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mcq on motion in Two Dimensions for NEET Students
A 20 kg Mass is rotating in a vertical circle using a string that is 4 meters long. The change in kinetic energy between the lowest and highest points of the circle is (g = 10 m/s²)
a) 1800 J
b) 1000 J
c) 1600 J
d) 2000 J
Explanation: In vertical circular motion, kinetic energy varies because gravitational potential energy changes as the Mass moves along the path. At the top, potential energy is higher, reducing kinetic energy, while at the bottom, kinetic energy is maximized.
The change in kinetic energy can be determined by calculating the gravitational potential energy difference between the top and bottom. The height difference equals twice the string length, and the relation ΔKE = m g h is applied. Understanding the conversion of potential to kinetic energy along the circular path is key.
Consider a roller coaster on a loop: the cars move slower at the top and faster at the bottom due to energy transformation between height and speed.
The energy change reflects the balance between potential and kinetic energy in vertical motion, demonstrating how gravity affects circular motion.
Option c – 1600 J
A helicopter moving horizontally at 500 km/h drops a Food packet from a height of 2000 m when it is at point P. The packet lands at point Q. What is the distance PQ?
a) 2.81 km
b) 3.46 km
c) 4.81 km
d) 1.23 km
Explanation: In projectile motion, horizontal and vertical motions are independent. The horizontal distance depends on horizontal speed and time of fall, while vertical motion is controlled by gravity.
The fall time is found using h = ½ g t2. Once the time is known, horizontal speed multiplied by fall time gives the horizontal distance. The independence of vertical and horizontal components allows separate calculations.
Dropping a coin from a moving train illustrates the principle: it moves forward with the train while falling due to gravity.
Horizontal distance results from combining constant horizontal motion with vertical free-fall, demonstrating the key separation of motion components in projectile analysis.
Option a – 2.81 km
By increasing the initial speed of a projectile, the maximum height is raised by 8% while keeping the angle constant. What is the percentage increase in the projectile’s total flight time?
a) 8%
b) 16%
c) 4%
d) 6.25%
Explanation: Maximum height in projectile motion depends on the vertical component of velocity. A change in speed increases vertical velocity, affecting both height and time to reach the peak.
The total flight time is twice the time taken to reach the maximum height. By analyzing how vertical speed influences rise and fall durations, the change in flight time can be estimated. The angle of projection remains constant, so horizontal motion does not alter the calculation.
This is similar to throwing a ball higher: it takes longer to reach the top and return, extending total travel time.
Flight duration depends on vertical velocity and gravitational acceleration, showing how speed changes impact motion timing.
Option d – 6.25%
Which statement about a projectile’s motion is correct?
a) Kinetic energy is highest at the peak of the path
b) Kinetic energy is lowest at the peak of the path
c) Gravitational acceleration is greatest at the peak
d) Gravitational acceleration is smallest at the peak
Explanation: In projectile motion, gravitational acceleration is constant throughout, and kinetic energy changes along the path. Kinetic energy is lowest at the top where vertical velocity is zero, while horizontal velocity remains constant.
The motion is governed by the independence of horizontal and vertical components. Vertical velocity changes due to gravity, affecting kinetic energy, while horizontal velocity does not. Gravitational acceleration remains the same everywhere, irrespective of the projectile’s height.
Think of a ball thrown in a parabolic arc: it slows as it rises and speeds up as it falls, with gravity constantly acting downward.
Understanding energy and acceleration distribution in projectile motion clarifies motion behavior without revealing specific numerical values.
Option b – Kinetic energy is lowest at the peak of the path
A 4 kg Mass moves in a circular path with a radius of 3 m on a smooth horizontal surface. If the string holding it can withstand up to 40 N, what is the highest speed in revolutions per minute?
a) 17 rev/min
b) 20 rev/min
c) 8 rev/min
d) 5 rev/min
Explanation: Circular motion requires centripetal force to maintain the path. The maximum speed is determined by the force the string can withstand, using F = m v2/R.
The speed is linked to the force and radius, and converting linear speed to revolutions per minute involves relating distance traveled per revolution to time. Knowing the maximum sustainable force allows determination of the maximum possible speed before the string would break.
