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Mechanics, SPEC2 2021 VCAA 5

A mass of `m_1` kilograms is placed on a plane inclined at 30° to the horizontal. It is connected by a light inextensible string to a second mass of `m_2` kilograms that hangs below a frictionless pulley situated at the top end of the incline, over which the string passes.
 


 

  1. Given that the inclined plane is smooth, find the relationship between `m_1` and `m_2` if the mass `m_1` moves down the plane at constant speed.  (2 marks)

The masses are now placed on a rough plane inclined at 30°, with the light inextensible string passing over a frictionless pulley in the same way, as shown in the diagram above. Let `N` be the magnitude of the normal force exerted on the mass `m_1` by the plane. A resistance force of magnitude `lambdaN` acts on and opposes the motion of the mass `m_1`.

  1. The mass `m_1` moves up the plane.
  2.   i. Mark and label all forces acting on this mass on the diagram above.  (1 mark)
  3.  ii. Taking the direction up the plane as positive, find the acceleration of the mass `m_1` in terms of `m_1`, `m_2` and `lambda`.  (2 marks)

Some time after the masses have begun to move, the mass `m_2` hits the ground at 4.5 ms`\ ^(-1)` and the string becomes slack. At this instant, the mass `m_1` is at the point `P` on the plane, which is 2 m from the pulley. Take the value of `lambda` to be 0.1

  1. How far from point `P` does the mass `m_1` travel before it starts to slide back down the plane?
  2. Give your answer in metres, correct to two decimal places.  (2 marks)
  3. Find the time taken, from when the string becomes slack, for the mass `m_1` to return to point `P`.
  4. Give your answer correct to the nearest tenth of a second.  (3 marks)
Show Answers Only
  1. `2m_2`
  2. i. 
       
     
  3. ii. `a = ((2m_2 – m_1 – lambda m_1 sqrt3)g)/(m_1 + m_2)`
  4. `s = 1.76\ text{m}`
  5. `1.7\ text{seconds}`
Show Worked Solution

a.   `m_1g sin30 – m_2g = (m_1 + m_2)a`

`text(S)text(ince)\ m_1\ text(moves at constant speed,)\ a = 0`

`m_1 g · 1/2 – m_2 g` `= 0`
`m_1` `= 2m_2`

 
b.i.   

 

b.ii.   `text(S)text(ince)\ m_1\ text(is moving up the slope)`

♦ Mean mark part (b)(ii) 43%.

`m_2g – m_1 g · 1/2 – lambdam_1 g · cos 30` `= (m_1 + m_2)a`
`m_2g – m_1 g · 1/2 – lambdam_1 g · cos sqrt3/2` `= (m_1 + m_2)a`

`:. a = ((2m_2 – m_1 – lambda m_1 sqrt3)g)/(2(m_1 + m_2))`

 

c.   `text(After)\ m_2\ text(hits the ground)`

♦♦♦ Mean mark part (c) 17%.
`m_1a` `=-m_1g*1/2 – lambda m_1 g sqrt3/2`
`a` `= -g/2(1 + lambda sqrt3)`
  `= -g/2(1 + 0.1 xx sqrt3)`

 
`text(By CAS, solve)\ \ v^2 = u^2 + 2as,\ text(for)\ \ s:`

`0 = 4.5^2 – g(1 + 0.1 xx sqrt3)s`

`s = 1.76\ text{m (to 2 d.p.)}`

 

d.   `u = 4.5\ \ text(ms)^(-1), s = 1.76\ text(m),\ a = -g/2(1 + 0.1sqrt3)`

♦♦♦ Mean mark part (d) 7%.

`text(Let)\ \ t_1 = text(Time travelling up slope until stopping)`

`s = ut_1 + 1/2at^2`

`1.76 = 4.5t_1 – 1/2 · g/2(1 + 0.1sqrt3)t_1^2`

`t_1 = 0.78\ text(seconds)`
 

`text(Let)\ t_2 =\ text(time travelling down the slope)`

`=>\ text(friction is reversed)`

`a` `= g/2(1 – 0.1sqrt3)`
`1.76` `= 0 xx t_2 + 1/2 · g/2(1 – 0.1sqrt3)  t_2^2`
`t_2` `= 0.93\ text(seconds)`

 

`:.\ text(Total time)` `= t_1 + t_2`
  `= 0.78 + 0.93`
  `= 1.7\ text{seconds (to 1 d.p.)}`

Mechanics, SPEC2 2021 VCAA 16 MC

An object of mass `m` kilograms slides down a smooth slope that is inclined at an angle of `theta^@` to the horizontal, where  `0^@ < theta^@ < 45^@`. The acceleration of the object down the slope is  `a\ text(ms)^(-2), a > 0`.

If the angle of inclination of the slope is doubled to `2theta^@`, then the acceleration of the object down the slope, in `text(ms)^(-2)`, is

  1. `2a`
  2. `(2a)/gsqrt(g^2 - a^2)`
  3. `(2a^2 - g^2)/g`
  4. `a/g sqrt(g^2 - a^2)`
  5. `2asqrt(g^2 - a^2)`
Show Answers Only

`B`

Show Worked Solution

`ma = mgsintheta`

♦ Mean mark 48%.
`sintheta` `= a/g`
`cos^2theta` `= 1 – (a^2)/(g^2)`
`costheta` `= sqrt(1 – (a^2)/(g^2))`

 
`text(If incline angle) = 2theta`

`ma` `= mgsin(2theta)`
`a` `= g*2sinthetacostheta`
  `= g *2* a/g sqrt(1 – (a^2)/(g^2))`
  `= (2a)/g sqrt(g^2 – a^2)`

 
`=>\ B`

Mechanics, SPEC2 2021 VCAA 15 MC

The diagram below shows a stationary body being acted on by four forces whose magnitudes are in newtons. The force of magnitude `F_1` newtons acts in the opposite direction to the force of magnitude 8 N.
 

The value of `F_1` is

  1. `8-2sqrt3`
  2. `2sqrt3`
  3. `8`
  4. `8 + 2sqrt3`
  5. `8-3sqrt3`
Show Answers Only

`A`

Show Worked Solution

`text(Resolve forces vertically:)`

`F_2sin60^@` `= 6sin30^@`
`F_2 sqrt3/2` `= 6 xx 1/2`
`F_2` `= 6/sqrt3`
  `= 2sqrt3`

 
`text(Resolve forces horizontally:)`

`F_1 + 6cos30^@` `= 2sqrt3 cos60^@ + 8`
`F_1 + 3sqrt3` `= sqrt3 + 8`
`F_1` `= 8 – 2sqrt3`

`=>\ A`

Mechanics, SPEC2 2020 VCAA 5

Two objects, each of mass `m` kilograms, are connected by a light inextensible strings that passes over a smooth pulley, as shown below. The object on the platform is initially at point A and, when it is released, it moves towards point C. The distance from point A to point C is 10 m. The platform has a rough surface and, when it moves along the platform, the object experiences a horizontal force opposing the motion of magnitude `F_1` newtons in the section AB and a horizontal force opposing the motion of magnitude `F_2` newtons when it moves in the section BC.
 

