A metre scale is moving with uniform velocity. This implies (a) the force acting on the scale is zero, but a t

NCERT Exemplar · Q2

Q. A metre scale is moving with uniform velocity. This implies

(a) the force acting on the scale is zero, but a torque about the centre of mass can act on the scale.

(b) the force acting on the scale is zero and the torque acting about centre of mass of the scale is also zero.

(d) neither the force nor the torque need to be zero.

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✓ Free question

Concept understanding — Newton Second Law

Newton's Second Law: The Law That Connects Force and Motion

Imagine you're pushing a shopping cart. If you push gently, it moves slowly. Push harder, and it speeds up faster. Now imagine the cart is full of groceries — even with the same push, it accelerates much more slowly than an empty cart. This everyday experience is exactly what Newton's Second Law captures.

The Intuition First

Two things matter when you push something:

How hard you push — the force you apply.
How heavy the object is — its mass.

The harder you push, the more the object speeds up. The heavier the object, the less it speeds up for the same push. So acceleration depends on both force and mass — and in opposite ways.

Note

"Acceleration" here means any change in velocity — speeding up, slowing down, or changing direction. It's not just "going faster."

The Precise Statement

Newton's Second Law says:

The acceleration of an object is directly proportional to the net force acting on it, and inversely proportional to its mass. The acceleration is in the same direction as the net force.

In one equation:

\vec{a} = \frac{\vec{F}_{\text{net}}}{m}

Or more commonly:

\vec{F}_{\text{net}} = m \vec{a}

Where:

$\vec{F}_{\text{net}}$ is the net force (the vector sum of all forces acting on the object) — measured in newtons (N)
$m$ is the mass of the object — measured in kilograms (kg)
$\vec{a}$ is the acceleration — measured in metres per second squared ( $\text{m/s}^2$ )

\vec{F}_{\text{net}} = m \vec{a}

What This Really Means

Force causes acceleration, not velocity. A constant net force produces constant acceleration — meaning the velocity keeps changing at a steady rate. If you stop pushing, the net force becomes zero, and acceleration becomes zero (the object continues at constant velocity — that's Newton's First Law).

Mass is a measure of inertia. The more mass an object has, the harder it is to change its motion. A truck needs a much larger force than a bicycle to achieve the same acceleration.

Direction matters. Force and acceleration are vectors — they point the same way. If you push north, the acceleration is north. If multiple forces act, you must add them as vectors to find the net force.

A Simple Example

A 2 kg block is pushed with a net force of 10 N to the right.

a = \frac{F_{\text{net}}}{m} = \frac{10\ \text{N}}{2\ \text{kg}} = 5\ \text{m/s}^2

The block accelerates at $5\ \text{m/s}^2$ to the right. Every second, its velocity increases by 5 m/s in that direction.

Watch out

A common mistake: thinking that a constant force means constant velocity. It doesn't — constant force means constant acceleration, so velocity keeps changing. Only when net force is zero does velocity stay constant.

Why This Law Is So Powerful

Newton's Second Law is the bridge between forces (the causes) and motion (the effects). It lets you:

Predict how an object will move if you know the forces on it
Calculate the force needed to produce a desired motion
Understand why heavier things are harder to accelerate

It applies everywhere — from a ball you throw to a rocket launching into space. The same law governs them all.

For quick revision, remember that Newton Second Law is drawn directly from the Laws of Motion coverage of the NCERT/CBSE Class 11 Physics syllabus and recurs often in JEE Main and NEET papers, which is exactly why "Newton Second Law important questions" shows up so often in Physics question banks. The clearest way to build exam confidence here is to combine this explanation with the NCERT Physics textbook's own solved examples and chapter-end questions.

A metre scale moving with uniform velocity implies that both its linear and angular accelerations are zero. Consequently, the net force and the net torque acting on it must both be zero. The correct option is (B).

When a rigid body is described as "moving with uniform velocity," it means that its entire state of motion is constant. This includes both its translational motion (the motion of its center of mass) and its rotational motion (its rotation about its center of mass). Let's break this down using Newton's laws.

Understanding "Uniform Velocity" for Translational Motion: "Uniform velocity" means that the velocity vector $\vec{v}$ of the centre of mass of the scale is constant in both magnitude and direction. If the velocity is constant, then the linear acceleration $\vec{a}$ of the centre of mass must be zero:

\vec{a} = \frac{d\vec{v}}{dt} = 0

Relating Zero Acceleration to Net Force: According to Newton's Second Law of Motion for translational motion, the net external force ( $\Sigma \vec{F}$ ) acting on an object is equal to the product of its mass ( $m$ ) and its acceleration ( $\vec{a}$ ):

\Sigma \vec{F} = m\vec{a}

Since we established that $\vec{a} = 0$ for uniform velocity, it follows that:

\Sigma \vec{F} = m(0) = 0

Therefore, the net force acting on the scale is zero.

3. Understanding "Uniform Velocity" for Rotational Motion:

For a rigid body, "uniform velocity" also implies that its rotational state is constant. This means its angular velocity vector $\vec{\omega}$ (if it's rotating) is constant in both magnitude and direction. If the angular velocity is constant, then the angular acceleration $\vec{\alpha}$ must be zero:

\vec{\alpha} = \frac{d\vec{\omega}}{dt} = 0

If the scale were undergoing angular acceleration, its rotational motion would not be uniform, and the simple phrase "moving with uniform velocity" would be insufficient to describe its state.