This is similar to spinning an object tied to a string: tighter swings allow faster rotation, but string strength limits speed.
Centripetal force and motion constraints dictate maximum rotational speed, ensuring the object remains in circular motion without exceeding the string’s tension.
Option a – 17 rev/min
Which of the following is true regarding projectile motion?
a) The minimum range is achieved at a 180° angle
b) The minimum range is achieved at a 90° angle
c) The minimum range is achieved at a 60° angle
d) The minimum range is achieved at a 45° angle
Explanation: Projectile motion exhibits independence of horizontal and vertical components. Horizontal velocity remains constant while vertical velocity changes under gravity. The minimum range occurs when launch angle is smallest relative to motion, and maximum range occurs at an optimal angle (usually 45°).
Analyzing projectile paths and understanding motion symmetry clarifies which statements correctly describe behavior. The dependence of vertical motion on gravity versus constant horizontal motion underpins projectile dynamics.
Throwing a ball at different angles shows varying ranges while horizontal speed remains unaffected.
Understanding the relationships between motion components is key to predicting range, height, and time of flight.
Option b – The minimum range is achieved at a 90° angle
A ball is thrown by one person and catches it by another in 4 seconds. What is the maximum height above the projection point? (g = 10 m/s²)
a) 20 m
b) 40 m
c) 15 m
d) 60 m
Explanation: The maximum height of a projectile is determined by the vertical component of velocity and the time to reach the peak. Half the total flight time corresponds to ascent, with gravity decelerating the upward motion.
Time to peak is used with the formula h = ½ g t2 for vertical motion. Understanding how gravity affects ascent allows estimation of maximum height without considering horizontal motion.
This is similar to tossing a ball straight up: it slows until it stops at the top before descending.
Maximum height depends on vertical speed and gravitational acceleration, highlighting the symmetry of upward and downward motion in projectile trajectories.
Option a – 20 m
When a ball is thrown upwards and follows a parabolic path back to the ground, which quantity stays unchanged?
a) Speed
b) Vertical velocity
c) Horizontal velocity
d) Kinetic energy
Explanation: In ideal projectile motion, horizontal velocity remains constant because no horizontal forces act, while vertical velocity changes due to gravity. Speed, vertical velocity, and kinetic energy vary along the path.
The independence of motion components ensures that horizontal motion does not affect vertical motion. While vertical velocity reduces upward and increases downward, horizontal velocity remains the same.
Dropping a coin from a moving vehicle shows the same principle: it keeps moving forward at constant speed while falling.
Understanding horizontal constancy is crucial for analyzing projectile motion without directly revealing numeric outcomes.
Option c – Horizontal velocity
Calculate the maximum height reached by a ball launched at a 60° angle with a horizontal speed of 20 m/s. (g = 10 m/s²)
a) 40 m
b) 30 m
c) 60 m
d) 80 m
Explanation: Maximum height depends on the vertical component of velocity, v sin θ. Gravity decelerates upward motion until vertical velocity becomes zero at the peak.
Height is calculated using h = (v sin θ)2 / (2 g), where only the vertical velocity component contributes. Horizontal motion does not affect height.
Throwing a ball upward at an angle demonstrates the same effect: vertical speed determines peak height, while horizontal speed affects range.
Maximum height results from balancing vertical velocity and gravitational pull, demonstrating energy conversion in projectile motion.
Option c – 60 m
Find the linear speed of a body completing 6 revolutions in a circular path of 4 cm radius in one minute.
a) 0.1 m/s
b) 0.01 m/s
c) 0.025 m/s
d) 0.05 m/s
Explanation: Linear speed in circular motion is calculated by multiplying the circumference of the circle by the number of revolutions per time unit: v = 2 π R × (revolutions / time).
The radius is small, and time conversion may be required to obtain speed in m/s. Understanding the relation between angular motion and linear speed is crucial.
Spinning a toy wheel demonstrates how circumference and rotation rate determine the distance traveled per unit time.
Linear speed depends on radius and revolutions, illustrating the connection between circular motion parameters.
