A particle of unit mass moves horizontally in a straight line. It experiences a resistive force proportional to \(v^2\), where \(v\) m s\(^{-1}\) is the speed of the particle, so that the acceleration is given by \(-k v^2\). Initially the particle is at the origin and has a velocity of 40 m s\(^{-1}\) to the right. After the particle has moved 15 m to the right, its velocity is 10 m s\(^{-1}\) (to the right). --- 8 WORK AREA LINES (style=lined) --- --- 4 WORK AREA LINES (style=lined) --- --- 8 WORK AREA LINES (style=lined) ---
Mechanics, EXT2 M1 2023 HSC 14c
A projectile of mass \(M\) kg is launched vertically upwards from the origin with an initial speed \(v_0\) m s\(^{-1}\). The acceleration due to gravity is \( {g}\) ms\(^{-2}\).
The projectile experiences a resistive force of magnitude \(kMv^2\) newtons, where \(k\) is a positive constant and \(v\) is the speed of the projectile at time \(t\) seconds.
- The maximum height reached by the particle is \(H\) metres.
- Show that \(H=\dfrac{1}{2 k} \ln \left(\dfrac{k v_0{ }^2+g}{g}\right)\). (3 marks)
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- When the projectile lands on the ground, its speed is \(v_1 \text{m} \ \text{s}^{-1}\), where \(v_1\) is less than the magnitude of the terminal velocity.
- Show that \(g\left(v_0^2-v_1^2\right)=k v_0^2 v_1^2\). (3 marks)
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Mechanics, EXT2 M1 2022 HSC 12c
A particle with mass 1 kg is moving along the `x`-axis. Initially, the particle is at the origin and has speed `u` m s-1 to the right. The particle experiences a resistive force of magnitude `v+3 v^2` newtons, where `v` m s-1 is the speed of the particle after `t` seconds. The particle is never at rest.
- Show that `(dv)/(dx)=-(1+3v)` (1 mark)
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- Hence, or otherwise, find `x` as a function of `v`. (2 marks)
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Mechanics, EXT2 M1 2022 HSC 8 MC
As a projectile of mass `m` kilograms travels through air, it experiences a frictional force. The magnitude of this force is proportional to the square of the speed `v` of the projectile. The constant of proportionality is the positive number `k`. The position of the particle at time `t` is denoted by `([x],[y])`. The acceleration due to gravity is `g \ text{m s}^(-2)`.
Based on Newton's laws of motion, which equation models the motion of this projectile?
- `([0],[-mg])+kv([overset.x],[overset.y])=m([overset{..}x],[overset{..}y])`
- `([0],[-mg])-kv([overset.x],[overset.y])=m([overset{..}x],[overset{..}y])`
- `([0],[-mg])+kv^(2)([overset.x],[overset.y])=m([overset{..}x],[overset{..}y])`
- `([0],[-mg])-kv^(2)([overset.x],[overset.y])=m([overset{..}x],[overset{..}y])`
Mechanics, EXT2 M1 2021 HSC 14b
An object of mass 5 kg is on a slope that is inclined at an angle of 60° to the horizontal. The acceleration due to gravity is `g\ text(m s)^(-2)` and the velocity of the object down the slope is `v\ text(m s)^(-1)`.
As well as the force due to gravity, the object is acted on by two forces, one of magnitude `2v` newtons and one of magnitude `2v^2` newtons, both acting up the slope.
- Show that the resultant force down the slope is `(5sqrt3)/2 g - 2v - 2v^2` newtons. (2 marks)
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- There is one value of `v` such that the object will slide down the slope at a constant speed.
- Find this value of `v` in `text(m s)^(-1)`, correct to 1 decimal place, given that `g = 10`. (2 marks)
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Mechanics, EXT2 M1 2021 HSC 15c
An object of mass 1 kg is projected vertically upwards with an initial velocity of `u` m/s. It experiences air resistance of magnitude `kv^2` newtons where `v` is the velocity of the object, in m/s, and `k` is a positive constant. The height of the object above its starting point is `x` metres. The time since projection is `t` seconds and acceleration due to gravity is `g` m/s².
- Show that the time for the object to reach its maximum height is `1/sqrt{gk} arctan (u sqrt{k/g})` seconds. (3 marks)
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- Find an expression for the maximum height reached by the object, in terms of `k`, `g`, and `u`. (3 marks)
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Mechanics, EXT2 M1 2020 HSC 14b
A particle starts from rest and falls through a resisting medium so that its acceleration, in m/s2, is modelled by
`a = 10 (1 - (kv)^2)`,
where `v` is the velocity of the particle in m/s and `k = 0.01`.