  1. On the diagram above, mark all forces that act on each object once the object on the platform has been released and the system is in motion.  (2 marks)

The force `F_1` is given by  `F_1 = kmg, \ k ∈ R^+`.

  1.  i. Show that an expression for the acceleration, in `text(ms)^(−2)`, of the object on the platform, in terms of `k`, as it moves from point A to point B is given by  `(g(1 - k))/2`.  (2 marks)
  2. ii. The system will only be in motion for certain values of `k`.
  3.     Find these values of `k`.  (1 mark)

Point B is midway between points A and C.

  1. Find, in terms of `k`, the time taken, is seconds, for the object on the platform to reach point B.  (2 marks)
  2. Express, in terms of `k`, the speed `v_B`, in `text(ms)^(−1)`, of the object on the platform when it reaches point B.  (2 marks)
  3. When the object on the platform is at point B, the string breaks. The velocity of the object at point B is  `v_B = 2.5\ text(ms)^(−1)`. The force that opposes motion from point B to point C is  `F_2 = 0.075 mg + 0.4 mv^2`, where `v` is the velocity of the object when it is a distance of `x` metres from point B. The object on the platform comes to rest before point C.
  4. Find the object's distance from point C when it comes to rest. Give your answer in metres, correct to two decimal places.  (4 marks)
Show Answers Only
  1.  
  2.  i. `text(See Worked Solutions)`
  3. ii. `k ∈ (0, 1), k ∈ R^+`
  4. `2sqrt(5/(g(1 – k)))`
  5. `v_text(B) = sqrt(5g(1 – k))`
  6. `3.15\ text(m)`
Show Worked Solution
a.   

 

b. i.   `text(Horizontally:)`

`ma = T – F_1 = T – kmg\ …\ (1)`

`text(Vertically:)`

`ma = mg – T\ …\ (2)`
 

`text(Add)\ \ (1) + (2) :`

`2ma = mg – kmg`

`:. a` `= (g – kg)/2`
  `= (g(1 – k))/2`
♦ Mean mark (b)(ii) 45%.

 

b. ii.   `text(System in motion when)\ a > 0`

`(g(1 – k))/2 > 0`

`:. k ∈ (0, 1), \ k ∈ R^+`

 

c.   `text(AB) = 5\ (text(given)), u = 0\ (text(given))`

`text(Find)\ t\ text(when)\ s = 5:`

`s = ut + 1/2at^2`

`5 = 0 + 1/2 · (g(1 – k))/2 · t^2`

`t^2` `= 20/(g(1 – k))`
`t` `= sqrt(20/(g(1 – k)))`
  `= 2sqrt(5/(g(1 – k)))`

 

d.   `text(At B,)\ s = 5`

`v_text(B)^2` `= u^2 + 2as`
  `= 0 + 2 · (g(1 – k))/2 · 5`
  `= 5g(1 – k)`

 
`:. v_text(B) = sqrt(5g(1 – k))`

 

e.   `text(Acceleration is against the direction of motion.)`

♦♦ Mean mark (e) 35%.
`a` `= −F/m`
  `= −0.075g – 0.4v^2`
  `= −0.4(0.1875g + v^2)`

 

`d/(dx)(1/2 v^2)` `= −0.4(0.1875g + v^2)`
`d/(dx)(v^2)` `= −0.8(0.1875g + v^2)`
`(dx)/(d(v^2))` `= −1.25(1/(0.1875g + v^2))`
`:. x` `= −1.25 int_(2.5^2)^0 1/(0.1875 + v^2)\ dv^2`
  `= 1.85\ text(m)`

 

`:.\ text(Distance from C)` `= 5 – 1.85`
  `= 3.15\ text(m)`

Mechanics, SPEC2 2020 VCAA 18 MC

A particle of mass `m` kilogram hangs from a string that is attached to a fixed point. The particle is acted on by a horizontal force of magnitude `F` newtons. The system is in equilibrium when the string makes an angle `alpha` to the horizontal, as shown in the diagram below. The tension in the string has magnitude `T` newtons.
 


 

The value of  `tan(alpha)`  is

  1. `(mg)/T`
  2. `T/(mg)`
  3. `T/F`
  4. `F/(mg)`
  5. `(mg)/F`
Show Answers Only

`E`

Show Worked Solution

`text(Resolving forces vertically:)`

`mg = Tsin(alpha)`
 

`text(Resolving forces horizontally:)`

`F = Tcos(alpha)`

`:. tan(alpha) = (mg)/F`

`=>E`

Mechanics, SPEC1 2020 VCAA 1

A 2 kg mass is initially at rest on a smooth horizontal surface. The mass is then acted on by two constant forces that cause the mass to move horizontally. One force has magnitude 10 N and acts in a direction 60° upwards from the horizontal, and the other force has magnitude 5 N and acts in a direction 30° upwards from the horizontal, as shown in the diagram below.
 


 

  1. Find the normal reaction force, in newtons, that the surface exerts on the mass.  (2 marks)
  2. Find the acceleration of the mass, in ms−2, after it begins to move.  (2 marks)
  3. Find how far the mass travels, in metres, during the first four seconds of motion.   (1 mark)
Show Answers Only
  1. `R = 2g – 5/2 – 5 sqrt 3\ text(N)`
  2. `a = 5/2 – (5 sqrt 3)/2\ text(ms)^(-2)`
  3. `20 – 10 sqrt 3\ text(m)`
Show Worked Solution
a.  

`text(Resolving forces vertically:)`

`2g` `= 5 sin 30 + 10 sin 60 + R`
`2g` `= 5/2 + 5 sqrt 3 + R`
`R` `= 2g – 5/2 – 5 sqrt 3\ text(N)`

 

b.    `2a` `= 10 cos 60 – 5 cos 30`
  `2a` `= 5 – (5 sqrt 3)/2`
  `:.a` `= 5/2 – (5 sqrt 3)/4\ text(ms)^(-2)`
Mean mark part (c) 51%.

 

c.    `text(Distance)` `= ut + 1/2 at^2`
    `= 0+ 1/2 (5/2 – (5 sqrt 3)/4) × 4^2`
    `= 20 – 10 sqrt 3\ text(m)`

Mechanics, SPEC2-NHT 2019 VCAA 5

A pallet of bricks weighing 500 kg sits on a rough plane inclined at an angle of  `α°` to the horizontal, where  `tan(α°) = (7)/(24)`. The pallet is connected by a light inextensible cable that passes over a smooth pulley to a hanging container of mass `m` kilograms in which there is 10 L of water. The pallet of bricks is held in equilibrium by the tension `T` newtons in the cable and a frictional resistance force of 50 `g` newtons acting up and parallel to the plane. Take the weight force exerted by 1 L of water to be `g` newtons.
  


 

  1. Label all forces acting on both the pallet of bricks and the hanging container on the diagram above, when the pallet of bricks is in equilibrium as described.   (1 mark)
  2. Show that the value of `m` is 80.  (3 marks)

Suddenly the water is completely emptied from the container and the pallet of bricks begins to slide down the plane. The frictional resistance force of 50 `g` newtons acting up the plane continues to act on the pallet.