Option c – 0.025 m/s
Determine the flight time of a ball thrown at a 30° angle with an initial speed of 20 m/s. (g = 10 m/s²)
a) 6 s
b) 1 s
c) 2 s
d) 4 s
Explanation: The total flight time of a projectile depends on the vertical component of initial velocity. Gravity slows upward motion and accelerates downward motion symmetrically.
Time to reach the peak is calculated using t = (v sin θ)/g. Total flight time is twice the time to peak. Horizontal velocity does not affect flight time, but understanding vertical motion is essential.
Throwing a ball at an angle shows that higher vertical speed results in a longer time in the air.
Total flight time reflects the balance between vertical speed and gravitational acceleration.
Option c – 2 s
In uniform circular motion, which physical quantity changes continuously?
a) Momentum
b) Speed
c) Mass
d) Kinetic energy
Explanation: In uniform circular motion, speed is constant, but the direction of velocity changes continuously, causing acceleration toward the center.
Centripetal acceleration is always perpendicular to velocity and points inward, ensuring the object remains on the circular path. Momentum Vector also changes direction, while Mass and speed remain constant.
Spinning a stone on a string demonstrates continuous change in direction even though speed is steady.
Direction changes create acceleration, highlighting the difference between uniform speed and changing velocity in circular motion.
Option a – Momentum
Calculate the frictional force acting on an 80 kg cyclist moving in a circle of radius 4 m while leaning at a 30° angle. (g = 10 m/s²)
a) 231 N
b) 462 N
c) 326 N
d) 481 N
Explanation: When a cyclist leans while turning, friction provides the centripetal force required to maintain the circular path.
The horizontal component of friction balances centripetal force, while the vertical component balances weight. Using F = m v2 / R and trigonometric components of the leaning angle allows calculation of friction.
Bicyclists leaning in curves illustrate how tilting generates necessary horizontal force to maintain circular motion.
Friction depends on speed, radius, Mass, and lean angle to prevent slipping during a turn.
Option b – 462 N
Find the safest speed for a 10 m wide road with a curvature radius of 50 m if the raised edge is 2 m higher than the inner edge. (g = 10 m/s²)
a) 15 m/s
b) 20 m/s
c) 5 m/s
d) 10 m/s
Explanation: Banked roads allow vehicles to negotiate curves without relying entirely on friction. Safe speed depends on the banking angle derived from height difference and road width.
Using tan θ = height / width, the banking angle determines the speed for which centripetal force is provided by normal force alone: v = √(R g tan θ).
Cars on racetracks show the effect: higher banks allow higher speeds without slipping.
Safe speed results from balancing gravitational force and centripetal acceleration along the banked curve.
Option d – 10 m/s
What is the angle of projection when an object is thrown vertically upwards?
a) 30°
b) 90°
c) 0°
d) 60°
Explanation: Vertical projection implies motion along the vertical axis with no horizontal component. Gravity decelerates upward motion until vertical velocity is zero at the peak.
Horizontal velocity is zero; only vertical velocity determines height and flight time. The angle of projection is measured from the horizontal, making it 90°.
Throwing a ball straight up demonstrates motion along a single axis, showing maximum height is achieved when horizontal displacement is zero.
Vertical motion is purely influenced by gravity, defining the projection angle as perpendicular to the horizontal.
Option c – 0°
Calculate the minimum coefficient of friction needed to prevent a car from skidding while moving at 80 km/h around a circular track of radius 80 m. (g = 10 m/s²)
a) 0.71
b) 0.62
c) 0.23
d) 0.54
Explanation: Friction provides the centripetal force needed to keep a car moving along a curve without slipping.
Using μ m g = m v2 / R, the coefficient of friction is obtained by dividing centripetal force requirement by the weight. Conversion of speed to m/s is necessary.
Cars on icy roads illustrate low friction causing slipping; higher friction ensures safe circular motion.
Minimum coefficient depends on speed, radius, and gravitational acceleration to prevent skidding.
Option b – 0.62
Assertion: At the peak of a projectile’s path, acceleration remains equal to ‘g’. Reason: The acceleration stays constant at all points of projectile motion and equals ‘g’.
a) Both the assertion and reason are correct, and the reason correctly explains the assertion.
b) Both the assertion and reason are correct, but the reason does not explain the assertion.
c) The assertion is correct, but the reason is incorrect.
d) The assertion is incorrect, but the reason is correct.