Find the velocity of the particle after 5 seconds. (4 marks)
Mechanics, EXT2 M1 EQ-Bank 4
A torpedo with a mass of 80 kilograms has a propeller system that delivers a force of `F` on the torpedo, at maximum power. The water exerts a resistance on the torpedo proportional to the square of the torpedo's velocity `v`.
- Explain why `(dv)/(dt) = 1/80 (F - kv^2)`
where `k` is a positive constant. (1 mark)
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- If the torpedo increases its velocity from `text(10 ms)\ ^(−1)` to `text(20 ms)\ ^(−1)`, show that the distance it travels in this time, `d`, is given by
`d = 40/k log_e((F - 100k)/(F - 400k))` (3 marks)
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Mechanics, EXT2 M1 EQ-Bank 2
A particle with mass `m` moves horizontally against a resistance force `F`, equal to `mv(1 + v^2)` where `v` is the particle's velocity.
Initially, the particle is travelling in a positive direction from the origin at velocity `T`.
- Show that the particle's displacement from the origin, `x`, can be expressed as
`x = tan^(-1)((T - v)/(1 + Tv))` (2 marks)
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- Show that the time, `t`, when the particle is travelling at velocity `v`, is given by
`t = 1/2 log_e ((T^2(1 + v^2))/(v^2(1 + T^2)))` (4 marks)
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- Express `v^2` as a function of `t`, and hence find the limiting values of `x` and `v`. (2 marks)
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Mechanics, EXT2 M1 EQ-Bank 1
A canon ball of mass 9 kilograms is dropped from the top of a castle at a height of `h` metres above the ground.
The canon ball experiences a resistance force due to air resistance equivalent to `(v^2)/500`, where `v` is the speed of the canon ball in metres per second. Let `g=9.8\ text(ms)^-2` and the displacement, `x` metres at time `t` seconds, be measured in a downward direction.
- Show the equation of motion is given by
`ddotx = g - (v^2)/4500` (1 mark)
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- Show, by integrating using partial fractions, that
`v = 210((e^(7/75 t) - 1)/(e^(7/75 t) + 1))` (5 marks)
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- If the canon hits the ground after 4 seconds, calculate the height of the castle, to the nearest metre. (3 marks)
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Mechanics, EXT2 M1 2018 HSC 14b
A falling particle experiences forces due to gravity and air resistance. The acceleration of the particle is `g - kv^2`, where `g` and `k` are positive constants and `v` is the speed of the particle. (Do NOT prove this.)
Prove that, after falling from rest through a distance, `h`, the speed of the particle will be `sqrt(g/k (1 - e^(−2kh)))`. (3 marks)
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Mechanics, EXT2 M1 2017 HSC 13c
A particle is projected upwards from ground level with initial velocity `1/2 sqrt(g/k)\ text(ms)^(-1)`, where `g` is the acceleration due to gravity and `k` is a positive constant. The particle moves through the air with speed `v\ text(ms)^(-1)` and experiences a resistive force.
The acceleration of the particle is given by `ddot x = -g - kv^2\ text(ms)^(-2)`. Do NOT prove this.
The particle reaches a maximum height, `H`, before returning to the ground.
Using `ddot x = v (dv)/(dx)`, or otherwise, show that `H = 1/(2k) log_e (5/4)` metres. (4 marks)
Mechanics, EXT2 M1 2015 HSC 15a
A particle `A` of unit mass travels horizontally through a viscous medium. When `t = 0`, the particle is at point `O` with initial speed `u`. The resistance on particle `A` due to the medium is `kv^2`, where `v` is the velocity of the particle at time `t` and `k` is a positive constant.
When `t = 0`, a second particle `B` of equal mass is projected vertically upwards from `O` with the same initial speed `u` through the same medium. It experiences both a gravitational force and a resistance due to the medium. The resistance on particle `B` is `kw^2`, where `w` is the velocity of the particle `B` at time `t`. The acceleration due to gravity is `g`.
- Show that the velocity `v` of particle `A` is given by
`1/v = kt + 1/u.` (2 marks)
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- By considering the velocity `w` of particle `B`, show that
`t = 1/sqrt(gk) (tan^-1(u sqrt(k/g)) - tan^-1 (w sqrt(k/g))).` (3 marks)
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- Show that the velocity `V` of particle `A` when particle `B` is at rest is given by
`1/V = 1/u + sqrt(k/g) tan^-1 (u sqrt (k/g)).` (1 mark)
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- Hence, if `u` is very large, explain why
`V ~~ 2/pi sqrt(g/k).` (1 mark)
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Mechanics, EXT2 M1 2007 HSC 6b
A raindrop falls vertically from a high cloud. The distance it has fallen is given by
`x = 5 log_e ((e^(1.4 t) + e^(-1.4 t))/2)`
where `x` is in metres and `t` is the time elapsed in seconds.
