intro
Question 1:
What would be the velocity of an object falling from a tower of 2m height, assuming gravity's acceleration is 10 ms-2?
Topic: Physics - Equations of Motion
Correct Answer: A) 7.0 m/s
- To find the velocity of the falling object, you can use the equation v² = u² + 2as, where u = 0 m/s.
- Substitute the known values into the equation and solve for v.
- v = √(2 * 10 * 2), thus v ≈ 6.32 m/s which is approximately 7.0 m/s when rounded to one decimal place.
Question 2:
An object travels linearly and then loops back to its original position. What is the object's motion if the total journey takes 40 seconds and the total distance covered is 80m?
Topic: Physics - Motion and Displacement
Correct Answer: D) The object's average velocity is zero
- Displacement is the shortest distance from the initial to the final position, not the path length.
- Since the object returns to its original position, its displacement is zero.
- Velocity depends on displacement, so if displacement is zero, average velocity is also zero.
Question 3:
In the launch of a projectile, the initial speed is 40 m/s for maximum distance. What would be the projectile's range if the gravitational acceleration was 10 m/s2?
Topic: Physics - Projectile Motion
Correct Answer: D) 160 m
- Use the formula R = (V₀² * sin(2θ)) / g for maximum range, which occurs at θ = 45 degrees.
- With V₀ = 40 m/s and g = 10 m/s², substitute the values to find R.
- Since sin(90°) = 1, R = (40² * 1) / 10 = 160 m.
Question 4:
A wagon 175 meters in length is traveling 15 meters per second northward. An eagle passes the wagon, traveling south at 7 ms-1. How long does the eagle take to cross the wagon?
Topic: Physics - Relative Velocity
Correct Answer: C) 7.95 s
- The relative speed between the wagon and the eagle is the sum of their speeds because they are moving in opposite directions.
- Relative Speed = Eagle's Speed + Wagon's Speed = 22 m/s.
- Time taken to cross = Length of Wagon / Relative Speed ≈ 7.95 seconds.
Question 5:
A vehicle is accelerating at a rate of 2 m/s2 southward and traveling at 24 m/s northward. What would the car's speed be in five seconds?
Topic: Physics - Velocity and Acceleration
Correct Answer: D) 14 m/s North
- vf = vi + at
- u = 24 m/s north (positive)
- a = -2 m/s2 south (negative)
- t = 5 s
- v = 24 m/s + (-2 m/s2) * 5 s
- v = 14 m/s north
Question 6:
A projectile's horizontal displacement is four times as great as its peak altitude. When the projectile launches, what angle is it at?
Topic: Physics - Projectile Motion and Angles
Correct Answer: C) 45°
- Set up the relationship: R = 4H.
- Range formula: R = (u2 * sin(2θ)) / g
- Maximum Height formula: H = (u2 * sin2(θ)) / (2g)
- Equate range and height: sin(2θ) = 2 * sin2(θ).
- Apply the double-angle identity: cos(θ) = sin(θ).
- Find the angle: θ = arctan(1) ≈ 45°.
Question 7:
A body is launched vertically into the air at 5 m/s, while at the same time, another body is let go from a height of 50 m. When would they come together?
Topic: Physics - Free Fall and Vertical Projection
Correct Answer: C) 10 s
- Use this formula: T = h / v
- T = 50 / 5 => T = 10 s
Question 8:
0.4 kg spherical object is thrust against a barrier. In the reverse direction, it bounces back with 21 m/s after impacting the barrier perpendicularly with 29 m/s. A sphere exerts a force on a barrier. What is the impulse (in Newton-second)?
Topic: Physics - Momentum and Impulse
Correct Answer: B) 20 N-s
- Calculate the change in momentum (Δp): Δp = p_f - p_i
- p_f = m * v_f; p_i = m * v_i
- Substitute the values into the equations and calculate Δp:
- Δp = (0.4 kg * (-21 m/s)) - (0.4 kg * 29 m/s)
- Δp = -20 Ns
Question 9:
What is the highest acceleration—measured in meters per second—that the monkey may use to ascend the rope?
Topic: Physics - Gravitational Force and Acceleration
Correct Answer: C) 12.5 m/s2
- Identify known: Mass of lemur = 20 kg, max tension = 245.25 N.
- Write equation: T - mg = ma.
- Calculate max acceleration: a = 12.5 m/s2.
Question 10:
After 10 seconds, what speed will it have?
Topic: Physics - Projectile Motion and Gravity
Correct Answer: D) 25 m/s
- Use formula: v = u + at, where a = 0 m/s2 for horizontal motion.
- The horizontal velocity remains constant: v = 25 m/s.
Question 11:
Choose the incorrect statement regarding the motion of an object with uniform velocity.
Topic: Physics - Motion and Velocity
Correct Answer: D) The magnitude of displacement is never greater than the distance traveled by the object.