  1. Find the distance, in metres, travelled by the pallet after 10 seconds.  (3 marks)
  2. When the pallet reaches a velocity of  `3\ text(ms)^-1`, water is poured back into the container at a constant rate of 2 L per second, which in turn retards the motion of the pallet moving down the plane. Let  `t`  be the time, in seconds, after the container begins to fill. 
  3.   i. Write down, in terms of  `t`, an expression for the total mass of the hanging container and the water it contains after `t` seconds. Give your answer in kilograms.  (1 mark)
  4.  ii. Show that the acceleration of the pallet down the plane is given by  `(text(g)(5 - t))/(t + 290)\ text(ms)^-2`  for  `t ∈[0, 5)`.  (2 marks)
  5. iii. Find  the velocity of the pallet when  `t = 4`. Give your answer in metres per second, correct to one decimal place.  (2 marks)
Show Answers Only
  1.  
    `qquad`
  2. `text(Proof(Show Worked Solution))`
  3. `(25 text(g))/(29)`
  4.   i. `80 + 2t`
     ii. `text(Proof (Show Worked Solution))`
    iii. `3.4\ text(ms)^-1`
Show Worked Solution

a.

 
b.   `text(Resolving vertical forces on container:)`

`T – (m + 10)g = 0 \ …\ (1)`

`text(Resolving forces on plane:)`
 


 

`tan α = (7)/(24) \ => \ sin α = (7)/(25)`
 

`text(Solve for m:)`

`(m + 10)g` `= 500 text(g) · (7)/(25) – 50 text(g)`
`m + 10` `= 140 – 50`
`:. \ m` `= 80`

 

c.   `text(Resolving vertical forces on container:)`

`T – 80 g = 80 a \ …\ (1)`

`text(Resolving forces on plane:)`

`500 g sin α – (T + 50 g) = 500 a`

`90 g – T = 500 a \ …\ (2)`

`text(Add) \ (1) + (2)`

`10 g` `= 580 a`
`a` `= (g)/(58)`
`s` `= ut + (1)/(2) at^2`
  `= 0 + (1)/(2) · (g)/(58) + 10^2`
  `= (25g)/(29)`

 

d.i.   `m = 80 + 2t`
 

d.ii.   `text(Resolving vertical forces on container:)`

`T – (80+2t)g = (80+2t)a \ …\ (1)`

`text(Resolving forces on plane:)`

`90g – T = 500a \ …\ (2)`

`text(Add)\ (1) + (2)`

`(90 – 80 – 2t)g` `= (500 + 80 + 2t)a`
`(10 – 2t)g` `=(580 + 2t)a`
`a` `= (g(5 – t))/(t + 290) ms^-2`

 

d.iii.   `(dv)/(dt) = (g(5 – t))/(t + 290)`

`v = int (dv)/(dt)\ dt = 295 log_e ((t + 290)/(290)) – g t + c`

 
`text(When)\ t = 0, v = 3\ \ text{(given)} \ => \ c = 3`

`:. \ v = 295 log_e ((t+290)/(290)) – g t + 3`

`:. \ v(4) = 3.4\ text(ms)^-1`

Mechanics, SPEC1-NHT 2019 VCAA 1

A 10 kg mass is placed on a rough plane that inclined at 30° to the horizontal, as shown in the diagram below. A force of 40 N is applied to the mass up the slope and parallel to the slope. There is also a frictional resistance force of magnitude `F` that opposes  the motion of the mass.
 


 

  1. Find the magnitude of the frictional resistance force, in newtons, acting up the slope if the force is just sufficient to stop the mass from sliding down the slope.  (2 marks)
     
  2. An additional force of magnitude `P` newtons is applied to the mass  up the slope and parallel to the slope. The sum of the additional force and the frictional resistance force of magnitude `F` that now acts down the slope is such that it is just sufficient to stop the mass from sliding up the slope.  (2 marks)
Show Answers Only
  1. `9 \ text(N)`
  2. `18 \ text(N)`
Show Worked Solution

a.       

`40 + F` `= 10g sin30°`
`F` `= 98 xx 0.5 – 40`
  `= 9\ text(N)`

 

b.   `text(Frictional force)\ F\ text(acts down slope)`

`40 + P` `= 10g sin30° + F`
`P` `= 5g + F – 40`
  `= 49 + 9 – 40`
  `= 18\ text(N)`

Mechanics, SPEC2 2019 VCAA 5

A mass of  `m_1`  kilograms is initially held at rest near the bottom of a smooth plane inclined at `theta` degrees to the horizontal. It is connected to a mass of  `m_2`  kilograms by a light inextensible string parallel to the plane, which passes over a smooth pulley at the end of the plane. The mass  `m_2`  is 2 m above the horizontal floor.

The situation is shown in the diagram below.
 


 

  1. After the mass  `m_1`  is released, the following forces, measured in newtons, act on the system:

     

    • weight forces  `W_1`  and  `W_2`
    •  the normal reaction force  `N`
    •  the tension in the string  `T`

     

    On the diagram above, show and clearly label the forces acting on each of the masses.  (1 mark) 

  2. If the system remains in equilibrium after the mass  `m_1`  is released, show that  `sin(theta) = (m_2)/(m_1)`.  (1 mark)
  3. After the mass  `m_1`  is released, the mass  `m_2`  falls to the floor.

     

    1. For what values of  `theta`  will this occur? Express your answer as an inequality in terms of  `m_1`  and  `m_2`.  (1 mark)
    2. Find the magnitude of acceleration, in ms−2, of the system after the mass  `m_1`  is released and before the mass  `m_2`  hits the floor. Express your answer in terms of  `m_1, \ m_2`  and  `theta`.  (2 marks)
  4. After the mass  `m_1`  is released, it moves up the plane.
    Find the maximum distance, in metres, that the mass  `m_1`  will move up the plane if  `m_1 = 2m_2`  and  `sin(theta) = 1/4`.  (5 marks)
Show Answers Only
  1.   
  2. `text(See Worked Solutions)`
    1. `theta < sin^(−1)\ ((m_2)/(m_1)), theta ∈ (0, pi/2)`
    2. `a = (g(m_2 – m_1 · sin(theta)))/(m_1 + m_2)`
  3. `10/3 \ text(m)`
Show Worked Solution
a.   

