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Force & Motion Past Papers
Solved past paper MCQs for Force & Motion from official UHS, NUMS, SZABMU, DUHS, and KMU examinations. Includes verified distractor autopsies and step-by-step cognitive explanations.
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Official printable A4 booklet containing all 115 verified past paper MCQs with unbroken options, 5-column rapid answer key, and Propolis cognitive error autopsies.
The slope of velocity-time graph gradually decreases, then the body is said to be moving with
A
Negative acceleration
B
Positive acceleration
C
Uniform velocity
D
Variable acceleration
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
The slope of a velocity-time graph is identically the acceleration.
Solution:
If the slope is "gradually decreasing," it means the value of the slope is changing at every point.
Because slope = acceleration, a changing slope strictly dictates a changing, non-constant acceleration.
Therefore, the body is moving with variable acceleration.
Why other options are incorrect:
"Negative acceleration" just tells you the sign of a constant slope. "Uniform velocity" would mean zero acceleration. Only "Variable" directly captures the changing nature of the slope.
If you stops your car quickly by wearing seat belts, chance of injury is greatly reduced because seat belt applied
A
Extra force
B
Perpendicular force
C
Opposite force
D
Zero force
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
Seatbelts utilize the impulse-momentum theorem and Newton's Third Law to protect passengers.
Solution:
During a sudden stop, your body wants to continue forward due to inertia.
The seatbelt stretches slightly, applying a restraining force directly backward (opposite) against your chest to safely reduce your momentum over a longer time span.
This opposite force counters the forward inertia, preventing a high-force impact with the windshield.
Why other options are incorrect:
Extra force is vague. Perpendicular force would push you sideways. Zero force means the seatbelt did absolutely nothing to restrain you.
Historical paradigm of classical mechanics before Einstein's relativity.
Solution:
Both Galileo Galilei and Isaac Newton constructed their entire frameworks of classical kinematics upon the rigid assumption that time is a universal, absolute constant.
In their universe, one second ticking on Earth was mathematically identical to one second ticking anywhere else, perfectly independent of an observer's speed or gravity.
It wasn't until Albert Einstein's Special Relativity that time was proven to be relative.
Why other options are incorrect:
Selecting only one ignores that classical mechanics as a whole was founded on absolute time.
A 40kg body starting from rest falls through a vertical distance of 125cm to ground. The velocity of the body just before it hits the ground is:
A
\( 250 \text{ m/s} \)
B
\( (250) \text{ m/s} \)
C
\( 25 \text{ m/s} \)
D
\( 5 \text{ m/s} \)
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
Energy conservation or the third equation of kinematics, heavily relying on strict unit conversion.
Formula:
$$ v = \sqrt{2gh} $$
Solution:
Mass is irrelevant for free fall speed.
CRITICAL: Convert height from cm to meters. \( h = 125 \text{ cm} = 1.25 \text{ meters} \).
Use gravity \( g = 10 \text{ m/s}^2 \) (standard estimation).
$$ v = \sqrt{2 \times 10 \times 1.25} $$
$$ v = \sqrt{25} = 5 \text{ m/s} $$
Why other options are incorrect:
If a student forgets to convert cm to meters, they calculate \( \sqrt{2 \times 10 \times 125} = \sqrt{2500} = 50 \). If they didn't square root properly, they might land on 25. The 40kg mass is purely a distractor.
Momentum is product of mass and velocity, then momentum and velocity are:
A
Parallel
B
Perpendicular
C
Antiparallel
D
Independent
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
Multiplying a vector by a strictly positive scalar preserves its direction entirely.
Formula:
$$ \vec{p} = m\vec{v} $$
Solution:
Mass (\( m \)) is an intrinsically positive scalar quantity.
When the velocity vector (\( \vec{v} \)) is multiplied by mass, the resulting momentum vector (\( \vec{p} \)) is scaled up, but its geometric direction is untouched.
Therefore, momentum ALWAYS points exactly in the same direction as velocity. Vectors in the same direction are parallel.
Why other options are incorrect:
Antiparallel would require a negative mass, which does not exist in classical physics.
Neglecting the effect of air resistance how long a stone takes to dropped of a 125 m high building lands on the ground, take g = \( 10 \text{ m/s}^2 \):
A
3 sec
B
4 sec
C
18 sec
D
5 sec
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
The time taken for an object dropped from rest to hit the ground strictly depends on the height and gravity.
