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C8.1 Force and acceleration AQA AS Physics P8 Newton’s Law Of Motion Kerboodle Answers: Page No. 140
1.Ans-
a the acceleration of the car
0.24ms-2
b the force on the car that produced this acceleration
190N
c the ratio of the accelerating force to the weight of the car.
0.024
2.Ans-a the deceleration of the aeroplane
2.4 m s-2
b the braking force on the aircraft.
12000N
3.Ans-a 360N
b 23s
4.Ans- a the deceleration Of the bullet
-1.3 x 105 ms-2
b the impact force on the bullet.
260N
8.2 Using F= ma AQA AS Physics P8 Newton’s Law Of Motion Kerboodle Answers : Page No. 143
1.a the weight of the rocket
5400N
b the thrust of the rocket engines,
7700N
2.a the tension in the tow bar
60N
b the engine force.
270N
3.
a stationary
11.8kN
b ascending at constant speed
11.8kN
c ascending at a constant acceleration of 0.4m s-2
12.3kN
d descending at a constant deceleration of 0.4 ms-2
12.3kN
4 a the acceleration of the brick
1.0 m s-2
b the frictional force on the brick due to the roof.
12.5 N
8.3 Terminal speed AQA AS Physics P8 Newton’s Law Of Motion Kerboodle Answers : Page No. 145
1. a0.04ms-1
b
1.5N
3.a its maximum acceleration from rest,
0.14 ms-2
b .520m
4.On Downhill there is an added component of acceleration due to gravity, so increased acceleration will give you eventually an increased top speed.
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8.4 On the road AQA AS Physics P8 Newton’s Law Of Motion Kerboodle Answers : Page No. 148
1.a
7.2m
ii the braking distance.
33.7 m
b 4.1 m
.a the deceleration of the vehicle from this speed to a standstill over this distance
6.75ms-2
b the frictional force on a Vehicle Of mass 1000 kg On this road as it stops.
6750N
3. a
Braking distance – The braking distance is the distance taken to stop once the brakes are applied.
b During braking, the engine is usually not actively applying torque to the shaft, so the forward force is gone. On the other hand, the brake pads are applying force on the rotor, slowing it down. This has the opposite effect on the wheel rotation: they are not forced to turn faster, but slower. This (+ friction) gives rise to decelerating force at the contact point with the road.
4. a show that the maximum deceleration on this road is 5.grns-?
b calculate the braking distance on this road for a speed of 30 ms-2.
76 m
8.5 Vehicle safety AQA AS Physics P8 Newton’s Law Of Motion Kerboodle Answers : Page No. 151
1.a the acceleration cf the car, in terms of g
2.0 g
b the impact force on the car.
24kN
2.a the impact time if it hits a wall at 20 ms-I
80ms
b the impact force.
375kN
3.a the deceleration Of the car from 3.0m to rest in 0.40s
7.5ms-2
b the impact force on the car.
6750N
4.a the deceleration of the passenger in terms of g,
6.4g
b the resultant force on the passenger.
4250N
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Practice questions: Page No. 152-153
1 (i)
(ii) To find the force F on the child, use
F = ma = 15 × 5.0 = 75 N
(b)She would have continued to move forward as the car slowed down and she would have collided with the back of a front seat or with the front windscreen of the car.
The impact time would have been much less so the impact force would have been much greater.
(c) The stopping distance would be much less at 20 mph than at 30 mph.
This is because the braking distance depends on the square of the speed so it would be reduced by more than half i.e. (20/30)2.
Fewer accidents would occur because cars travelling at 20 mph would stop in a much shorter distance than cars at 30 mph.
2 .(a) • when at constant speed, resultant force is zero (or forces are balanced)
• weight and tension must have equal magnitudes
• but act in opposite directions
• a correct application of either Newton’s first or second law of motion
b.(i)
maximum acceleration is produced when resultant force is a maximum;
Fmax = 12.0 + 8.0 = 20 N
use of F = ma gives
(ii) minimum acceleration is produced when resultant force is a minimum;
Fmin = 12.0 − 8.0 = 4.0 N
use of F = ma gives
3.
(a)
(b) (i)
(ii) resistive force = 2.0 × 103 (because resultant force on car is zero)
4.
(i) the acceleration of the car
(ii) the speed of the car after 8.0s
use of v = u + a t gives
v = 0 + (2.0 × 8.0) = 16 m s−1
(iii)
(b) In practice the resultant force on the car changes with time. Air resistance is one factor that affects the resultant force acting on the vehicle.
(i) resultant force decreases
because air resistance increases as the car’s speed increases
(ii) • eventually the propulsive force and the resistive force are equal in magnitude
• resultant force is zero
• F = 0 means there is no acceleration (or speed remains constant)
• a correct application of either Newton’s first or second law of motion
5.
(a) (i) Calculate the Iolal force of the engines acting on the aircraft.
270 × 4 = 1080 kN
(ii)
(b) (i) use of v = u + a t gives 90 = 0 + 3.38 t
∴ time to reach take-off speed = 27 s
(ii)
(c)
(d) using v = u + a t
gives 260 = 90 + 2.0 t
time to cruising speed t = 85 s
(e) (vertically) lift = weight, so flight is level
• (horizontally) thrust = drag, so no acceleration
• no resultant force either vertically or horizontally
6.
(a) component of weight parallel to ramp
= W sin θ = 7.2 × 103 sin 30°
= 3.6 × 103 N
(b)
(c)
(d) .
frictional forces act on car and passengers
these increase the resultant force acting down the ramp
therefore the deceleration is greater
energy is lost as heat
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References:
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