 

b.   `T = m_2g\ \ … \ (1)`

`T = m_1sin(theta)\ \ … \ (2)`

`text{Solve:  (1) = (2)}`

`m_1g sin(theta)` `= m_2g`
`sin(theta)` `= (m_2)/(m_1)`

 

c.i.   `m_2g > m_1gsin(theta)`

`sin(theta) < (m_2)/(m_1)`

`theta < sin^(−1)\ ((m_2)/(m_1)), \ \ theta ∈ (0, pi/2)`

 

c.ii.   `text(Net force)\ (F) = m_2g – m_1gsin(theta)`

`(m_1 + m_2) · a` `= g(m_2 – m_1 sin(theta))`
`:. a` `= (g(m_2 – m_1 sin(theta)))/(m_1 + m_2)`

 

d.   `text(Motion:)\ m_1\ text(will accelerate up the plane for 2 m.)`

 `m_1\ text(will then decelerate up plane until)\ v = 0.`

`text(Find)\ v_(m_1)\ \ text(given)\ \ s_(m_1) = 2, \ u = 0, \ m_1=2m_2`

`a = (g(m_2 – m_1  sintheta))/(m_1 + m_2) = (g(m_2 – 2m_2 · 1/4))/(2m_2 + m_2) = g/6`
 

`text(Using)\ \ v^2 = u^2 + 2as,`

`v_(m_1)^2 = 0 + 2 · g/6 · 2 = (2g)/3`

`text(Find distance)\ (s_2)\ text(for)\ m_1\ text(to decelerate until)\ v = 0:`

`a = −gsintheta = −g/4, \ u = sqrt((2g)/3)`

`0` `= (2g)/3 – 2 · g/4 · s_2`
`s_2` `= (2g)/3 xx 2/g`
  `= 4/3`

 

`:.\ text(Maximum distance)` `= 2 + 4/3`
  `= 10/3\ text(m)`

Mechanics, SPEC2 2019 VCAA 17 MC

A particle is held in equilibrium by three coplanar forces of magnitudes  `F_1, F_2`  and  `F_3`.

The angles between these forces are  `alpha, beta`  and  `gamma`  as shown in the diagram below.
 

If  `beta = 2alpha`, then  `(F_1)/(F_2)`  is equal to

  1. `1/2 sin(alpha)`
  2. `2sin(alpha)`
  3. `1/2text(cosec)(alpha)`
  4. `1/2cos(alpha)`
  5. `1/2sec(alpha)`
Show Answers Only

`E`

Show Worked Solution

`text(Using Lami’s theorem:)`

`(F_1)/(sin alpha)` `= (F_2)/(sinbeta)`
`(F_1)/(F_2)` `= (sin alpha)/(sin beta)`
  `= (sin alpha)/(sin 2alpha)`
  `= (sin alpha)/(2sin alphacos alpha)`
  `= 1/2 sec alpha`

 
`=>E`

Mechanics, SPEC2 2019 VCAA 14 MC

A 4 kg mass is held at rest on a smooth surface. It is connected by a light inextensible string that passes over a smooth pulley to a 2 kg mass, which in turn is connected by the same type of string to a 1 kg mass. This is shown in the diagram below.
 


 

When the 4 kg mass is released, the tension in the string connecting the 1 kg and 2 kg masses is `T` newtons. The value of `T` is

  1. `(4g)/7`
  2. `(3g)/7`
  3. `g/7`
  4. `(6g)/7`
  5. `g`
Show Answers Only

`A`

Show Worked Solution

`text(Considering the whole system:)`

`text(Total mass = 7 kg)`

`text(Net Force ↓ =)\ 3g`

`F` `= ma`
`3g` `= 7a`
`:.a` `= (3g)/7`

 
`text(Consider the forces on the 1 kg mass:)`

`g – T` `= a`
`g – T` `= (3g)/7`
`T` `= g – (3g)/7`
  `= (4g)/7`

 
`=>A`

Mechanics, SPEC1 2019 VCAA 9

  1. A light inextensible string is connected at each end to a horizontal ceiling. A mass of `m` kilograms hangs in equilibrium from a smooth ring on the string, as shown in the diagram below. The string makes an angle `alpha` with the ceiling.
      
    `qquad qquad`
     
    Express the tension, `T` newtons, in the string in terms of `m`, `g` and `alpha`.  (1 mark)
  2. A different light inextensible sting is connected at each end to a horizontal ceiling. A mass of `m` kilograms hangs from a smooth ring on the string. A horizontal force of `F` newtons is applied to the ring. The tension in the sting has a constant magnitude and the system is in equilibrium. At one end the string makes an angle `beta` with the ceiling and at the other end the string makes an angle `2beta` with the ceiling, as shown in the diagram below.
     

     
    Show that  `F = mg((1 - cos(beta))/(sin(beta)))`.  (3 marks)
Show Answers Only
  1. `T = (mg)/(2sinalpha)`
  2. `text(See Worked Solutions)`
Show Worked Solution
a.   
`2 xx Tsinalpha` `= mg`
`:.T` `= (mg)/(2sinalpha)`

 

b.   

 
`text(Resolving forces vertically:)`

`Tsin(beta) + Tsin(2beta)` `= mg`
`T` `= (mg)/(sin(beta) + sin(2beta))`

 
`text(Resolving forces horizontally:)`

`F + Tcos(2beta)` `= Tcos(beta)`
`F` `= Tcos(beta) – Tcos(2beta)`
  `= T(cos(beta) – cos(2beta))`
  `= T[cos(beta) – (2cos^2beta – 1)]`
  `= T(−2cos^2(beta) + cos(beta) + 1)`
  `= T(−2cos(beta) – 1)(cos(beta) – 1)`
  `= (mg(1 -cos(beta))(2cosbeta + 1))/(sin(beta) + 2sin(beta)cos(beta))`
  `= (mg(1 – cos(beta))(2cos(beta) + 1))/(sin(beta)(1+2cos(beta)))`
  `= mg((1 – cos(beta))/(sin(beta)))`

Mechanics, SPEC2 2012 VCAA 22 MC

A 12 kg mass is suspended in equilibrium from a horizontal ceiling by two identical light strings. Each string makes an angle of 60° with the ceiling, as shown.
 


 

The magnitude, in newtons, of the tension in each string is equal to

A.     `6 g`

B.   `12 g`

C.   `24 g`

D.   `4 sqrt 3 g`

E.   `8 sqrt 3 g`

Show Answers Only

`D`

Show Worked Solution

`T/(sin 30^@)` `= (12 g)/(sin 120^@)`
`2T` `= (24 g)/sqrt3`
`T` `= (12 g)/sqrt3`
  `= 4g sqrt 3`

 
`=> D`

Mechanics, SPEC2 2012 VCAA 20 MC

Particles of mass 3 kg and `m` kg are attached to the ends of a light inextensible string that passes over a smooth pulley, as shown.
 


 

If the acceleration of the 3 kg mass is  `4.9\ text(m/s)^2`  upwards, then

A.   `m` = 4.5

B.   `m` = 6.0

C.   `m` = 9.0

D.   `m` = 13.5

E.   `m` = 18.0

Show Answers Only

`C`

Show Worked Solution

`sum F` `=\ text(total mass × acceleration)`
`mg – 3g` `= (m + 3) xx 4.9`
`9.8m-3xx9.8` `= 4.9m + 14.7`
`:. m` `= 9`

 
`=> C`

Mechanics, SPEC2 2012 VCAA 14 MC

A particle is acted on by two forces, one of 6 newtons acting due south, the other of 4 newtons acting in the direction N60° W.

The magnitude of the resultant force, in newtons, acting on the particle is

  1. `10`
  2. `2sqrt7`
  3. `2sqrt19`
  4. `sqrt(52 - 24sqrt3)`
  5. `sqrt(52 + 24sqrt3)`
Show Answers Only

`B`

Show Worked Solution

Mean mark 51%.

