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Newton's Laws

Q1. The mass of fuel in a racing car decreases during a race.

As a result the lap time decreases.

Which of the following could explain this decrease?

A There is less friction on the race track.
B The maximum speed of the car has increased.
C The maximum acceleration and deceleration are greater.
D The engine is more efficient.

 

The mass of the whole car has decreased.

F= ma - therefore the force from the engine will produce greater acceleration and deceleration.

 

Q2. What is represented by the area under a force–displacement graph?

A
 rate of change of kinetic energy
B
change in momentum
C
work done
D
 acceleration

The area under a force displacement graph will have the units Nm

Work done is force x distance moved (units Nm)

 

Q3. Which of the following is not a unit of power?

A
 N m s–1
B
 J s
C
 W
D
kg m2 s –3

A watt is a J/s - so Js is wrong....

That is a quick one to answer - but here goes the other choices:

energy is work done (force x distance) , so power is work done/time taken - units N m s–1

power is measured in watts, natch!

force = mass x acceleration (kg m s –2), so force x distance (work done) has units kg m2 s –2 and work done/time taken (power) has units kg m2 s –3

 

Q4. In a test a 500 kg car travelling at 10 m s–1 hits a wall.

The front 0.30 m of the car crumples as the car is brought to rest.

What is the average force on the car during the impact?

A
 830 N
B
 7.5 kN
C
 8.3 kN
D
 83 kN

m = 500 kg

F = ma

a = ?

s = 0.30

v = 0 m/s

u = 10 m/s

v2 = u2 + 2as

02 = 102 + 2a x 0.30

0 = 100 + 0.60a

a = -100/0.60 = -167 m/s2

F = 500 x (-167) = 83,333 N

F = 83 kN

Therefore choice D is correct

Q5. A rocket has a mass of 12,000 kg.

It accelerates vertically upwards from the surface of the Earth at 1.4 m s−2.

What is the thrust of the rocket?

A
 1.7 × 104 N
B
 1.0 × 105 N
C
 1.3 × 105 N
D
 1.6 × 105 N

 

15The thrust of rocket T makes the rocket accelerate upwards with acceleration 'a' and overcomes the weight (mg) of the rocket.

T = ma + mg

T = 12,000 (1.4 + 9.81)

T = 1.2 x 104 x 11.21

T = 1.39 x 105 N

Q6. What is a correct unit for the area under a force−time graph?

A
 N m
B
 kg m s−1
C
 kg m s−2
D
 N s−1

 

Force x time units are:Ns

But we know that F=ma

So F units are also kg m s-2

Therefore Ft units must be kg m s-2s = kg m s-1

Q7. A mass of 2.5 kg is released from rest at X and slides down a ramp, of height 3.0 m, to point Y as shown.

When the mass reaches Y at the bottom of the ramp it has a velocity of 5.0 ms–1 .

What is the average frictional force between the mass and the ramp?

A
8.5 N
B
10.6 N
C
14.7 N
D
24.5 N

 

The component of weight acting down the slope is W cos 53.1o

This is opposed by friction

The net force of (W cos 53.1o - F) results in a final velocity of 5 m/s

v = 5 m/s

s = 5 m

u = 0 m/s

a = ?

v2 = u2 + 2 as

25 = 10 a

a = 2.5 m/s2

This acceleration is caused by the net force of W cos 53.1o - F

Now, F = ma and W = mg

mg cos 53.1o - F = 2.5m

F = mg cos 53.1o - 2.5m

F = 2.5 (gcos 53.1o - 2.5)

F = 8.5 N

Choice A

Q8. Two bodies of different masses undergo an elastic collision in the absence of any external force.

Which row gives the effect on the total kinetic energy of the masses and the magnitudes of the forces exerted on the masses during the collision?

Total kinetic energy
Magnitudes of forces
A
remains unchanged
same on both masses
B
remains unchanged
greater on the smaller mass
C
decreases
same on both masses
D
decreases
greater on the smaller mass

 

In an elastic collision the kinetic energy does not change.

In a collision the forces experienced by the two interacting objects will be equal in magnitude and opposite in direction - Newton III

Choice A