and, therefore, require directions.

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

3 marks

(3 x 1 mark)

3.3

Option 1

Upwards as positive

At the highest point vf is zero:

$\mathrm{v}_{\mathrm{f}}^2=\mathrm{v}_{\mathrm{i}}^2+2\mathrm{a}\Delta\mathrm{y}$

$0=(7,5)^2+(2)(-9,8)\Delta y$

$\Delta y=2,87(2,869)m$

This is higher than the height needed to reach point T (2,1 m) therefore the ball will pass point T.

Downwards as positive

At the highest point vf is zero:

$\mathrm{v}_{\mathrm{f}}^2=\mathrm{v}_{\mathrm{i}}^2+2\mathrm{a}\Delta\mathrm{y}$

$0=(-7,5)^2+(2)(-9,8)\Delta y$

$\Delta y=-2,87(-2,869)m$

This is higher than the height needed to reach point T (2,1 m) therefore the ball will pass point T.

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

5 marks

(5 x 1 mark)

Marking criteria:

Formula mark
Substitution mark
Mark for height/distance
Mark for comparison
Mark for conclusion

3.4

Tips:

When bulleted things are listed to be indicated on a graph, these must be shown as this is usually how the marks are allocated, and if any of the requirements are left out, you will lose marks.
Since the motion is free fall, the gradient of the line will be constant and equal to 9,8 m·s^-2.

2 marks

(2 x 1 mark)

Marking criteria:

Initial velocity and time for final velocity are shown
Correct straight line (including orientation) drawn

Option 2

Upwards as positive:

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

$\Delta y=(7,5)(0,77)+1/2(-9,8)(0,77)^2$

$\Delta y=2,87\mathrm{~m}(2,86\mathrm{~m})$

This is higher than the height needed to reach point T (2,1 m), therefore, the ball will pass point T.

Downwards as positive:

$\Delta y=v_{i}\Delta t+1/2a\Delta t^2$

$\Delta y=(-7,5)(0,77)+1/2(9,8)(0,77)^2$

$\Delta y=-2,87m(2,869m)$

This is higher than the height needed to reach point T (2,1 m), therefore, the ball will pass point T.

Option 3

$\left.\begin{array}{l}\left(E_{\text{mech }}\right)_{\text{Top }}=\left(E_{\text{mech }}\right)_{\text{Ground}}\\ \left(E_{P}+E_{K}\right)_{\text{Top }}=\left(E_{P}+E_{K}\right)_{\text{Bottom }}\\ \left(mgh+1/2mv^2\right)_{\text{Top }}=\left(mgh+1/2m^2\right)_{\text{Bottom }}\end{array}\right\}Anyone$

(9,8)(h)+0=0+(1/2)(7,5)^2

$\mathrm{h}=2,87\mathrm{~m}(2,869\mathrm{~m})$

This is higher than the height needed to pass the target (2,1 m), therefore, the ball will pass the target.

Option 4

$\mathrm{W}_{\text{net }}=\Delta\mathrm{E}_{\mathrm{K}}$

$mg\Delta x\cos\theta=1/2mv_{f}^2-1/2mv_{i}^2$

$(9,8)\Delta x\cos180^{\circ}=0-1/2(7,5)^2$

$\Delta x=2,87\mathrm{~m}(2,869\mathrm{~m})$

This is higher than the point height needed to pass point T (2,1 m), therefore, the ball will pass point T.

Option 5

Upwards as positive:

If the highest point is y_f then ∆y = (y_f – y_1,6). At the highest point, v_f is zero.

$v_{f}^2=v_{i}^2+2a\Delta y$

$0=\left[(7,5)^2+(2)(-9,8)\left(y_{f}-1,6\right)\right]$

$y_{f}=4,47(4,469)m$

Yes.

This point (4,47m) is higher than point T (or even the required height of 2,1 m), therefore, the ball will pass point T.

Downwards as positive:

If the highest point is y_f then Δy = (y_f – y_1,6). At the highest point, v_f is zero.

$v_{f}^2=v_{i}^2+2a\Delta y$

$0=\left[(-7,5)^2+(2)(9,8)\left\{y_{f}-(-1,6)\right\}\right]$

$y_{f}=-4,47(-4,469)m$

Height is 4,47 m.

This point (4,47 m) is higher than point T (or even the required height of 2,1 m), therefore, the ball will pass point T.

Option 6

Upwards as positive:

If the highest point is y_f then Δy = (y_f – y_1,6) At the highest point, v_f is zero.

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

$\left(y_{f}-1,6\right)=(7,5)(0,77)+1/2(-9,8)(0,77)^2$

$\mathrm{y}_{\mathrm{f}}=4,47\mathrm{~m}(4,469\mathrm{~m})$

This point (4,47m) is higher than point T (or even the required height of 2,1 m), therefore, the ball will pass point T.

Downwards as positive:

If the highest point is y_f then Δy = (y_f – y_1,6). At the highest point, v_f is zero.