Formula:
$$ S = v_it + \frac{1}{2}gt^2 $$
Solution:
Because it is dropped, initial velocity \( v_i = 0 \).
Take the square root: \( t = 5 \text{ seconds} \).
Why other options are incorrect:
Failing to take the square root results in 25. Ignoring the \( 1/2 \) multiplier gives roughly 3.5s. Simple arithmetic errors lead to the other values.
What will be the value of acceleration, if a body of mass 0.5 Kg is acted upon by a force of 10 N?
A
40 m/s²
B
5 m/s²
C
10 m/s²
D
20 m/s²
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
Newton's Second Law directly relates force, mass, and acceleration.
Formula:
$$ F = ma \implies a = \frac{F}{m} $$
Solution:
Force \( F = 10 \text{ N} \).
Mass \( m = 0.5 \text{ kg} \).
Substitute values: $$ a = \frac{10}{0.5} $$
$$ a = 20 \text{ m/s}^2 $$
Note: The options provided in the source use 'N' as the unit, which is a typographical error in the original paper (it should be \( \text{m/s}^2 \)). We select 20 based on the numerical value.
Why other options are incorrect:
Multiplying 10 by 0.5 yields 5. Mathematical errors in division yield 40 or 10.
Due to sudden rain, which of the following is true for the car moving on the road?
A
Force of engine decreases
B
Balance will not be maintained
C
Direction of motion will be effected
D
Linear momentum increases
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
Friction is heavily dependent on the nature of the surfaces in contact. Water acts as a lubricant.
Solution:
Rain introduces water between the car's tires and the asphalt, significantly reducing the coefficient of friction.
Without sufficient friction, the tires lose their grip, causing the car to slide (hydroplane) and lose its steady balance and control.
Why other options are incorrect:
Rain does not directly decrease the engine's mechanical power. Direction isn't necessarily affected unless a turn is attempted. Momentum does not suddenly increase just because the road is wet.
Two bodies of mass 10 kg and 40 kg are dropped from the same height at the same time. Value of which to the following remains the same during the motion of the two bodies?
A
Acceleration
B
Kinetic energy
C
Potential energy
D
Power
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
In the absence of air resistance, all bodies strictly fall at the identical rate of acceleration regardless of their mass.
Solution:
Galileo famously demonstrated that acceleration due to gravity (\( g \approx 9.8 \text{ m/s}^2 \)) is entirely independent of an object's mass.
Therefore, both the 10kg and 40kg bodies experience the exact same uniform downward acceleration.
Why other options are incorrect:
Kinetic and potential energies strictly depend on mass (\( \frac{1}{2}mv^2 \) and \( mgh \)). Since their masses are different, their energies are vastly different. Power also depends on mass.
A force 2F acting on a particle of mass 10kg produces an acceleration of \( 30 \text{ m/s}^2 \). a force 5F acting on a particle of mass M, produces an acceleration of \( 25 \text{ m/s}^2 \). What is the mass?
A
30 kg
B
21 kg
C
4.8 kg
D
3.3 kg
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
Use Newton's Second Law sequentially to link the two scenarios.
Formula:
$$ F_{net} = ma $$
Solution:
Scenario 1: $$ 2F = 10 \times 30 = 300 $$
This implies \( F = 150 \text{ N} \).
Scenario 2: A force of 5F acts on mass M. Force applied = \( 5 \times 150 = 750 \text{ N} \).
Set up equation 2: $$ 750 = M \times 25 $$
Solve for M: $$ M = \frac{750}{25} = 30 \text{ kg} $$
Why other options are incorrect:
Algebraic errors solving for F or dividing incorrectly result in the other distractor values.
The rate of change of displacement with respect to time is the direct physical definition of velocity.
Why other options are incorrect:
Acceleration is the slope of velocity-time. Distance is the magnitude of a purely scalar path length. Displacement is read directly from the y-axis itself.
The mathematical gradient (or slope) of a line on a Cartesian plane is the ratio of the change in the y-axis variable to the change in the x-axis variable.
In a velocity-time graph, the y-axis represents velocity (\( v \)) and the x-axis represents time (\( t \)).
The rate of change of velocity with respect to time (\( \frac{\Delta v}{\Delta t} \)) is the direct definition of acceleration.