Explanation: In projectile motion, acceleration is due to gravity only and is directed vertically downward throughout the motion.
At the peak, vertical velocity is zero, but acceleration continues to act downward. Gravity does not change with position or velocity, ensuring constant acceleration.
Throwing a ball in a parabola shows the topmost point still experiences gravitational pull, pulling it down.
Acceleration remains constant at all points, influencing vertical velocity and overall projectile trajectory.
Option a – Both the assertion and reason are correct, and the reason correctly explains the assertion.
A shell is fired from a cannon at a speed of 250 m/s making a 30° angle with the ground. What horizontal distance will it cover? (g = 10 m/s²)
a) 1.98 km
b) 2.64 km
c) 6.23 km
d) 5.41 km
Explanation: Horizontal distance (range) depends on horizontal velocity and total flight time, which is determined by vertical motion.
Time of flight is twice the time to reach the peak: t = 2 (v sin θ) / g. Horizontal distance is v cos θ × t. Independence of motion components allows calculation of range without interference from vertical motion.
Firing a cannon at an angle illustrates how both horizontal and vertical velocities combine to determine landing position.
Range depends on initial speed, angle, and gravitational acceleration.
Option d – 5.41 km
A car is moving along a circular track with a radius of 200 m at a speed of 40 m/s, and its speed is increasing at a rate of 4 m/s². What is its total acceleration?
a) 12 m/s²
b) 9 m/s²
c) 5 m/s²
d) 10 m/s²
Explanation: Total acceleration in non-uniform circular motion has two components: tangential acceleration (along the path) and centripetal acceleration (toward center).
Centripetal acceleration is ac = v2/R, and tangential acceleration is along the direction of increasing speed. Total acceleration is the Vector sum of these two perpendicular components.
A car speeding up on a curve demonstrates both inward pull and tangential push.
Total acceleration combines centripetal and tangential contributions, giving the overall rate of change of velocity.
Option b – 9 m/s²
What is the ratio of the angular speed of the minute hand to that of the hour hand in a clock?
a) 1 : 6
b) 12 : 1
c) 6 : 1
d) 1 : 12
Explanation: Angular speed is the angle rotated per unit time. The minute hand completes one full rotation in 60 minutes, the hour hand in 12 hours.
Using ω = θ / t, the ratio is obtained by comparing rotation rates. Angular displacement per rotation is 2 π radians for both hands, but time differs significantly.
Clock hands illustrate how rotation speed affects angular displacement over time.
Angular speed ratio highlights the difference in rotation rates of the clock hands over the same angular displacement.
Option b – 12 : 1
A ball rolls off a flat table at 5 m/s and hits the ground after 0.5 seconds. What is the height of the table? (g = 10 m/s²)
a) 1.5 m
b) 3.28 m
c) 2.50 m
d) 1.25 m
Explanation: Vertical motion of the ball is independent of horizontal motion. The fall time is determined solely by the vertical distance and gravity.
Using h = ½ g t2, the height can be calculated. Horizontal speed does not affect vertical distance. Understanding the independence of motion components is essential in projectile analysis.
Dropping a coin while walking forward shows the same effect: it falls straight down while moving horizontally.
Height is determined by vertical displacement due to gravity during the fall time.
Option d – 1.25 m
Which of the following statements is incorrect regarding angular displacement?
a) Angular displacement has physical dimensions
b) Finite angular displacement has both magnitude and direction
c) Finite angular displacement is not a Vector quantity
d) Very small angular displacement is a Vector
Explanation: Angular displacement can be treated as a Vector for small angles, and finite displacement has both magnitude and direction. However, very large or finite angular displacements are sometimes treated differently in scalar terms.
Understanding Vector properties and dimensions helps identify correct and incorrect descriptions. Angular displacement is measured in radians, a dimensionless quantity, but its Vector representation depends on context.
Rotating a fan blade shows how direction matters for angular displacement.
Recognizing Vector properties of angular displacement clarifies its behavior in rotational motion.