`x^2` `= 6^2 + 4^2 – 2(6)(4) cos 60^@`
  `= 28`
   
`:. x` `= sqrt 28`
  `= 2 sqrt 7`

 
`=> B`

Mechanics, SPEC1 2011 VCAA 7

A flowerpot of mass `m` kg is held in equilibrium by two light ropes, both of which are connected to a ceiling. The first rope makes an angle of 30° to the vertical and has tension `T_1` newtons. The second makes an angle of 60° to the vertical and has tension `T_2` newtons.
 

VCAA 2011 spec 7b
 

  1. Show that `T_2 = T_1/sqrt 3.`  (1 mark)
  2. The first rope is strong, but the second rope will break if the tension in it exceeds 98 newtons.

     

    Find the maximum value of `m` for which the flowerpot will remain in equilibrium.  (3 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `m = 20\ text(kg)`
Show Worked Solution

a.   `text(Forces in equilibrium:)`

`tan 30^@` `=T_2/T_1`
`T_2` `= T_1 xx tan30^@`
`T_2` `= T_1/sqrt3\ \ text(.. as required)`

 

b.    `sin30^@` `= (T_2)/(mg)`
  `T_2` `= (mg)/2`
  `98` `=(m_text(max) xx 9.8)/2`
  `m_text(max)` `=(2 xx 98)/9.8`
    `=20\ text(kg)`

Mechanics, SPEC2 2013 VCAA 16 MC

Forces of magnitude 5 N, 7 N and `Q` N act on a particle that is in equilibrium, as shown in the diagram below.
 

SPEC2 2013 VCAA 16 MC
 

The magnitude of `Q`, in newtons, can be found by evaluating

A.   `sqrt(5^2 + 7^2 - 2 xx 5 xx 7 cos(70^@))`

B.   `5^2 + 7^2 - 2 xx 5 xx 7 cos(110^@)`

C.   `sqrt(5^2 + 7^2 - 2 xx 5 xx 7 cos(110^@))`

D.   `5^2 + 7^2 - 2 xx 5 xx 7 cos(70^@)`

E.   `sqrt(5^2 + 7^2 - 2 xx 5 xx 7 cos(20^@))`

Show Answers Only

`A`

Show Worked Solution

`text(Using cosine rule:)`
 


 

`Q^2` `= 5^2 + 7^2 – 2(5)(7)cos(70^@)`
`Q` `= sqrt(5^2 + 7^2 – (5)(7)cos(70^@))`

 
`=> A`

Mechanics, SPEC1 2013 VCAA 1

A body of mass 10 kg is held in place on a smooth plane inclined at 30° to the horizontal by a tension force, `T` newtons, acting parallel to the plane.

  1. On the diagram below, show all other forces acting on the body and label them.  (1 mark)
     

     

                   VCAA 2013 spec 1a
     

  2. Find the value of `T.`  (2 marks)
Show Answers Only
  1.  

  2. `49\ \ text(N)`
Show Worked Solution
a.   

 

b.    `T – 10g\ sin30^@` `= 0`
  `T – (10g)/2` `= 0`
  `T` `= (10g)/2`
  `T` `= 5g`
    `=49\ \ text(N)`

Mechanics, SPEC2 2014 VCAA 20 MC

Particles of mass 3 kg and 5 kg are attached to the ends of a light inextensible string that passes over a fixed smooth pulley, as shown above. The system is released from rest.

Assuming the system remains connected, the speed of the 5 kg mass after two seconds is

A.     4.0 m/s

B.     4.9 m/s

C.     9.8 m/s

D.   10.0 m/s

E.   19.6 m/s

Show Answers Only

`B`

Show Worked Solution
`sum F` `= 5g – 3g = 2g`
  `=2 xx 9.8`
  `=19.6\ \ text(N)`
`(5+3)a` `= 19.6`
`a` `=2.45`

  
`u = 0, quad t = 2, quad a = 2.45`

`v` `= u + at`
  `= 0 + 2.45 xx 2`
  `=4.9\ \ text(m/s)`

 
`=> B`

Mechanics, SPEC2 2014 VCAA 18 MC

A body on a horizontal smooth plane is acted upon by four forces, `underset ~F_1`, `underset ~F_2`, `underset ~F_3` and `underset ~F_4` as shown.

The force `underset ~F_1` acts in a northerly direction and the force `underset ~F_4` acts in a westerly direction.
 

 

Given that  `|\ underset ~F_1\ | = 1`, `|\ underset ~F_2\ | = 2`, `|\ underset ~F_3\ | = 4` and `|\ underset ~F_4\ | = 5`, the motion of the body is such that it

A.   is in equilibrium.

B.   moves to the west.

C.   moves to the north.

D.   moves in the direction 30° south of west.

E.   moves to the east.

Show Answers Only

`E`

Show Worked Solution

`text(Resolving forces horizontally:)`

♦ Mean mark 45%.

`sum F` `= underset ~F_2 cos30 + underset ~F_3 sin60 –  underset ~F_4`
  `= 2 xx sqrt3/2 + 4 xx sqrt3/2 -5`
  `~~0.2\ \ text(East)`

 
`text(Resolving forces vertically:)`

`sum F` `= underset ~F_1 + underset ~F_2 sin30 – underset ~F_3 cos 60`
  `= 1 + 2 xx 1/2 -4 xx 1/2`
  `= 0`

 
`:.\ text(Body moves to the east.)`

`=> E`

Mechanics, SPEC2 2015 VCAA 19 MC

A light inextensible string passes over a smooth pulley, as shown below, with particles of mass 1 kg and `m` kg attached to the ends of the string.
 

SPEC2 2015 VCAA 19 MC

If the acceleration of the 1 kg particle is 4.9 `text(ms)^(-2)` upwards, then `m` is equal to

A.   1

B.   2

C.   3

D.   4

E.   5

Show Answers Only

`C`

Show Worked Solution

`T – (9.8 xx 1) = 4.9 xx 1`

`T=14.7`
 

`m xx 9.8 – T` `=m xx 4.9`
`4.9m` `= 14.7“
`:. m` `= 14.7/4.9`
  `= 3`

`=> C`

Mechanics, SPEC2 2015 VCAA 16 MC

SPEC2 2015 VCAA 16 MC

The diagram above shows a mass suspended in equilibrium by two light strings that make angles of `60^@` and `30^@` with a ceiling. The tensions in the strings are `T_1` and `T_2`, and the weight force acting on the mass is `underset~W`. The correct statement relating the given forces is

A.   `underset~T_1 + underset~T_2 + underset~W = underset~0`

B.   `underset~T_1 + underset~T_2 - underset~W = underset~0`

C.   `underset~T_1 xx 1/2 + underset~T_2 xx sqrt3/2 = underset~0`

D.   `underset~T_1 xx sqrt3/2 + underset~T_2 xx 1/2 = underset~W`

E.   `underset~T_1 xx 1/2 + underset~T_2 xx sqrt3/2 = underset~W`

Show Answers Only

`A`

Show Worked Solution

`text(S)text(ince equilibrium exists:)`

♦♦♦ Mean mark 26%.