Why other options are incorrect:
The area under the graph gives displacement/distance. Force requires mass (Newton's 2nd Law). Speed is a scalar magnitude of velocity, not its gradient over time.
Because only angles that satisfy the trigonometric identity \( \sin(2(90^\circ-\theta)) = \sin(180^\circ-2\theta) = \sin(2\theta) \) will yield the same range.
Two cars travelling on straight road in opposite direction with speed \( 70 \text{ kmh}^{-1} \) and \( 60 \text{ kmh}^{-1} \), their relative velocity will be:
A
\( 10 \text{ kmh}^{-1} \)
B
\( 130 \text{ kmh}^{-1} \)
C
\( 65 \text{ kmh}^{-1} \)
D
\( 5 \text{ kmh}^{-1} \)
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
Relative velocity measures how fast one object is moving from the perspective of another. When objects move in opposite directions, their approach (or separation) rate is the sum of their individual speeds.
Formula:
$$ \vec{v}_{rel} = \vec{v}_1 - \vec{v}_2 $$
Solution:
Let the first car's velocity be \( \vec{v}_1 = +70 \text{ km/h} \).
Since the second car moves in the opposite direction, \( \vec{v}_2 = -60 \text{ km/h} \).
The relative velocity magnitude is \( |\vec{v}_{rel}| = |70 - (-60)| = 130 \text{ km/h} \).
Why other options are incorrect:
Subtracting the magnitudes (10 km/h) applies only if they are travelling in the same direction. Averaging them (65 km/h) has no physical meaning here.
Substitute this into the equation for the first body: $$ v_1' = \left( \frac{m - m}{m + m} \right)v_1 + \left( \frac{2m}{2m} \right)v_2 = (0)v_1 + (1)v_2 = v_2 $$
Similarly, for the second body, \( v_2' = v_1 \).
Therefore, they simply swap velocities.
Why other options are incorrect:
Options A and B represent a specific subset where one of the bodies is initially at rest. Option C suggests they bounce back with their own reversed velocities, which only happens if they hit a massive immovable wall.
Distance is a scalar quantity that measures the total length of the path covered by a moving object. Because it is a cumulative total, it can only increase or remain the same.
Solution:
The slope of a distance-time graph represents speed (\( \frac{\Delta d}{\Delta t} \)).
Since distance cannot physically decrease over time, \( \Delta d \) must always be greater than or equal to zero.
Therefore, the slope can be positive (moving) or zero (stationary), but it can mathematically and physically never be negative.
Why other options are incorrect:
Displacement-time graphs CAN have a negative slope (if returning toward the origin), but distance is fundamentally additive.
Projectile motion combines uniform horizontal motion and uniformly accelerated vertical motion due to gravity.
Solution:
Throughout the entire flight, the horizontal velocity \( v_x \) remains constant because there is no horizontal force (ignoring air resistance).
The vertical velocity \( v_y \) decreases as the object rises, reaching exactly zero at the absolute highest point of the trajectory, before becoming negative as it falls back down.
Therefore, at the peak, \( v_y = 0 \) and \( v_x = \text{constant} \).
Why other options are incorrect:
If \( v_x \) was also zero (Option A), the projectile would drop straight down like a stone. Gravity ensures \( v_y \) is never constant (Option C, D).
In Newton's first law of motion which quantity remains constant:
A
Velocity
B
Angular displacement
C
Amplitude
D
Amount of work
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
Newton's First Law (Law of Inertia) states that an object will remain at rest or in uniform motion in a straight line unless acted upon by a net external force.
Solution:
"Uniform motion in a straight line" is the exact definition of constant velocity.
If the net force is zero, the acceleration is zero.
Zero acceleration strictly means the velocity vector (both magnitude and direction) remains absolutely constant.
Why other options are incorrect:
Angular displacement, amplitude, and work are not the fundamental quantities preserved by translational inertia. An object can have constant velocity while its displacement continuously changes.
The Law of Inertia (Newton's First Law) applies exclusively when the net external force on a system is strictly zero.
Solution:
When the sum of all forces acting on an object is zero (\( \Sigma F = 0 \)), the object is in a state of translational equilibrium.
Because the law of inertia describes motion precisely under this \( \Sigma F = 0 \) condition, it inherently satisfies the first condition of equilibrium.
Why other options are incorrect:
Variable forces or contact forces induce acceleration, violating the premise of inertia. Mass conservation is a separate principle altogether.