- Show that the velocity of the raindrop, `v` metres per second, is given by
`v = 7 ((e^(1.4 t) - e^(-1.4 t))/(e^(1.4 t) + e^(-1.4 t)))` (2 marks)
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- Hence show that `v^2 = 49 (1 - e^(-(2x)/5)).` (2 marks)
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- Hence, or otherwise, show that `ddot x = 9.8 - 0.2v^2.` (2 marks)
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- The physical significance of the 9.8 in part (iii) is that it represents the acceleration due to gravity.
What is the physical significance of the term `–0.2 v^2?` (1 mark)
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- Estimate the velocity at which the raindrop hits the ground. (1 mark)
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Mechanics, EXT2 M1 2011 HSC 6a
Jac jumps out of an aeroplane and falls vertically. His velocity at time `t` after his parachute is opened is given by `v(t)`, where `v(0) = v_0` and `v(t)` is positive in the downwards direction. The magnitude of the resistive force provided by the parachute is `kv^2`, where `k` is a positive constant. Let `m` be Jac’s mass and `g` the acceleration due to gravity. Jac’s terminal velocity with the parachute open is `v_T.`
Jac’s equation of motion with the parachute open is
`m (dv)/(dt) = mg - kv^2.` (Do NOT prove this.)
- Explain why Jac’s terminal velocity `v_T` is given by
`sqrt ((mg)/k).` (1 mark)
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- By integrating the equation of motion, show that `t` and `v` are related by the equation
`t = (v_T)/(2g) ln[((v_T + v)(v_T - v_0))/((v_T - v)(v_T + v_0))].` (3 marks)
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- Jac’s friend Gil also jumps out of the aeroplane and falls vertically. Jac and Gil have the same mass and identical parachutes.
Jac opens his parachute when his speed is `1/3 v_T.` Gil opens her parachute when her speed is `3v_T.` Jac’s speed increases and Gil’s speed decreases, both towards `v_T.`
Show that in the time taken for Jac's speed to double, Gil's speed has halved. (3 marks)
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Mechanics, EXT2 M1 2012 HSC 13a
An object on the surface of a liquid is released at time `t = 0` and immediately sinks. Let `x` be its displacement in metres in a downward direction from the surface at time `t` seconds.
The equation of motion is given by
`(dv)/(dt) = 10 − (v^2)/40`,
where `v` is the velocity of the object.
- Show that `v = (20(e^t − 1))/(e^t + 1)`. (4 marks)
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- Use `(dv)/(dt) = v (dv)/(dx)` to show that
`x = 20\ log_e(400/(400 − v^2))` (2 marks)
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- How far does the object sink in the first 4 seconds? (2 marks)
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Mechanics, EXT2 M1 2013 HSC 15d
A ball of mass `m` is projected vertically into the air from the ground with initial velocity `u`. After reaching the maximum height `H` it falls back to the ground. While in the air, the ball experiences a resistive force `kv^2`, where `v` is the velocity of the ball and `k` is a constant.
The equation of motion when the ball falls can be written as
`m dot v = mg - kv^2.` (Do NOT prove this.)
- Show that the terminal velocity `v_T` of the ball when it falls is
`sqrt ((mg)/k).` (1 mark)
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- Show that when the ball goes up, the maximum height `H` is
`H = (v_T^2)/(2g) ln (1 + u^2/(v_T^2)).` (3 marks)
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- When the ball falls from height `H` it hits the ground with velocity `w`.
Show that `1/w^2 = 1/u^2 + 1/(v_T^2).` (2 marks)
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Mechanics, EXT2 M1 2014 HSC 14c
A high speed train of mass `m` starts from rest and moves along a straight track. At time `t` hours, the distance travelled by the train from its starting point is `x` km, and its velocity is `v` km/h.
The train is driven by a constant force `F` in the forward direction. The resistive force in the opposite direction is `Kv^2`, where `K` is a positive constant. The terminal velocity of the train is 300 km/h.
- Show that the equation of motion for the train is
`m ddot x = F[1 − (v/300)^2]`. (2 marks)
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- Find, in terms of `F` and `m`, the time it takes the train to reach a velocity of 200 km/h. (4 marks)
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