`sumunderset~F = underset~(T_1) + underset~(T_2) + underset~W = underset~0`

 
`=> A`

Mechanics, SPEC2 2016 VCAA 14 MC

Two light strings of length 4 m and 3 m connect a mass to a horizontal bar, as shown below

The strings are attached to the horizontal bar 5 m apart.
 


 

Given the tension in the longer string is  `T_1`  and the tension in the shorter string is  `T_2`,  the ratio of the tensions  `T_1/T_2`  is

A.   `3/5`

B.   `3/4`

C.   `4/5`

D.   `5/4`

E.   `4/3`

Show Answers Only

`B`

Show Worked Solution

`text{Strings make up 3-4-5 (right-angled) triangle:}`

♦ Mean mark 41%.

`text(Let)\ \ theta =\ text(angle between 3m string and horizontal bar)`

`T_1/T_2` `= cot theta`
  `= 3/4`

  
`=>  B`

Mechanics, SPEC2-NHT 2017 VCAA 5

A 5 kg mass is initially held at rest on a smooth plane that is inclined at 30° to the horizontal. The mass is connected by a light inextensible string passing over a smooth pulley to a 3 kg mass, which in turn is connected to a 2 kg mass.

The 5 kg mass is released from rest and allowed to accelerate up the plane.

Take acceleration to be positive in the directions indicated.
 

  1. Write down an equation of motion, in the direction of motion, for each mass.   (3 marks)
  2. Show that the acceleration of the 5 kg mass is  `g/4\ text(ms)^(-2)`.  (1 mark)
  3. Find the tensions  `T_1`  and  `T_2`  in the string in terms of  `g`.  (2 marks)
  4. Find the momentum of the 5 kg mass, in kg ms`­^(-1)`, after it has moved 2 m up the plane, giving your answer in terms of `g`.  (2 marks)
  5. A resistance force  `R`  acting parallel to the inclined plane is added to hold the system in equilibrium, as shown in the diagram below.
     

     

    `qquad`
     

     

    Find the magnitude of  `R`  in terms of  `g`.  (2 marks)

Show Answers Only
  1. `2a = 2g – T_2`

     

    `3a = 3g + T_2 – T_1`

     

    `5a = T_1 – (5g)/2`

  2. `text(Proof)\ text{(See Worked Solutions)}`
  3. `T_1 = (15g)/4`

     

    `T_2 = (3g)/2`

  4. `p = 5 sqrt g`
  5. `R = (5g)/2`
Show Worked Solution
a.    `2\ text(kg): \ 2a` `= 2g – T_2`
  `3\ text(kg): \ 3a` `= 3g + T_2 – T_1`
  `5\ text(kg): \ 5a` `= T_1 – 5g sin (30^@)`
  `5a` `= T_1 – (5g)/2`

 

 

b.   `sum F = 3g + 2g – 5g sin 30^@ = (5 + 3 + 2)a`

`5g – (5g)/2` `= 10a`
`a` `= (5g)/(2 xx 10)`
`:. a` `= g/4\ text(ms)^(-2)`

 

c.   ` T_1 – (5g)/2` `= 5a`
`:. T_1` `= (5g)/2 + 5(g/4)`
  `= (15g)/4`

 

`2g – T_2` `= 2a`
`:. T_2` `= 2g – 2(g/4)`
  `= (3g)/2`

 

d.   `u = 0,\ \ a = g/4,\ \ s = 2`

`text(Find)\ \ v\ \ text(when)\ \ s=2:`

`v^2` `= u^2 + 2as`  
  `=0 + 2 (g/4) xx 2`  
  `=g`  
`v` `=sqrtg\ \ \ (v>0)`  

 
`:. p = 5 sqrt g`

 

e.   `sum F = 2g + 3g – 5g sin 30^@ – R = 0`

`:. R` `= 2g + 3g – 5g sin 30^@`
  `= 5g – (5g)/2`
  `= (5g)/2`

Mechanics, SPEC2-NHT 2018 VCAA 3

A 200 kg crate rests on a smooth plane inclined at `theta` to the horizontal. An external force of `F` newtons acts up the plane, parallel to the plane, to keep the crate in equilibrium.

  1. On the diagram below, draw and label all forces acting on the crate.  (1 mark)

 

 

  1. Find `F` in terms of `theta`.  (1 mark)

The magnitude of the external force `F` is changed to 780 N and the plane is inclined at  `theta = 30^@`.

    1. Taking the direction down the plane to be positive, find the acceleration of the crate.  (2 marks)
    2. On the axes below, sketch the velocity–time graph for the crate in the positive direction for the first four seconds of its motion.  (1 mark)
      `qquad`
       

       
       
    3. Calculate the distance the crate travels, in metres, in its first four seconds of motion.  (1 mark)

Starting from rest, the crate slides down a smooth plane inclined at  `alpha`  degrees to the horizontal.

A force of  `295 cos(alpha)`  newtons, up the plane and parallel to the plane, acts on the crate.

  1. If the momentum of the crate is 800 kg ms¯¹ after having travelled 10 m, find the acceleration, in ms¯², of the crate.  (2 marks)
  2. Find the angle of inclination, `alpha`, of the plane if the acceleration of the crate down the plane is 0.75 ms¯².  Give your answer in degrees, correct to one decimal place.  (2 marks)
Show Answers Only
  1. `text(See Worked Solutions)`
  2. `F = 200\ text(kg)\ sin(theta)`
    1. `a = 1\ text(ms)^(-2)`
    2. `text(See Worked Solutions)`
    3. `8 text(m)`
  4. `a = 0.8\ text(ms)^(-2) quad text(down the incline)`
  5. `a ~~ 12.9^@`
Show Worked Solution
a.   

 

b.   `F – 200g sin (theta) = 0`

`:. F = 200g sin (theta)`

 

c.i.   `sum F` `= -F + 200g sin (30^@)`
  `200a` `= -780 + 200g xx 1/2`
    `=- 780 + 980`
  `:.a` `=1\ text(ms)^(-1)`

 

c.ii.  

 

c.iii.  `text(Distance travelled in 4 seconds)`

`=\ text(Area under graph between)\ \ t=0 and t=4`

`=1/2 xx 4 xx 4`

`= 8\ text(m)`
 

d.   `p` `=mv`
  `800` `=200v`
  `:.v` `=4`

 
`text(Find)\ \a\ \ text(given)\ \ x = 10,\ \ v=4:`

`v^2` `= u^2 + 2ax`  
`16` `=20a`  
`:.a` `=0.8\ \ text(ms)^(-2) quad text(down the incline)`  

 

e.  `200g sin(alpha) – 295 cos (alpha) = 200 xx 0.75`

`alpha ~~ 12.9^@\ \ \ text{(by CAS)}`

Mechanics, SPEC2 2018 VCAA 5

Luggage at an airport is delivered to its owners via a 15 m ramp that is inclined at 30° to the horizontal. A 20 kg suitcase, initially at rest at the top of the ramp, slides down the ramp against a resistance of `v` newtons per kilogram, where `v\ text(ms)^(-1)` is the speed of the suitcase.