A projectile is launched, its velocity is maximum at:
A
Point of projection
B
Highest point
C
Between launching and highest point
D
All points
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
The total velocity of a projectile is the vector sum of its horizontal and vertical components. As it rises against gravity, its kinetic energy decreases and converts to potential energy.
Solution:
At the point of projection, the projectile possesses its full initial vertical velocity (\( v_{iy} \)) and horizontal velocity (\( v_x \)).
As it moves upward, \( v_x \) remains constant, but \( v_y \) strictly decreases, meaning the overall magnitude \( v = \sqrt{v_x^2 + v_y^2} \) decreases.
The velocity is at its absolute minimum at the highest point (where \( v_y = 0 \)). It returns to its maximum magnitude right as it hits the ground (landing point). Assuming a ground-to-ground launch, projection and landing share the maximum speed.
Why other options are incorrect:
The highest point has the minimum velocity. Velocity is not constant across all points.
Dividing the total net displacement covered during an entire journey by the total elapsed time yields a single summarizing value.
This macro-level calculation is the formal definition of average velocity.
Why other options are incorrect:
Instantaneous velocity is measured at a precise moment (\( \Delta t \to 0 \)). Uniform and variable velocities are states of motion, not defined merely by a bulk division formula.
The velocity time graph of a motion starting from rest with uniform acceleration is a straight line:
A
Not passing through origin
B
Parallel to time axis
C
Parallel to velocity axis
D
Passing through origin
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
A velocity-time graph maps velocity on the y-axis against time on the x-axis. The slope of this graph is the acceleration.
Formula:
$$ v = v_i + at $$
Solution:
The object starts from rest, meaning initial velocity \( v_i = 0 \).
Substituting this gives the equation \( v = at \).
This is a linear equation matching the form \( y = mx + c \), where \( c = 0 \) (the y-intercept).
Because the y-intercept is exactly 0, the straight line must strictly pass through the origin (0,0).
Why other options are incorrect:
If it did not pass through the origin, \( v_i \) would not be zero. Parallel to time axis means zero acceleration. Parallel to velocity axis is physically impossible.
A projectile is thrown at an angle of \( 45^\circ \) with horizontal and its range is \( R_1 \). Another projectile is thrown at an angle of \( 45^\circ \) with vertical and its range is \( R_2 \). The relation between \( R_1 \) and \( R_2 \) is:
A
\( R_2 = 2R_1 \)
B
\( R_1 = 2R_2 \)
C
\( R_1 = R_2 \)
D
\( 3R_1 = R_2 \)
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
Standard kinematics uses the angle relative to the horizontal to calculate range.
Solution:
Projectile 1: Angle with horizontal is \( \theta_1 = 45^\circ \).
Projectile 2: Angle with the vertical is \( 45^\circ \). Because the axes are orthogonal (\( 90^\circ \)), the angle with the horizontal is \( 90^\circ - 45^\circ = 45^\circ \). So, \( \theta_2 = 45^\circ \).
Since both projectiles are effectively launched at the exact same horizontal angle (\( 45^\circ \)), assuming equal initial velocities, their ranges will be absolutely identical.
Therefore, \( R_1 = R_2 \).
Why other options are incorrect:
Any factor scaling implies the angles resulted in different trigonometric outputs, which is false here.
Air friction (drag) is a vector force that always opposes the instantaneous velocity vector of a moving object.
Solution:
A projectile travels in a 2D plane, meaning its velocity vector has both horizontal (\( v_x \)) and vertical (\( v_y \)) components.
Because drag strictly opposes the direction of motion, it creates both a backward horizontal force (slowing down horizontal speed) and a vertical force (opposing upward rise or slowing downward fall).
Therefore, air friction continuously affects both dimensions of the motion.
Why other options are incorrect:
Drag does not affect the initial launch velocity (which is determined by the launcher). It cannot selectively target only one axis because the projectile moves diagonally through the air.
Which of the following pairs has the same direction always?
A
Force, displacement
B
Force, velocity
C
Force, acceleration
D
Force, momentum
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
According to Newton's Second Law of motion, the relationship between net force and acceleration is rigidly governed by a scalar mass.
Formula:
$$ \vec{F}_{net} = m\vec{a} $$
Solution:
Mass (\( m \)) is a strictly positive scalar value.