Option a – Angular displacement has physical dimensions
A 5 kg object is tied with a nylon cord outside a spacecraft and moves in a circular path at 6 m/s under a force of 100 N. What is the radius of its path?
a) 0.8 m
b) 2.8 m
c) 3.8 m
d) 1.8 m
Explanation: Circular motion requires centripetal force, F = m v2 / R. Knowing Mass, speed, and force allows solving for radius.
Radius determines the path size while force ensures the object remains in circular motion. High-speed rotation or heavier Mass requires greater centripetal force for the same radius.
Swinging a stone on a string shows how faster speed or stronger pull affects the circular path size.
The radius is dictated by balance between speed, Mass, and centripetal force.
Option d – 1.8 m
A 4 kg stone is tied to a 2 m string and spun in a vertical circle at 6 m/s. What is the highest tension in the string?
a) 112 N at the lowest point
b) 112 N at the highest point
c) 226 N at the lowest point
d) 226 N at the highest point
Explanation: Tension varies in vertical circular motion because centripetal force is combined with weight. At the lowest point, tension is maximum as it must counteract weight and provide centripetal acceleration.
Using T = m g + m v2 / R, maximum tension occurs at the bottom. Top tension is lower because weight and centripetal force partly cancel.
Swinging a bucket of water in a loop illustrates the need for stronger force at the bottom.
Maximum tension arises from combination of weight and centripetal requirement in vertical circular motion.
Option a – 112 N at the lowest point
A Mass moves in a circle with a 2 m radius on a smooth table. The string can withstand a maximum of 40 N. If the maximum speed is 60 revolutions per minute, what is the mass?
a) 0.51 kg
b) 2 kg
c) 1 kg
d) 2.52 kg
Explanation: Centripetal force F = m v2 / R limits mass for given speed and string strength. Linear speed is derived from revolutions per minute, then mass can be found.
Convert rotational speed to linear velocity using v = 2 π R × (revolutions / time). Use maximum tension to determine mass.
Spinning a toy on a string demonstrates how mass and speed affect tension.
Maximum mass depends on speed, radius, and string tension to maintain circular motion.
Option a – 0.51 kg
A 6.2 kg particle moves in a vertical circle using a string. What is the difference in string tension between the bottom and top positions?
a) 372 N
b) 146 N
c) 128 N
d) 209 N
Explanation: Tension at top and bottom differ due to combination of weight and centripetal force requirements. Bottom tension includes weight plus centripetal need; top tension is reduced because weight partially assists centripetal force.
ΔT = Tbottom − Ttop = 2 m g. Understanding forces along vertical paths helps estimate variation without knowing speeds.
A roller coaster at loop top and bottom experiences similar force differences.
Tension difference arises from weight contributing or opposing centripetal requirement at top and bottom points.
Option a – 372 N
What is the total acceleration of a motorcycle traveling along a circular path with a radius of 200 m at 20 m/s?
a) 4 m/s²
b) 3 m/s²
c) 2 m/s²
d) 1.8 m/s²
Explanation: For uniform circular motion, total acceleration equals centripetal acceleration a = v2/R, directed toward the center of the circle.
There is no tangential acceleration if speed is constant. Acceleration changes direction continuously but maintains constant magnitude.
Motorcycles turning on a circular track illustrate acceleration pointing inward despite constant speed.
Total acceleration is determined entirely by centripetal requirement for circular motion.
Option c – 2 m/s²
How is the average velocity of an object calculated?
a) Total distance divided by total time
b) The average of initial and final velocities
c) Both a and b
d) None of the above
Explanation: Average velocity is defined as total displacement divided by total time. It differs from average speed, which uses total distance.
Displacement considers starting and ending points, taking direction into account. Average velocity provides insight into overall motion over a time interval regardless of path taken.
Driving 10 km north and returning 5 km south illustrates how average velocity accounts for NET displacement.
Average velocity combines displacement and time to describe overall motion direction and rate.
Option b – The average of initial and final velocities
What does negative acceleration mean?
a) The object’s speed is increasing
b) The object’s speed is decreasing
c) The object’s speed remains unchanged
d) The object comes to rest
Explanation: Negative acceleration indicates that the velocity of an object is decreasing along the direction of motion.
It does not imply motion in the opposite direction; rather, it is deceleration. Magnitude of velocity reduces with time due to forces acting opposite to motion.