  1.  On the diagram below, show all forces acting on the suitcase during its motion down the ramp.  (1 mark)
     
         

  2.  i. By resolving forces parallel to the ramp, write down an equation of motion for the 20 kg suitcase.  (1 mark)
  3. ii. Hence, show that the magnitude of the acceleration, `a\ text(ms)^(-2)`, of the suitcase down the ramp is given by  `a = (g - 2v)/2`. (1 mark)
  4. By expressing `a` in an appropriate form, find the distance `x` metres that the suitcase has slid as a function of `v`. Give your answer in the form  `x = bv + c log_e(c/(c - v))`, where  `b, c in R`.  (2 marks)
  5. Find the velocity of the suitcase just before it reaches the end of the ramp. Give your answer in `text(ms)^(-1)`, correct to two decimal places.  (1 mark)
  6.  i. Write down a definite integral that gives the time taken for the suitcase to reach a speed of `4.5\ text(ms)^(-1)`.  (1 mark)
  7. ii. Find the time taken for the suitcase to reach a speed of `4.5\ text(ms)^(-1)`. Give your answer in seconds, correct to two decimal places.  (1 mark)
Show Answers Only
  1.    

    1. `10g – 20v = 20a`
    2. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `x = -v + 4.9 ln ((4.9)/(4.9 – v))`
  3. `v ~~4.81\ text(ms)^(-1)`
    1. `t(4.5) = int_0^4.5 2/(g – 2v) dv`
    2. `t(4.5) ~~ 2.51\ text(seconds)`
Show Worked Solution
a.   

♦ Mean mark part (a) 44%.

 

b.i.    `20g sin 30^@ – 20v` `= sumF`
  `10g – 20v` `= 20a`

 

b.ii.    `(10g – 20v)/20` `= a`
  `g/2 – v` `= a`
  `:.a` `= (g – 2v)/2`

 

♦ Mean mark part (c) 48%.

c.   `v *(dv)/(dx)` `= (g – 2v)/2`
  `(dv)/(dx)` `= (g – 2v)/(2v)`
  `(dx)/(dv)` `= (2v)/(g – 2v)`
  `(dx)/(dv)` `= -(2v)/(2v – g)`
    `= – ((2v – g + g))/(2v – g)`
    `= -1 – g/(2v – g)`

 

`x` `= int_0^v – 1 – g/(2v – g)\ dv`
  `= int_0^v – 1 – g/2 (2/(2v – g))\ dv`
  `= [-v – 4.9ln\ |2v – g|]_0^v`

 
`text(When)\ \ x=0, v=0:`

`2v – g < 0\ \ =>\ \ |2v – g| = g – 2v`

 

`x` `= [-v – 4.9 ln (g – 2v)]_0^v`
  `= -v – 4.9 ln (g – 2v) – (0 – 4.9 ln (g))`
  `= -v – 4.9 ln (g – 2v) + 4.9 ln (g)`
  `= -v + 4.9 ln (g/(g – 2v))`
  `=-v + 4.9 ln(9.8/(9.8-2v))`
`:. x` `= -v + 4.9 ln ((4.9)/(4.9 – v))`

 

d.   `text(Find)\ \ v\ \ text(when)\ \ x=15:`

`15 = -v + 4.9 ln (4.9/(4.9 – v))`

♦♦ Mean mark part (d) 26%.

`:. v ~~ 4.81\ text(ms)^(-1)\ \ \ (v>0)`

 

♦ Mean mark 41%.

e.i    `(dv)/(dt)` `= (g – 2v)/2`
  `(dt)/(dv)` `= 2/(g – 2v)`
  `:.t` `= int_0^4.5 2/(g – 2v)\ dv`

 

♦ Mean mark 41%.

e.ii.  `t=[-ln (9.8 -2v)]_0^4.5`

`=> t ~~ 2.51 text(s)`

Mechanics, SPEC1-NHT 2017 VCAA 1

A 5 kg mass on a smooth plane inclined at 30° is held in equilibrium by a horizontal force of magnitude `P` newtons, as shown in the diagram below.
 


 

  1. On the diagram above, show all other forces acting on the mass and label them. (1 mark)
  2.  Find  `P`.  (2 marks)
Show Answers Only
  1.  `text(See Worked Solutions)`
  2. `P = (5g sqrt 3)/3`
Show Worked Solution
a.   

 

b.    `P cos 30^@` `= 5g sin 30^@`
  `(P sqrt 3)/2` `= (5g)/2`
  `P sqrt 3` `= 5g`
  `P` `= (5g)/sqrt 3`
  `P` `= (5g sqrt 3)/3`

Mechanics, SPEC1 2016 VCAA 1

A taut rope of length `1 2/3` m suspends a mass of 20 kg from a fixed point `O`. A horizontal force of `P` newtons displaces the mass by 1 m horizontally so that the taut rope is then at an angle of  `theta`  to the vertical.

  1. Show all the forces acting on the mass on the diagram below.  (1 mark)

 

 

  1. Show that  `sin (theta) = 3/5`.  (1 mark)
  2. Find the magnitude of the tension force in the rope in newtons.  (2 marks)
Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(Proof)\ text{(See Worked Solutions)}`
  3. `T = 245\ text(N)`
Show Worked Solution
a.   

 

b.    `sin(theta)` `= 1/(1 2/3)`
    `= 1/((5/3)`
    `= 3/5`

 

c.   

`cos theta = (20g)/T`

`sin theta = 3/5\ \ text{(using part b)}`

`underbrace{(400g^2)/T^2 + 9/25}_(cos^2 theta + sin^2 theta = 1)` `= 1`
`(400g^2)/T^2` `= 16/25`
`(20g)/T` `=4/5`
`T/(20g)` `= 5/4`
`T` `= 25g`
  `=245\ text(N)`

Mechanics, SPEC1 2014 VCAA 8

A body of mass 5 kg is held in equilibrium by two light inextensible strings. One string is attached to a ceiling at `A` and the other to a wall at `B`. The string attached to the ceiling is at an angle `theta` to the vertical and has tension `T_1` newtons, and the other string is horizontal and has tension `T_2` newtons. Both strings are made of the same material.

  1. i. Resolve the forces on the body vertically and horizontally, and express `T_1` in terms of  `theta`.  (2 marks)
  2. ii. Express `T_2` in terms of  `theta`.  (1 mark)
  3.  Show that  `tan (theta) < sec (theta)`  for  `0 < theta < pi/2`.  (1 mark)
  4.  The type of string used will break if it is subjected to a tension of more than 98 N.

     

     Find the maximum allowable value of  `theta`  so that neither string will break.  (3 marks)

Show Answers Only
  1.  i. `T_1 = 5g sec theta`
  2. ii. `T_2 = 5g tan theta`
  3.  `text(Proof)\ text{(See Worked Solutions)}`
  4.  `theta_max = pi/3`
Show Worked Solution
a.i.   

`text(Resolve vertically): \ 5g = T_1 cos theta`

`text(Resolve horizontally): \ T_1 sin theta = T_2`

`:. T_1 = 5g sec theta`

 

a.ii.    `T_2` `= 5g sec theta sin theta`
    `= 5g tan theta`

 

b.    `cos theta in (0, 1), \ theta in (0, pi/2)`
  `sin theta in (0, 1), \ theta in (0, pi/2)`

`:. sin theta < 1, \ theta in (0, pi/2)`

♦ Mean mark part (b) 40%.