When a vector (\( \vec{a} \)) is multiplied by a positive scalar, the resulting vector (\( \vec{F} \)) MUST point in the exact same direction.
Therefore, the net force and the acceleration it produces are permanently and flawlessly aligned in direction.
Why other options are incorrect:
Displacement and velocity vectors tell you where an object is going, not where it is being pushed. For example, a car braking has forward velocity but backward force.
Newton's Third Law states: To every action, there is always an equal and opposite reaction.
Solution:
The profound implication of this law is that a solitary, isolated force physically cannot exist in the universe.
Forces result entirely from interactions between two separate objects.
Therefore, forces ALWAYS occur in action-reaction pairs. The third law fundamentally declares the existence of this dual-pairing nature of forces.
Why other options are incorrect:
The effect of force is explained by the Second Law (\( F=ma \)). The existence of a single force conceptually traces back to the First Law. "Two forces" is vague, while "pair of forces" explicitly covers the action-reaction mutual relationship.
Forgetting the square root gives 16s. Forgetting the \( 1/2 \) in the formula gives \( \sqrt{8} \approx 2.8 \text{s} \). Calculating linearly without squared time gives 8s.
4.9 comes from dividing by 2 (confusing with distance formula). 9.4 is a typo for 9.8 (1 second of fall). The decimal point shifts in Options A, B, and C represent calculation errors.
A bullet is fired horizontally with 20m/s. In the absence of air friction, its horizontal velocity after 2 seconds will be:
A
\( 10 \text{ ms}^{-1} \)
B
\( 5 \text{ ms}^{-1} \)
C
\( 20 \text{ ms}^{-1} \)
D
\( 40 \text{ ms}^{-1} \)
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
Newton's First Law of motion states an object will maintain constant velocity unless acted on by an external force.
Solution:
A projectile in flight experiences only one force: gravity, which acts strictly vertically downwards.
Because there is zero air friction, there are absolutely no horizontal forces acting on the bullet.
Therefore, horizontal acceleration is zero, and the horizontal velocity remains exactly \( 20 \text{ m/s} \) for the entire duration of the flight (Option C).
Why other options are incorrect:
Options A and B incorrectly assume deceleration. Option D wrongly multiplies velocity by time or assumes horizontal acceleration exists.
A box of mass m = 6kg slides with speed v = 4m/s across a frictionless floor, it suddenly explodes into two pieces. One piece, with mass \( m_1 \) = 2kg moves in the same direction with speed \( v_1 \)=8 m/s. The velocity of the second piece is:
A
2 m/sec
B
3 m/sec
C
4 m/sec
D
6 m/sec
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
An explosion is governed by internal forces. Therefore, the total momentum of the isolated system is perfectly conserved before and after the explosion.
Formula:
$$ P_{initial} = P_{final} $$
$$ mv = m_1v_1 + m_2v_2 $$
Solution:
Total initial momentum: \( 6 \text{ kg} \times 4 \text{ m/s} = 24 \text{ kg m/s} \).
After explosion, Piece 1 mass \( m_1 = 2 \text{ kg} \) and \( v_1 = 8 \text{ m/s} \). Its momentum = \( 2 \times 8 = 16 \text{ kg m/s} \).
A helicopter of weight 'W' is ascending with zero acceleration. The upward force 'F' on it is:
A
\( F > W \)
B
\( F = \sqrt{W} \)
C
\( F = 0 \)
D
\( F = W \)
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
According to Newton's Second Law, acceleration is directly proportional to net force. Zero acceleration implies zero net force.
Formula:
$$ \Sigma F = ma $$
$$ F_{up} - F_{down} = m(0) $$
Solution:
The upward force is the lift \( F \).
The downward force is the weight \( W \).
Since acceleration is exactly zero (it is moving at a steady, uniform velocity), the forces must perfectly balance.
$$ F - W = 0 \implies F = W $$
Why other options are incorrect:
If \( F > W \), the helicopter would be accelerating upward (speeding up). If \( F < W \), it would be decelerating (slowing down). Zero force implies no lift, meaning it would be in freefall.
When a body moves with uniform velocity than it's:
A
Average velocity is zero
B
Instantaneous velocity is greater than average velocity
C
Instantaneous velocity is zero
D
Instantaneous velocity is equal to average velocity
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
Uniform velocity means the object's speed and direction remain absolutely constant for the entire duration of motion.