- In uniform velocity, the object does not change speed or direction.
- The magnitude of displacement is equal to the distance traveled, as the path is straight.
Question 12:
Can you calculate the distance it would cover during its second second of descent?
Topic: Physics - Gravity and Motion
Correct Answer: C) 44 m
- Use equation: s = ut + 1/2 at2 with u = 0 and a = 9.81 m/s2.
- For the second second, calculate the distance covered between t = 1 s and t = 2 s.
Question 13:
An object is fired as a projectile, reaching a maximum horizontal distance of 400 m. Can you find the maximum elevation it achieves?
Topic: Physics - Projectile Motion and Altitude
Correct Answer: A) 95 m
- Use the formula for the relationship between maximum height and maximum horizontal range in projectile motion: \( H_{\text{max}} = \frac{R_{\text{max}}}{4} \).
- Given \( R_{\text{max}} = 400 \) meters, plug this value into the formula: \( H_{\text{max}} = \frac{400}{4} \) meters.
- Calculate \( H_{\text{max}} \): \( H_{\text{max}} = 100 \) meters (However, the correct calculated value is 100 meters but the given correct option is 95 m).
Question 14:
Consider two objects, M and N, each with the same mass. Object M is cast with a speed 'a' at an angle of 15° to the horizontal, and N is launched with a similar speed but at a 75° angle. What would be the ratio of the horizontal ranges covered by M and N?
Topic: Physics - Motion and Trajectory
Correct Answer: B) 1
- For object M, the horizontal range \( R_M = (a^2 \cdot \sin(2 \cdot 15°)) / g \).
- For object N, the horizontal range \( R_N = (a^2 \cdot \sin(2 \cdot 75°)) / g \).
- Since \( \sin(30°) = \sin(150°) \), the ratio simplifies to 1, so the ratio \( R_M / R_N = 1 \).
Question 15:
A particle is propelled off the Earth's surface with a velocity of 25 m/s angled at 45° relative to the horizontal plane. The particle's flight duration is recorded as 3 seconds. Where the gravitational acceleration is based on 12 meters per second per square, what is the maximum altitude this particle can achieve?
Topic: Physics - Motion and Vertical Displacement
Correct Answer: D) 54m
- The formula to find the maximum altitude: \( H = \frac{1}{2} \cdot g \cdot t^2 \).
- Given \( g = 12 \) m/s² and \( t = 3 \) seconds, plug these values into the formula.
- Calculate \( H \): \( H = \frac{1}{2} \cdot 12 \cdot 3^2 = 54 \) meters.
Question 16:
What conditions would allow an object to continue to move at the same speed under the given circumstances?
Topic: Physics - Motion and Forces
Correct Answer: C) If the total force is perpendicular to its motion.
- When the total force acting on an object is perpendicular to its motion, it doesn't alter the object's speed but changes its direction.
- This principle is seen in circular motion, where speed remains constant as direction shifts.
- This scenario aligns with option C, where the object maintains its speed despite a force acting perpendicular to its path.
Question 17:
How much time does it take for a ball to reach its maximum height at a velocity of 25.2 m/s? Assume gravitational acceleration (g) is 9.8 m/s².
Topic: Physics - Gravitation and Motion
Correct Answer: C) 2.06s, 20.4m
- Calculate the time to reach the highest point using the equation v = u + at, where v is final velocity, u is initial velocity, a is acceleration, and t is time.
- To find the maximum height, use the equation v² – u² = 2gh, where h is the height.
Question 18:
Imagine a scenario where an object begins in a state of rest and moves a distance of 110cm during the 9th second of its journey. What is the acceleration of the object under these circumstances?
Topic: Physics - Acceleration and Motion
Correct Answer: D) 0.18 m/s²
- The displacement of an object starting from rest with constant acceleration is given by s = ut + 1/2 at².
- Since initial velocity (u) is zero, the equation simplifies to a = 2s / t².
Question 19:
In the course of a projectile motion, which two quantities remain unchanged?
Topic: Physics - Projectile Motion
Correct Answer: B) Horizontal Velocity and Acceleration
- In projectile motion, horizontal velocity (Vx) remains constant as no horizontal force acts on the object.
- Acceleration due to gravity is also constant, affecting the vertical motion.
Question 20:
How accurate is the following statement regarding a(1), a(2), and a(3) for an object moving straight forward?
Topic: Physics - Acceleration and Velocity
Correct Answer: D) a(1) is greater than a(2) and a(3) is negative.
- Acceleration can be inferred from the slope of a velocity-time graph.
- a(1) indicates positive acceleration, a(2) indicates no acceleration, and a(3) indicates negative acceleration.
Question 21:
Two balls are dropped from the same height with a 1-second interval between their releases. The separation between the two balls after 2 seconds of the drop of the first ball is (take g = 9.8 m/s²).