Slowing a car while moving forward demonstrates negative acceleration.
Negative acceleration reflects a decrease in speed in the direction of motion over time.
Option b – The object’s speed is decreasing
In distance-time graphs:
a) Distance is plotted on the X-axis
b) Time is plotted on the Y-axis
c) A straight line shows uniform motion
d) A straight line shows non-uniform motion
Explanation: Distance-time graphs plot distance along the vertical axis and time along the horizontal axis.
A straight line indicates uniform motion with constant speed; curved lines indicate changing speed. Graph shape visually represents motion characteristics, allowing understanding of uniformity or variability.
Walking at constant pace produces a straight line, while accelerating produces a curve.
Distance-time graphs reveal motion patterns based on slope and curvature.
Option c – A straight line shows uniform motion
In velocity-time graphs:
a) Velocity is plotted on the Y-axis and time on the X-axis
b) A straight line indicates constant acceleration
c) A straight line parallel to the X-axis shows uniform speed
d) All of the above
Explanation: Velocity-time graphs plot velocity along the vertical axis and time along the horizontal axis.
A straight line shows constant acceleration; a line parallel to the time axis indicates constant speed. The slope of the graph represents acceleration. Curved lines indicate changing acceleration.
A car accelerating uniformly produces a straight sloped line, while cruising at constant speed produces a horizontal line.
Velocity-time graphs provide visual insight into acceleration and speed trends.
Option d – All of the above
When an object moves in a circular path at constant speed: ( Motion in Two Dimensions mcq )
a) Velocity stays constant and acceleration is zero
b) Both velocity and acceleration continuously change direction
c) Both velocity and acceleration are increasing
d) Velocity is constant but acceleration direction changes
Explanation: Even if speed is constant, the velocity Vector changes direction continuously, causing centripetal acceleration toward the circle’s center.
Acceleration is always perpendicular to velocity. Both direction of velocity and acceleration change, while speed magnitude remains constant.
Spinning a stone on a string shows that motion direction changes, resulting in acceleration despite constant speed.
Constant-speed circular motion involves continuous directional change producing inward acceleration.
Option b – Both velocity and acceleration continuously change direction
If the velocity-time graph of a car is a straight line, what does it indicate? ( Motion in Two Dimensions mcq )
a) The car started from rest
b) The car had constant acceleration
c) The car’s acceleration kept increasing
d) The car moved at a constant speed
Explanation: A straight line in a velocity-time graph indicates constant acceleration. The slope of the line equals the acceleration magnitude.
The intercept indicates initial velocity, and uniform slope shows acceleration does not change with time.
A car smoothly increasing speed demonstrates a straight line in a velocity-time graph.
Straight velocity-time lines reveal constant rate of change of velocity.
Option b – The car had constant acceleration
A car is moving north on a straight path and suddenly applies brakes. What happens next?
a) The car experiences no acceleration
b) The car gains acceleration towards the south
c) The car gains acceleration towards the north
d) The car accelerates either east or west
Explanation: Braking creates acceleration opposite to the car’s motion, reducing speed.
Acceleration Vector points south, opposite to northward motion, because forces applied oppose velocity. Magnitude depends on braking strength.
Stopping a bicycle quickly demonstrates acceleration opposite to motion direction.
Applying brakes produces acceleration in the direction opposite to movement, slowing the vehicle.
Option b – The car gains acceleration towards the south
An object will move at a constant velocity when ( Motion in Two Dimensions mcq )
a) The forces acting on it are balanced
b) No external force is applied
c) Both a and b
d) Either a or b
Explanation: Constant velocity occurs when NET external forces are zero or balanced. No unbalanced force allows the object to maintain both speed and direction.
Both Newton’s first law and equilibrium conditions apply. Any unbalanced force changes speed or direction.
A sliding puck on frictionless ice continues in a straight line at constant speed.
Constant velocity requires balanced forces or absence of external forces.
Option c – Both a and b
A car accelerates from 12 m/s to 30 m/s in 15 seconds. What is its acceleration? ( Motion in Two Dimensions mcq )
a) 2.4 m/s²
b) 1.2 m/s²
c) 2 m/s²
d) 5.2 m/s²
Explanation: Acceleration is the rate of change of velocity with respect to time.