Solution:
If a car drives strictly at \( 60 \text{ km/h} \) East without ever speeding up, slowing down, or turning, its velocity at any given split-second (instantaneous) is \( 60 \text{ km/h} \) East.
If you calculate the total displacement over total time (average), it will also perfectly yield \( 60 \text{ km/h} \) East.
Therefore, for uniform motion, the instantaneous velocity is permanently identical to the average velocity.
Why other options are incorrect:
Differences between average and instantaneous only arise if the velocity fluctuates (acceleration/deceleration) over the journey.
Kinetic is conserved but total energy is not conserved
C
Momentum is conserved but kinetic energy is not conserved
D
Both kinetic energy and momentum are not conserved
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
Collisions in isolated systems always conserve linear momentum. How they handle kinetic energy dictates their specific classification.
Solution:
By universal law, total momentum is conserved in all types of collisions.
An "elastic" collision is strictly defined as a perfect bounce where absolutely zero macroscopic kinetic energy is converted into heat, sound, or permanent deformation.
Therefore, both kinetic energy and momentum remain exactly equal before and after the impact.
Why other options are incorrect:
Option C defines an inelastic collision. Option B violates the fundamental First Law of Thermodynamics (total energy is always conserved universally). Option D violates basic momentum conservation.
At 90 degrees, the projectile shoots straight up and lands at the start (Range = 0). 30 and 60 degrees provide equal ranges, but they are smaller than the peak range at 45 degrees.
Newton's Third Law governs action and reaction force pairs.
Solution:
Forces can mathematically cancel each other out ONLY if they are applied to the exact same object.
Newton's Third Law pairs (Action/Reaction) ALWAYS strictly apply to two different interacting bodies (A pushes B, B pushes A).
Because they act on different bodies, they dictate the respective motion of those different bodies independently. They can never cancel each other. Therefore, Option D is a false statement.
Why other options are incorrect:
Options A, B, and C are entirely true properties of Newton's Third Law.
Two busses moving at 100 km/h and 80 km/h respectively cross each other while moving in opposite direction. Velocity of one bus relative to other bus is:
A
100 km/h
B
20 km/h
C
80 km/h
D
180 km/h
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
When two objects move in directly opposing directions, the speed at which the gap between them closes is the arithmetic sum of their respective speeds.
Formula:
$$ \vec{v}_{rel} = \vec{v}_1 - \vec{v}_2 $$
Solution:
Assign a directional axis. Let Bus 1 move forward: \( \vec{v}_1 = +100 \text{ km/h} \).
The explosion of explosive material is application of:
A
Law of conservation of energy
B
Law of conservation of mass
C
Law of conservation of momentum
D
Newton's third law of motion
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
An explosion is a dramatic event driven entirely by internal forces within a closed system.
Solution:
Because no external forces act on the bomb at the exact instant of explosion, the total vector momentum of the system must remain constant.
A stationary bomb (zero momentum) will explode into fragments that fly off in all directions. The vector sum of all fragment momentums mathematically cancels out to exactly zero.
This makes explosions a classic demonstration of the Law of Conservation of Linear Momentum.
Why other options are incorrect:
Kinetic energy massively increases during an explosion (converting chemical potential to kinetic), making "energy conservation" a tricky descriptor for kinematic modeling here. Third law applies to specific pair fragments, but momentum conservation governs the entire system.
If two objects of equal masses 'm' are moving towards each other with the same speeds 'v' then what will be the total final momentum after elastic head-on collision?
A
\( -mv \text{ kg/s} \)
B
\( 2 mv \text{ kg/s} \)
C
\( mv \text{ kg m/s} \)
D
\( 0 \text{ kg m/s} \)
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
Conservation of vector momentum.
Solution:
Total initial \( P = mv - mv = 0 \). Thus final is 0.
The range of the projectile depends upon the velocity of the projection and the angle of the projection i.e \( 45^\circ \). For a fixed velocity, when the angle of projection is larger than \( 45^\circ \). Which of the following is correct?
A
Both the height and the range attained by the projectile will be less
B
The height attained by the projectile will be less but the range is more
C
Both the height and the range attained by the projectile will be more
D
The height attained by the projectile will be more but the range is less
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
Steeper angles increase vertical velocity but decrease efficiency of horizontal range past \( 45^\circ \).
Solution:
Height increases, Range decreases.