$\Delta y=v_{i}\Delta t+1/2a\Delta t^2$

$\left\{y_{f}-(-1,6)\right\}=(-7,5)(0,765)+1/2(9,8)(0,765)^2$

$y_{f}=-4,47m(-4,469m)$

Height is 4,47 m.

This point (4,47m) is higher than point T (or even the required height of 2,1 m), therefore, the ball will pass point T.

Option 7

Upwards as positive:

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

$=\left(\frac{0+7,5}{2}\right)(0,77)$

$=2,89\mathrm{~m}$

This is higher than the height needed to pass the target (2,1 m), therefore, the ball will pass the target.

Downwards as positive:

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

$=\frac{0-7,5}{2}(0,77)$

$=-2,89\mathrm{~m}$

Height is 2,98 m.

This is higher than the height needed to pass the target (2,1 m), therefore, the ball will pass the target.

Option 8

Upwards as positive:

At the highest point, v_f is zero:

$v_{f}^2=v_{i}^2+2a\Delta y$

$0=v_{i}^2-(2)(9,8)(2,1)$

$\mathrm{v}_{\mathrm{i}}=6,42\mathrm{~m}\cdot\mathrm{s}^{-1}$

This is the actual velocity needed to reach the target. The given velocity is greater than the actual velocity needed. The ball will pass the target.

Downwards as positive:

At the highest point, v_f is zero:

$\mathrm{v}_{\mathrm{f}}^2=\mathrm{v}_{\mathrm{i}}^2+2\mathrm{a}\Delta\mathrm{y}$

$0=v_{i}^2+(2)(9,8)(-2,1)$

$v_{i}=6,42\mathrm{~m}\cdot\mathrm{s}^{-1}$

This is the actual velocity needed to pass the target. The given velocity is greater than the actual velocity needed. The ball will reach the target.

Option 9

$\mathrm{W}_{\mathrm{nc}}=\Delta\mathrm{E}_{\mathrm{p}}+\Delta\mathrm{E}_{\mathrm{k}}$

$0=mgh_{f}-mgh_{i}+1/2mv_{f}^2-1/2mv_{j}^2$

$0=(9,8)h_{f}-(9,8)(1.6)+1/2(0)^2-1/2(7,5)^2$

$0=(9,8)h_{f}-43,805$

$\therefore\mathrm{h}_{\mathrm{f}}=4,47\mathrm{~m}$

$\therefore\text{ The ball will pass point }\mathrm{T}$

Option 10

Upwards as positive:

$\Delta\mathrm{t}(\mathrm{max}\text{. height})=0,77\mathrm{s}$

$\Delta y=v_{i}\Delta t+1/2a\Delta t^2$

$2,1=(7,5)\Delta t+1/2(-9,8)\Delta t^2$

$\therefore\Delta t=0,36\mathrm{~s}$

$\therefore\Delta\mathrm{t}\text{ \lparen max height, }0,77\mathrm{~s})>\Delta\mathrm{t}\text{ \lparen to reach point }\mathrm{T},0,36\mathrm{~s})$

$\therefore\text{ The ball passed point }\mathrm{T}$

Downwards as positive:

$\Delta\mathrm{t}(\max\text{ height })=0,77\mathrm{~s}$

$\Delta y=v_{i}\Delta t+1/2a\Delta t^2$

$2,1=(7,5)\Delta t+1/2(-9,8)\Delta t^2$

$\therefore\Delta\mathrm{t}=0,36\mathrm{~s}$

$\therefore\Delta\mathrm{t}\text{ \lparen max height, }0,77\mathrm{~s})>\Delta\mathrm{t}\text{ \lparen to reach point }\mathrm{T},0,36\mathrm{~s})$

$\therefore\text{ The ball passed point }\mathrm{T}$

Option 11

Upwards as positive:

$\Delta y=v_{i}\Delta t+1/2a\Delta t^2$

$(3,7-1,6)=7,5\Delta t+1/2(-9,8)\Delta t^2$

$\Delta t=0,375\mathrm{~s}$

The time to pass point T is less than the time to reach maximum height. The ball will pass point T.

Downwards as positive:

$\Delta y=v_{i}\Delta t+1/2a\Delta t^2$

$(3,7-1,6)=-7,5\Delta t+1/2(9,8)\Delta t^2$

$\Delta\mathrm{t}=0,375\mathrm{~s}$

The time to reach point T is less than the time to reach maximum height. The ball will pass point T.

Option 12

Upwards as positive:

$v_{f}^2=v_{i}^2+2a\Delta y$

$v_{f}^2=(7,5)^2+2(-9,8)(2,1)$

$\mathrm{v}_{\mathrm{f}}=3,88\mathrm{~m}\cdot\mathrm{s}^{-1}$

Velocity at T is 3,88 m·s^-1, therefore, the ball still moving towards its maximum height.

Downwards as positive:

$v_{f}^2=v_{i}^2+2a\Delta y$

$v_{f}^2=(-7,5)^2+2(9,8)(-2,1)$

$\mathrm{v}_{\mathrm{f}}=-3,88\mathrm{~m}\cdot\mathrm{s}^{-1}$

Velocity at T is -3,88 m·s^-1, therefore, the ball is still moving towards its maximum height.