Using a = (vf − vi)/t, where vf is final velocity, vi is initial velocity, and t is time. Units must be consistent.
A car speeding up on a straight road illustrates constant acceleration over a time interval.
Acceleration reflects how quickly an object’s speed changes over a given time.
Option b – 1.2 m/s²
The change in an object’s position with reference to a fixed point is called
a) Origin
b) Starting point
c) Ending point
d) Distance
Explanation: The difference between initial and final position relative to a reference point is called displacement.
Displacement includes direction, whereas distance measures total path length regardless of direction. Vector quantity characteristics apply.
Walking from home to a shop shows how displacement is measured from start to end.
Displacement describes NET positional change from a reference point.
Option a – Origin
Displacement is defined as ( Motion in Two Dimensions mcq )
a) The shortest distance between the starting and ending points
b) The total distance traveled between the starting and ending points
c) The distance covered by the moving object
d) The distance covered per unit time
Explanation: Displacement is the shortest straight-line distance between initial and final positions, including direction.
It differs from distance, which is scalar and path-dependent. Displacement is a vector quantity important in motion analysis.
A hiker completing a circular path returns to start; displacement is zero despite total distance traveled.
Displacement reflects NET change in position, incorporating both magnitude and direction.
Option a – The shortest distance between the starting and ending points
If an object travels 10 km in 15 minutes, its displacement can be ( Motion in Two Dimensions mcq )
a) 10 km
b) Zero
c) More than 10 km
d) Any of the above
Explanation: Displacement depends on initial and final positions. Depending on path, it can equal distance, be less than distance, or even zero.
Traveling in a straight line yields displacement equal to distance; circular or back-and-forth paths reduce NET displacement.
A runner circling a track returns to start: distance is covered, but displacement is zero.
Displacement reflects NET positional change, independent of total path traveled.
Option d – Any of the above
When an object covers the same distance in the same amount of time repeatedly, it is said to be in ( Motion in Two Dimensions mcq )
a) Circular motion
b) Uniform motion
c) Oscillatory motion
d) Non-uniform motion
Explanation: Repeated equal distance in equal time defines uniform motion, with constant speed along the path.
Velocity is uniform if direction remains unchanged. Non-uniform motion occurs when speed or direction varies.
A train moving on a straight track at steady speed demonstrates uniform motion.
Uniform motion is characterized by consistent distance coverage per unit time.
Option b – Uniform motion
Two balls, A and B, each of the same mass, are thrown from the top of a building. Ball A is thrown upwards and ball B is thrown downwards, both with the same speed. What happens when they reach the ground?
a) A hits the ground with more speed than B
b) B hits the ground with more speed than A
c) Both A and B reach the ground at the same speed
d) None of the above
Explanation: The final speed of a falling object depends on initial velocity and gravitational acceleration.
Both balls experience the same gravitational acceleration over the same vertical distance. Ball A slows first while going upward, then accelerates downward; Ball B accelerates downward immediately. Using energy or kinematics, both reach the ground with the same speed.
Dropping two balls from a roof, one thrown up and one down, illustrates that initial direction does not affect final speed for equal height and gravity.
Gravity acts equally on both, making their impact speeds identical.
Option c – Both A and B reach the ground at the same speed
A ball is dropped from a tower of height h. It takes T seconds to reach the ground. Where will the ball be after T/3 seconds?
a) h/9 meters above the ground
b) 7h/9 meters above the ground
c) 8h/9 meters above the ground
d) 17h/18 meters above the ground
Explanation: Vertical displacement under gravity is s = ½ g t2.
At t = T/3, s = ½ g (T/3)2 = 1/9 of total height. The ball is h − h/9 = 8h/9 above the ground.
Dropping an object illustrates that displacement scales with the square of elapsed time.
Position at any fraction of total fall time is determined using proportionality of t2 to distance.
Option c – 8h/9 meters above the ground
When a bus turns suddenly, passengers seem to be thrown outward because of ( Motion in Two Dimensions mcq )
a) Inertia of motion
b) Acceleration
c) Speed of motion
d) Both b and c
Explanation: Inertia causes passengers to continue in straight-line motion while the bus changes direction.