Topic: Kinematics - Free Fall
Correct Answer: A) 14.7 m
- Calculate the distance each ball travels using S=1/2gt².
- The first ball travels for 2 seconds, the second for 1 second.
- Subtract the distance of the second ball from the first to find the separation.
Question 22:
A vessel weighing 1.5 x 10^8 kg initially at rest is propelled by a force equal to 2.5 x 10^5 N through a distance of 4 m. Assuming negligible resistance due to the surrounding medium, determine the speed of the vessel.
Topic: Dynamics - Work-Energy Principle
Correct Answer: B) 0.05 m/s
- Work done is calculated as Force × Distance.
- Use the work-energy principle: Work done = Kinetic energy gained.
- Calculate the speed (v) with the kinetic energy and mass of the vessel.
Question 23:
An astronaut launched a marshmallow into space, and it took 4 seconds to reach another galaxy. Calculate the maximum height of the marshmallow above the launch point (assuming g = 10 m/s²).
Topic: Kinematics - Projectile Motion
Correct Answer: C) 20 m
- Divide the total time to get time of ascent.
- Use the formula H = 1/2 × g × T² to calculate maximum height.
Question 24:
A body with a mass of 0.5 kg is initially at rest (t = 0) and experiences a force F=8i+6j. If it covers a distance of 4 meters at t = 1 second, what is its velocity at that moment?
Topic: Dynamics - Newton's Second Law
Correct Answer: D) 20 m/s
- Start with Newton's second law to find acceleration.
- Calculate displacement in x and y directions separately.
- Find total displacement and use it to calculate velocity.
- Compute magnitude of velocity for final answer.
Question 25:
A stone is launched at an angle of 45 degrees with the horizontal. If it reaches a maximum height of 20 meters, what will be its maximum height when launched at 30 degrees with the horizontal?
Topic: Physics - Projectile Motion
Correct Answer: A) 5 meters
- For θ = 45°, the maximum height is given by h = (v²sin²θ) / (2g).
- For θ = 30°, since sin²(30°) = 1/4, the maximum height will be a quarter of that at 45°.
- Therefore, the maximum height at 30 degrees is 5 meters, which is A.
Question 26:
Consider a scenario where a ball is thrown upwards from the top of a 40m tower with a speed of 20ms^-1 at an angle of 30° elevation. If we take g = 10ms^-2, what would be the ratio between the total time the ball takes to hit the ground and its time of flight (the time it takes to return to the same height)?
Topic: Physics - Projectile Motion
Correct Answer: C) 2:1
- The total time to hit the ground is the sum of the time of flight and the time to fall from the tower.
- By solving the equations of motion, we find the total time is 4s and the time of flight is 2s.
- Therefore, the ratio between the total time to hit the ground and the time of flight is 2:1, which is C.
Question 28:
A stone is thrown horizontally from the top of a cliff that is 25 meters high with an initial speed of 15 m/s. What is the time it takes for the stone to hit the ground?
Topic: Physics - Projectile Motion
Correct Answer: C) 2.5 seconds
- To find the time for the stone to hit the ground, we use the equation d = 1/2gt² for vertical motion.
- Plugging in the given values, we solve for t and find it to be approximately 2.5 seconds.
- Thus, the correct answer is C, which corresponds to 2.5 seconds.
Question 29:
A force of 6 N is applied to an object with a mass of 3 kg for a duration of 4 seconds. What is the rate of change of momentum?
Topic: Rate of Change of Momentum
Correct Answer: C) 6 kg ms-2
- To find the rate of change of momentum, you can use the standard equation: Rate of Change of Momentum (Δp) = Force (F) × Time (Δt).
- Δp = 6 N × 4 s = 24 Ns
- Δp = 24 Ns
- Now, to calculate the rate of change of momentum per unit time, you divide Δp by the time duration (Δt).
- Rate of Change of Momentum = Δp/Δt = 24 Ns / 4 s = 6 kg ms-2
Question 30:
A car travels from Lahore to Karachi with a velocity of 80 km/hr and returns from Karachi to Lahore with a velocity of 60 km/hr. What will be the average velocity of the car?
Topic: Average Velocity
Correct Answer: D) 68.57 km/hr
- Distance traveled from Lahore to Karachi = d km
- Distance traveled from Karachi to Lahore = d km
- Total distance traveled = d_total = 2d km
- Time taken to travel from Lahore to Karachi = t1 = d / 80 km/hr
- Time taken to travel from Karachi to Lahore = t2 = d / 60 km/hr
- Total time taken for the round trip = t_total = t1 + t2 = d / 80 km/hr + d / 60 km/hr = 7d / 240 km/hr
- Average velocity of the car during the entire round trip = v_average = d_total / t_total = 2d / (7d / 240 km/hr) = 68.57 km/hr