KINEMATICS
What is the difference between distance and displacement?
Distance is total path travelled; displacement is change in position including direction.
State Newton's First Law of Motion. and si unit for force
An object remains at rest or in uniform motion unless acted upon by an unbalanced external force. and N
A projectile is launched at an angle. What two independent components can its initial velocity be resolved into?
Answer: Horizontal and vertical components.
What is the direction of centripetal acceleration for an object undergoing uniform circular motion?
Answer: Towards the centre of the circular path.
what is -9.8 ms-2 stands for
gravity
A car travels 20 m east, then 5 m west. What is its displacement?
15 m east.
A 70 kg person stands in an elevator accelerating upwards at 2 m/s².
Calculate the normal force acting on the person.
N − mg = ma
N = m(g + a)
= 70(9.8 + 2)
= 826 N
A ball is launched at 20 m/s at an angle of 30° above the horizontal. Calculate its initial vertical velocity.
Answer:
vy=vsinθv_y = v\sin\theta
vy=20sin30°v_y = 20\sin30°
10 m/s
A car travels around a circular track with a radius of 50 m at 20 m/s. Calculate its centripetal acceleration.
Answer:
ac=v2ra_c=\frac{v^2}{r}
ac=20250a_c=\frac{20^2}{50}
8.0 m/s²
Using the relationship between gravitational force and centripetal force, derive an expression for the orbital velocity of the satellite.
ravitational force provides centripetal force:
GMmr2=mv2r\frac{GMm}{r^2}=\frac{mv^2}{r}
Cancel m:
GMr2=v2r\frac{GM}{r^2}=\frac{v^2}{r}
Therefore:
v2=GMrv^2=\frac{GM}{r} v=rGM
What does the gradient of a displacement-time graph represent?
Velocity.
A 5 kg block is pulled across a horizontal surface with a force of 30 N. The coefficient of friction is 0.2.
Calculate its acceleration.
Friction:
Ff = μN
= 0.2(5)(9.8)
= 9.8 N
Net force:
F = 30 − 9.8 = 20.2 N
a = 20.2/5
= 4.04 m/s²
A projectile is launched horizontally at 15 m/s from a cliff 20 m high. Calculate the time taken to hit the ground.
Answer:
s=12gt2s=\frac{1}{2}gt^2
20=4.9t220=4.9t^2
t=2.02 s
State Kepler's First Law of Planetary Motion.
Answer: Planets orbit the Sun in elliptical paths with the Sun located at one focus.
Explain why an object moving at constant speed in a circular path is still accelerating.
Although its speed remains constant, its velocity continuously changes because its direction changes. Since acceleration is the rate of change of velocity, the object experiences centripetal acceleration towards the centre of the circle.
An object accelerates from 5 m/s to 25 m/s in 4 seconds. Calculate its acceleration.
5 m/s².
A 1200 kg car travels around a flat circular bend of radius 50 m at 15 m/s.
Calculate the minimum coefficient of friction required to prevent the car from sliding.
Answer:
Friction provides centripetal force:
μmg = mv²/r
μ = v²/rg
= 225/(50 × 9.8)
= 0.46
A projectile is launched at 25 m/s at an angle of 40°. Calculate its horizontal component of velocity.
Answer:
vx=25cos40°v_x=25\cos40°
19.2 m/s
Using Kepler's Third Law, what happens to a planet's orbital period if its orbital radius increases?
Answer: The orbital period increases.
T2∝r3
Explain why astronauts in orbit experience apparent weightlessness even though Earth's gravitational force is acting on them.
The astronauts and spacecraft are both continuously in free fall towards Earth. Because they are falling together, there is no normal contact force supporting the astronauts, causing them to experience apparent weightlessness.
A car travelling at 10 m/s accelerates uniformly at 3 m/s² for 6 seconds. Calculate its final velocity.
28 m/s.
A 10 kg object rests on a slope inclined at 30°.
Calculate:
a) The component of weight parallel to the slope.
b) The normal force.
Answer:
Parallel:
F = mg sin θ
= 10 × 9.8 × sin30°
= 49 N
Normal:
N = mg cos θ
= 10 × 9.8 × cos30°
= 84.9 N
Explain why a projectile launched at 45° does not always produce the maximum horizontal range.
Answer: 45° only gives maximum range when the projectile lands at the same height from which it was launched and air resistance is negligible.
Explain what escape velocity means and why an object does not need continuous propulsion once it reaches escape velocity.
Answer: Escape velocity is the minimum initial speed required for an object to escape a gravitational field without further propulsion. At this speed, its initial kinetic energy is sufficient to overcome its gravitational potential energy.
Explain how a banked road allows a car to travel around a curve with less reliance on friction.
On a banked road, the normal force acts at an angle. Its horizontal component provides some or all of the centripetal force required to keep the car moving in a circular path, reducing the dependence on friction between the tyres and road.