Gravity and Motion Practice Questions
ATAR (Australia) · ATAR Physics · 149 free MCQs with instant results and detailed explanations.
149
Total
54
Easy
74
Medium
21
Hard
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Sample Questions from Gravity and Motion
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Q1
Easy
What is the acceleration due to gravity on Earth?
A.
9.8 m/sยฒ
B.
10 m/sยฒ
C.
8.0 m/sยฒ
D.
9.0 m/sยฒ
Show Answer & Explanation
Correct Answer: A
The standard acceleration due to gravity on Earth is approximately 9.8 m/sยฒ, which is a fundamental value used in physics calculations.
Q2
Easy
What is the primary force that causes an object to fall towards the Earth?
A.
Magnetic force
B.
Electrostatic force
C.
Gravitational force
D.
Frictional force
Show Answer & Explanation
Correct Answer: C
The gravitational force is the force that attracts two bodies towards each other, which is why an object falls towards the Earth.
Q3
Easy
If a satellite is in orbit around the Earth, which force keeps it in motion around the planet?
A.
Centrifugal force
B.
Tension force
C.
Gravitational force
D.
Normal force
Show Answer & Explanation
Correct Answer: C
The gravitational force is what keeps satellites in orbit around the Earth by providing the necessary centripetal force for circular motion.
Q4
Medium
Two objects, A and B, are in free fall from the same height. Object A has a mass of 2 kg and object B has a mass of 5 kg. Which object will hit the ground first?
A.
Object A
B.
Object B
C.
Both will hit at the same time
D.
It depends on their shapes
Show Answer & Explanation
Correct Answer: C
In the absence of air resistance, all objects fall at the same rate due to gravity, regardless of their mass. Therefore, both objects will hit the ground at the same time.
Q5
Medium
A car accelerates uniformly from rest to a speed of 25 m/s over a distance of 200 m. What is the acceleration of the car?
A.
1.56 m/sยฒ
B.
3.125 m/sยฒ
C.
5 m/sยฒ
D.
7.5 m/sยฒ
Show Answer & Explanation
Correct Answer: B
Using the equation vยฒ = uยฒ + 2as, where v is the final speed, u is the initial speed, a is acceleration, and s is distance. Plugging in values gives 25ยฒ = 0 + 2a(200), solving this yields a = 3.125 m/sยฒ.
Q6
Medium
A satellite is in a circular orbit around Earth at a height of 500 km. What is the orbital speed of the satellite? (Use g = 9.81 m/sยฒ and Earth's radius = 6371 km)
A.
7.0 km/s
B.
7.5 km/s
C.
8.0 km/s
D.
8.5 km/s
Show Answer & Explanation
Correct Answer: B
The orbital speed (v) can be calculated using the formula v = โ(g * (R + h)), where R is the Earth's radius and h is the height above the Earth's surface. Here, R = 6371 km and h = 500 km: v = โ(9.81 * (6371 + 0.5)) = 7.5 km/s.
Q7
Medium
An astronaut on the Moon throws a rock with a velocity of 10 m/s at an angle of 30 degrees to the horizontal. How far will the rock travel horizontally before it hits the ground? (Use g = 1.62 m/sยฒ)
A.
5.5 m
B.
10.0 m
C.
15.0 m
D.
20.0 m
Show Answer & Explanation
Correct Answer: B
The horizontal range (R) can be calculated using R = (vยฒ * sin(2ฮธ)) / g. Here, v = 10 m/s, ฮธ = 30 degrees, and g = 1.62 m/sยฒ. R = (10ยฒ * sin(60ยฐ)) / 1.62 = 10 m.
Q8
Hard
A 2 kg object is dropped from a height of 80 m. Calculate the speed of the object just before it hits the ground, ignoring air resistance.
A.
12.6 m/s
B.
16.0 m/s
C.
20.0 m/s
D.
14.0 m/s
Show Answer & Explanation
Correct Answer: B
Using the formula for gravitational potential energy converted to kinetic energy, we have mgh = 0.5mv^2. Thus, v = sqrt(2gh) = sqrt(2*9.81*80) = 16.0 m/s.
Q9
Hard
A satellite orbits Earth at a height where the gravitational field strength is 6 N/kg. If the mass of the satellite is 500 kg, what is the weight of the satellite in this orbit?
A.
3000 N
B.
500 N
C.
1500 N
D.
6000 N
Show Answer & Explanation
Correct Answer: A
Weight is calculated by multiplying mass by gravitational field strength. Thus, Weight = Mass x Gravitational field strength = 500 kg x 6 N/kg = 3000 N.
Q10
Hard
A ball of mass 2 kg is thrown vertically upward with an initial speed of 20 m/s. What is the maximum height reached by the ball? (Assume g = 9.8 m/sยฒ)
A.
20.4 m
B.
10.2 m
C.
30.6 m
D.
15.5 m
Show Answer & Explanation
Correct Answer: A
Using the formula h = (vยฒ)/(2g), where v is the initial velocity and g is the acceleration due to gravity, we have h = (20 m/s)ยฒ / (2 * 9.8 m/sยฒ) = 400 / 19.6 = 20.4 m. Therefore, the maximum height reached by the ball is 20.4 m.
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Gravity and Motion โ ATAR (Australia) ATAR Physics Practice Questions Online
This page contains 149 practice MCQs for the chapter Gravity and Motion in ATAR (Australia) ATAR Physics. The questions are organized by difficulty โ 54 easy, 74 medium, 21 hard โ so you can choose the right level for your preparation.
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