Lesson 4C.4.1
4C.4.1 Newton's laws of motion for particles Quiz: Pearson Edexcel Further Maths, Unit 33
20 questions
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Lesson 4C.4.1, Newton's laws of motion for particles: 20 multiple choice questions for the Pearson Edexcel Further Maths (9FM0), Unit 33: Newton's laws and simple harmonic motion, written with Revision Ninja.
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The 20 questions
-
For a particle moving in one dimension under a variable force F(x), which equation is correct?
- m dx/dt = F(x)
- m v dv/dx = F(x)
- F(x) = m v^2
- m dv/dx = F(x)
-
What is the magnitude of the gravitational force between masses M and m a distance r apart?
- Gm/r^2
- GMm r^2
- GMm/r^2
- GMm/r
-
Which equation gives v^2 after integrating a = v dv/dx with respect to x?
- x = integral of v dv
- v = integral of a dx
- v^2 = 2 integral of a dx + constant
- v^2 = integral of a dt
-
What is the purpose of the constant of integration when finding velocity from acceleration?
- it is fixed by the initial conditions
- it equals the mass of the particle
- it equals the acceleration
- it equals the total distance travelled
-
Newton's first law states that a particle remains at rest or moves uniformly unless
- gravity acts on it
- acted on by a resultant force
- it has a non-zero mass
- its acceleration is constant
-
Newton's third law says that action and reaction forces are
- unequal in magnitude
- equal and in the same direction
- equal and acting on the same body
- equal and opposite, acting on different bodies
-
What is the SI unit of force?
- newton (kg m s^-2)
- newton per second
- joule (kg m^2 s^-2)
- kg m s^-1
-
A particle of mass 1 kg starts from rest at x = 0 and moves under a resultant force F = 3x newtons. What is its speed at x = 2 m?
- sqrt(6) m/s
- 6 m/s
- 2 sqrt(3) m/s
- 4 m/s
-
A particle is at distance 2R from the centre of a planet of radius R. Compared with its weight at the surface, its weight at this point is
- one eighth
- twice
- one half
- one quarter
-
A particle is released from rest at x = a and is attracted towards O with acceleration mu/x^2. What is its speed when x = a/2?
- sqrt(2mu/a)
- sqrt(mu/a)
- sqrt(4mu/a)
- sqrt(mu/(2a))
-
A particle of mass 2 kg has resultant force F = 4t newtons and starts from rest. What is its velocity at t = 3 s?
- 6 m/s
- 18 m/s
- 9 m/s
- 4.5 m/s
-
What distance from the centre of the Earth gives a weight equal to one ninth of the surface weight?
- 9R
- 4.5R
- R/3
- 3R
-
A particle of unit mass moves with resultant force -kv opposing its motion, starting with speed u. What is its speed at time t?
- u e^(kt)
- u(1 - kt)
- u - kt
- u e^(-kt)
-
A particle of mass 2 kg has resultant force F = 10 - 2x newtons along the x-axis. What is its acceleration at x = 2 m?
- 5 m/s^2
- 3 m/s^2
- 1.5 m/s^2
- 8 m/s^2
-
A particle of mass 1 kg starts at rest with resultant force F = 2t newtons. What is the impulse from t = 0 to t = 3 s?
- 18 N s
- 6 N s
- 9 N s
- 3 N s
-
A particle is projected vertically upwards from the surface of a planet of radius R with surface gravity g. What speed is needed to escape to infinity?
- sqrt(2gR)
- sqrt(gR/2)
- sqrt(2g/R)
- sqrt(gR)
-
A particle of mass m moves in a straight line with resistance mkv^2, starting with speed u. Does it stop in a finite distance?
- Yes, after distance ln(2)/k
- Yes, after distance u/k
- No, v = u e^(-kx), so the speed never reaches zero
- Yes, after distance 1/(ku)
-
A particle released from rest at distance 4a from the centre of an inverse square attraction with constant mu = GM reaches distance a. What is its speed there?
- sqrt(3mu/a)
- sqrt(2mu/a)
- sqrt(3mu/(2a))
- sqrt(mu/(2a))
-
Why is a = v dv/dx preferred to a = dv/dt when the force depends on position?
- Velocity is always constant with position
- Displacement is always constant
- dv/dt is always zero when the force depends on position
- It links velocity directly to displacement, so an equation in x can be integrated without needing time
-
A particle of mass 2 kg has resultant force 6 - 3v newtons, where v is its speed in m/s, and starts from rest. What is its terminal speed?
- 4 m/s
- 3 m/s
- 2 m/s
- 6 m/s
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