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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, 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, 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-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)
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  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`