 

`text(If)\ sin theta < 1 and cos theta > 0:`

`(sin theta)/(cos theta) < 1/(cos theta)`

`:.  tan theta < sec theta, quad 0 < theta < pi/2`

 

c.   `tan theta < sec theta \ \ =>\ \  T_2\ text(will be smaller) => \ T_1\ text(will break first)`

♦ Mean mark part (c) 46%.

`5text(g)\ sec theta_max` `= 98`
`sec theta_max` `= 98/(5 xx 9.8)`
`sec theta_max` `= 10/5`
`sec theta_max` `= 2`
`:.  theta_max` `= pi/3`

Mechanics, SPEC2 2017 VCAA 17 MC

Forces of 10 N and 8 N act on a body as shown below.
 


 

The resultant force acting on the body will, correct to one decimal place, have

  1. magnitude 15.6 N and act at 26.3° to the 10 N force.
  2. magnitude 9.2 N and act at 49.1° to the 10 N force.
  3. magnitude 15.6 N and act at 33.7° to the 10 N force.
  4. magnitude 9.2 N and act at 70.9° to the 10 N force.
  5. magnitude 15.6 N and act at 49.1° to the 10 N force.
Show Answers Only

`A`

Show Worked Solution

 
 

`|∑underset~F|^2` `= 10^2 + 8^2 – 2 xx 10 xx 8cos120°`
`|∑underset~F|` `= sqrt244`
  `~~ 15.6`

 

`(sin(theta))/8` `= (sin(120°))/15.62`
`sin(theta)` `= (4sqrt3)/15.62`
`:. theta` `~~ 26.3°`

 
`=> A`

Mechanics, SPEC2 2017 VCAA 16 MC

An object of mass 20 kg, initially at rest, is pulled along a rough horizontal surface by a force of 80 N acting at an angle of 40° upwards from the horizontal. A friction force of 20 N opposes the motion.

After the pulling force has acted for 5 seconds, the magnitude of the momentum, in kg ms−1, of the object is closest to

  1.     10
  2.     40
  3.   160
  4.   210
  5. 4100
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`D`

Show Worked Solution


 

`∑underset~F = 80cos(40°) – 20`

`ma` `= 80cos(40°) – 20`
`a` `=4 cos(40°) – 1`

 

`Deltav` `= 0+at`
  `= 5(4cos(40°) – 1)`

 

`m Deltav` `= 100(4cos(40°) – 1)`
  `~~ 206.4`

 
`=> D`

Mechanics, SPEC2 2017 VCAA 14 MC

Two particles with mass `m_1` kilograms and `m_2` kilograms are connected by a taut light string that passes over a smooth pulley. The particles sit on smooth inclined planes, as shown in the diagram below.
 


 

If the system is in equilibrium, then  `m_1/m_2`  is equal to

  1.  `(sec(theta))/2`
  2.  `2sec(theta)`
  3.  `2cos(theta)`
  4.  `1/2`
  5.  `1`
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`A`

Show Worked Solution

Mean mark 51%.

`∑F` `=m_1gsin(2theta) – m_2gsin(theta) = 0`
   
`m_1gsin(2theta)` `= m_2gsin(theta)`
`m_1/m_2` `= (sin(theta))/(sin(2theta))`
  `= (sin(theta))/(2sin(theta)cos(theta))`
  `= 1/(2 cos(theta))`
  `= 1/2 sec(theta)`

 
`=> A`

Mechanics, SPEC2-NHT 2018 VCAA 14 MC

The diagram above shows a particle at `O` in equilibrium in a plane under the action of three forces of magnitudes `P, Q` and `R`.

Which one of the following statements is false?

A.   `R = Q sin (60^@)`

B.   `Q = R sin (60^@)`

C.   `P = R sin(30^@)`

D.   `Q cos (60^@) = P cos (30^@)`

E.   `P cos (60^@) + Q cos (30^@) = R`

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`A`

Show Worked Solution

 

`text(By elimination:)`

`sin 60^@` `= Q/R`
`Q` `= R\ sin 60^@ \ \ text{(B correct)}`
`sin30^@` `= P/R`
`P` `= R\ sin30^@\ \ text{(C correct)}`

 

`text(Equating horizontal forces:)`

`Q\ cos 60^@ = P\ cos 30^@\ \ text{(D correct)}`

 

`text(Equating vertical forces:)`

`P\ cos 60^@ + Q\ cos 30^@ = R\ \ text{(E correct)}`

 
`=>  A`

Mechanics, SPEC1-NHT 2018 VCAA 1

A light inextensible string hangs over a frictionless pulley connecting masses of 3 kg and 7 kg, as shown below.
 


 

  1.  Draw all of the forces acting on the two masses on the diagram above. (1 mark)
  2.  Calculate the tension in the string.  (2 marks)
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  1.  `text(See Worked Solutions)`
  2. `(21g)/5 N`
Show Worked Solution
a.   

 

b.    `sum F = 7g – 3g = 4g` ` = (7 + 3)a`
  `a = (4g)/10= (2g)/5`  
  `sum F_3 = T – 3g` `= 3a`
  `T – 3g` `= (6g)/5`
  `T` `= 3g + (6g)/5`
    `= (21g)/5\ \ N`

Mechanics, SPEC2 2018 VCAA 16 MC

The diagram below shows a mass being acted on by a number of forces whose magnitudes are labelled. All forces are measured in newtons and the system is in equilibrium.
 

The value of `F_2` is

A.   `sqrt 2/2 (8 + 3 sqrt 3)`

B.   `(11 sqrt 2)/2`

C.   `(3 sqrt 2)/2`

D.   `7.78`

E.   `7.0`

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`B`

Show Worked Solution

`text(vertical:)\ \ sum F_y = 4 + 3 sin 30^@ +\ ^(−)F_2 sin 45^@ = 0`

`4 + 3/2 – F_2/sqrt 2` `= 0`
`F_2/sqrt 2` `= 4 + 3/2`
`:. F_2` `= sqrt 2 (4 + 3/2)`
  `= (11 sqrt 2)/2`

 
`=>  B`

Mechanics, SPEC1 2018 VCAA 1

Two objects of masses 5 kg and 8 kg are attached by a light inextensible string that passes over a smooth pulley. The 8 kg mass is on a smooth plane inclined at 30° to the horizontal. The 5 kg mass is hanging vertically, as shown in the diagram below.
 


 

  1. On the diagram above, show all forces acting on both masses.  (1 mark)
  2. Find the magnitude, in `text(ms)^(-2)`, and state the direction of the acceleration of the 8 kg mass.  (3 marks)
Show Answers Only
a.  

b.   `a = 9/13\ text(ms)^(-2)\ text(up the incline)`

Show Worked Solution
a.  

 

b.   
 `sum F` `= 5 text(g) – 8 text(g)\ sin (30^@) = (5 + 8)a\ \ text{(up slope → +)}`
  `5 text(g) – 4 text(g)` `= 13a`
  `:. a` `= 9/13\ text(ms)^(-2)\ text(up the incline)`