Why other options are incorrect:
Trigonometric fact of \( \sin^2\theta \) vs \( \sin(2\theta) \).
A cyclist is traveling at \( 15 \text{ ms}^{-1} \) she applies brakes so that she doesn't collide with the wall in front of her distance of 18m. Calculate the magnitude of deceleration.
A bullet of mass m moving with a velocity v is fired into large wooden block of mass M if the bullet remains embedded in the wooden block, the velocity of the system will be:
A projectile is launched at \( 45^\circ \) to the horizontal with initial K, Energy, E. Assuming air resistance to be negligible, what will be the kinetic energy of the projectile when it reaches its highest point?
A
\( 0.71 E \)
B
\( 0.50 E \)
C
\( 0.87 E \)
D
\( E \)
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
At max height, only horizontal velocity \( v_x = v_i \cos\theta \) exists.
Solution:
$$ K.E_{\text{top}} = \frac{1}{2}m(v_i \cos 45^\circ)^2 = E \cos^2(45^\circ) $$
$$ \cos^2(45^\circ) = 0.5 $$
So, \( K.E = 0.50 E \).
Why other options are incorrect:
0.71 E comes from forgetting to square the cosine.
One ball is thrown vertically upward with a velocity of 98 m/s. If the takes 10 seconds to reach the highest point, then the acceleration of the ball is
A
\( 9.8 \text{ ms}^{-2} \)
B
\( 980 \text{ ms}^{-2} \)
C
\( 98 \text{ ms}^{-2} \)
D
\( -9.8 \text{ ms}^{-2} \)
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
Acceleration opposes upward velocity.
Solution:
$$ a = \frac{0 - 98}{10} = -9.8 \text{ m/s}^2 $$
Why other options are incorrect:
Missing the negative sign ignores the directional reality of gravity.
If a force of 12N is applied on a body and its momentum is changed form \( 60 \text{ kgms}^{-1} \) to \( 36 \text{ kgms}^{-1} \), then find the time during which this force acts:
A
1 second
B
12 seconds
C
2 seconds
D
24 seconds
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
Force is the rate of change of momentum.
Formula:
$$ F = \frac{\Delta P}{\Delta t} $$
Solution:
\( \Delta P = 36 - 60 = -24 \text{ Ns} \).
The retarding force is \( F = -12 \text{ N} \).
\( \Delta t = \frac{-24}{-12} = 2 \text{ s} \).
Why other options are incorrect:
Arithmetic mistakes in subtracting momentum or dividing.
A body is moving with an initial velocity of \( 2 \text{ kms}^{-1} \). After a time of 50 secs its velocity becomes \( 1.5 \text{ kms}^{-1} \). Its acceleration will be:
A
\( 30 \text{ m s}^{-2} \)
B
\( 20 \text{ m s}^{-2} \)
C
\( 40 \text{ m s}^{-2} \)
D
\( 10 \text{ m s}^{-2} \)
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
Calculate the rate of change of velocity using proper SI units.
In elastic collision, when a massive body collides with light body at conditions \( m_1 \gg m_2 \) and \( v_2 = 0 \text{ ms}^{-1} \), then the change in velocity will be written as:
A
\( v_1' \approx -v_1 ; v_2' \approx v_1 \)
B
\( v_1' \approx v_1 ; v_2' \approx 2v_1 \)
C
\( v_1' \approx v_1 ; v_2' \approx 0 \)
D
\( v_1' \approx -v_1 ; v_2' \approx 0 \)
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
In elastic collisions where a massive object hits a stationary tiny object, the massive object is unaffected, and the tiny object flies off at twice the speed.
Solution:
Because \( m_1 \gg m_2 \), \( v_1' \approx v_1 \) and \( v_2' \approx 2v_1 \).
Why other options are incorrect:
Option A is if the light body hit a stationary massive wall.
A ball is just allowed to fall from the window of a moving train, it will hit the ground following:
A
Circular path
B
Hyperbolic
C
Straight line path
D
Parabolic path
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
When an object is dropped from a moving frame of reference, it retains the horizontal velocity of that frame due to inertia, while falling under gravity.
Solution:
The combination of constant horizontal velocity and constant vertical acceleration produces a parabolic trajectory.
Why other options are incorrect:
A straight line only occurs if there is no horizontal velocity. Circular and hyperbolic paths do not match projectile kinematics.
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