The apparent outward force is due to the bus’s acceleration toward the center of the turn, not an actual force acting on passengers. Centripetal acceleration of the vehicle causes relative motion perception.
Turning in a car at speed demonstrates how passengers lean outward due to inertia.
Inertia explains apparent outward motion during sudden changes in vehicle direction.
Option a – Inertia of motion
A thief runs west at 6 m/s after stealing a purse. A policeman starts chasing from 15 m to the east at 8.5 m/s. How far will the policeman run to catch the thief? ( Motion in Two Dimensions MCQ )
a) 12 m
b) 51 m
c) 61 m
d) 55 m
Explanation: Relative motion determines time and distance to catch the moving target.
Relative speed = 8.5 + 6 = 14.5 m/s. Time to catch = distance / relative speed = 15 / 14.5 s. Policeman’s distance = 8.5 × t.
Chasing games or cars illustrate how relative velocities simplify pursuit calculations.
Distance depends on relative speed and initial separation.
Option b – 51 m
A car moving at 24 m/s passes a stationary motorcycle. As the car passes, the motorcycle starts accelerating at 3.2 m/s². How long will it take for the motorcycle to catch up with the car?
a) 12 s
b) 14 s
c) 20 s
d) 18 s
Explanation: The motorcycle starts from rest, while car moves at constant speed.
Using s = ut + ½ a t2 for the motorcycle and s = v t for the car, SET distances equal to solve for time. Quadratic equation may be used.
Pursuit problems in Physics demonstrate how acceleration over time allows a slower start to catch a moving object.
Time to catch depends on equating distances traveled by both vehicles.
Option b – 14 s
A man walks 2.5 km to the market at 5 km/h. Finding the market closed, he returns immediately at 7.5 km/h. What is his average speed over the 40-minute trip?
a) 5 km/h
b) 25/4 km/h
c) 30/4 km/h
d) 45/8 km/h
Explanation: Average speed = total distance / total time.
Total distance = 2.5 + 2.5 = 5 km. Time to go = 2.5 / 5 = 0.5 h; time to return = 2.5 / 7.5 = 1/3 h. Total time = 5/6 h. Average speed = 5 ÷ (5/6) = 6 km/h.
A round trip with unequal speeds illustrates that average speed is not the arithmetic mean.
Average speed depends on total distance and total time, not individual speeds.
Option d – 45/8 km/h
A ball is dropped from a window 24 m high. How long will it take to hit the ground? ( Motion in Two Dimensions MCQ )
a) 2.2 s
b) 1.2 s
c) 4.5 s
d) 0.2 s
Explanation: Vertical motion is influenced by gravity: s = ½ g t2.
Substitute s = 24 m, g = 10 m/s² → t = √(2 s / g) = √(48 / 10) ≈ 2.2 s.
Dropping an object from a balcony demonstrates the effect of gravity on free fall.
Time to reach the ground is determined by vertical displacement and gravitational acceleration.
Option a – 2.2 s
Mohan cycles 3.2 km north in 20 minutes. What is his speed in km/h?
a) 8.1
b) 9.6
c) 1.2
d) 7.2
Explanation: Speed = distance / time.
Distance = 3.2 km, time = 20 min = 1/3 h → speed = 3.2 ÷ (1/3) = 9.6 km/h.
Cycling over a measured distance in a given time illustrates basic speed calculation.
Speed is calculated from distance covered divided by total time in consistent units.
Option b – 9.6
A pitcher throws a baseball horizontally at 42.1 m/s. The distance to the batter is 18.3 m. How far will the ball drop before it reaches the batter? ( Motion in Two Dimensions MCQ )
a) 0.8 m
b) 1.2 m
c) 2.2 m
d) 0.93 m
Explanation: Horizontal and vertical motions are independent. Horizontal motion: constant speed v = 42.1 m/s; vertical motion: s = ½ g t2.
Time to reach batter t = distance / horizontal speed = 18.3 / 42.1 s. Vertical drop = ½ g t2.
Throwing a ball demonstrates that horizontal motion does not affect vertical free-fall distance.
Vertical drop is determined by gravity acting during the horizontal travel time.
Option d – 0.93 m
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