4. Relating Zero Angular Acceleration to Net Torque:

According to Newton's Second Law of Motion for rotational motion, the net external torque ( $\Sigma \vec{\tau}$ ) acting on an object about its centre of mass is equal to the product of its moment of inertia ( $I$ ) about that axis and its angular acceleration ( $\vec{\alpha}$ ):

\Sigma \vec{\tau} = I\vec{\alpha}

Since we established that $\vec{\alpha} = 0$ for uniform velocity, it follows that:

\Sigma \vec{\tau} = I(0) = 0

Therefore, the net torque acting on the scale about its centre of mass is also zero.

> [!IMPORTANT]
> For a rigid body, "uniform velocity" implies both constant linear velocity (zero linear acceleration) and constant angular velocity (zero angular acceleration). This is a state of dynamic equilibrium.

Combining these two conclusions, both the net force and the net torque acting on the metre scale must be zero.

Let's evaluate the given options:

(A) the force acting on the scale is zero, but a torque about the centre of mass can act on the scale. (Incorrect, torque must also be zero)
(B) the force acting on the scale is zero and the torque acting about centre of mass of the scale is also zero. (Correct)
(C) the total force acting on it need not be zero but the torque on it is zero. (Incorrect, force must be zero)
(D) neither the force nor the torque need to be zero. (Incorrect, both must be zero)

✓Final answer

The correct option is (B), because uniform velocity implies both zero net force and zero net torque.

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
$8~\mathrm{N}$
›Reveal solutionSolution

The body's mass is fixed; find it from the first case, then apply $F=ma$ to the second. Answer: $8~\text{N}$ (C).

Step 1 — Find the mass from the first case.

Acceleration in the first case:
$a_1 = \frac{16-8}{4} = 2~\text{ms}^{-2}$
Using $F = ma$ :
$m = \frac{F}{a_1} = \frac{8}{2} = 4~\text{kg}$
Step 2 — Acceleration in the second case.
$a_2 = \frac{20-16}{2} = 2~\text{ms}^{-2}$
Step 3 — Required force.

The mass is the same body, so
$F_2 = m\,a_2 = 4 \times 2 = 8~\text{N}$
✓Final answer
$8~\text{N}$
TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
$Q. The force (F) acting on a particle in terms of its distance$
(x) from a fixed point is given by $F = \dfrac{A}{B + x^{1.5}}$ . If the dimensional formula of AB is $\left[ M^a L^b T^c \right]$ , then the value of $a + b + c$ is (A) 2 (B) 3 (C) 4 (D) 5
›Reveal solutionSolution
The key idea is that the denominator $B + x^{1.5}$ must be dimensionally consistent, so $B$ has the same dimensions as $x^{1.5}$ . Using the given force equation, we find the dimensions of $A$ and $B$ , then of $AB$ , and sum the exponents to get $a+b+c = 4$ .

We are given $F = \dfrac{A}{B + x^{1.5}}$ , where $F$ is force, $x$ is distance, and $A$ and $B$ are constants. The dimensional formula of $AB$ is $[M^a L^b T^c]$ , and we need $a+b+c$ .

Concept and intuition:

The fundamental rule in dimensional analysis is that you can only add or subtract quantities that have the same dimensions. Here, $B$ and $x^{1.5}$ are added, so they must share the same dimensional formula. That lets us find the dimensions of $B$ directly from $x$ . Then, since the whole fraction equals force, we can solve for the dimensions of $A$ . Multiplying the dimensions of $A$ and $B$ gives the dimensions of $AB$ , and we simply add the exponents.

Step-by-step reasoning:

Identify dimensions of known quantities.

Force $F$ has dimensions $[M L T^{-2}]$ .

Distance $x$ has dimensions $[L]$ .

Use the addition rule for $B + x^{1.5}$ .

Since $B$ and $x^{1.5}$ are added, they must have the same dimensions.

$x^{1.5}$ has dimensions $[L]^{1.5} = [L^{3/2}]$ .

Therefore, $B$ also has dimensions $[L^{3/2}]$ .

Find dimensions of $A$ from the force equation.

The equation $F = \dfrac{A}{B + x^{1.5}}$ implies that the denominator has the same dimensions as $A$ divided by $F$ .

So, $[A] = [F] \times [B + x^{1.5}]$ .

Since $B + x^{1.5}$ has dimensions $[L^{3/2}]$ , we have:

$[A] = [M L T^{-2}] \times [L^{3/2}] = [M L^{1 + 3/2} T^{-2}] = [M L^{5/2} T^{-2}]$ .

Determine dimensions of $AB$ .

$[AB] = [A] \times [B] = [M L^{5/2} T^{-2}] \times [L^{3/2}] = [M L^{(5/2 + 3/2)} T^{-2}] = [M L^{4} T^{-2}]$ .

So $a = 1$ , $b = 4$ , $c = -2$ .

Compute $a + b + c$ .

$a + b + c = 1 + 4 + (-2) = 3$ .

Watch out

A common mistake is to forget that $x^{1.5}$ has dimensions $[L^{3/2}]$ , not $[L^{1.5}]$ as a number — but dimensionally, it's the same: $[L^{3/2}]$ . Also, do not treat $B$ as dimensionless; the addition forces it to have the same dimensions as $x^{1.5}$ .

Tip

You can shortcut: since $B$ has dimensions of $x^{1.5}$ , and $A$ has dimensions of $F \times x^{1.5}$ , then $AB$ has dimensions of $F \times (x^{1.5})^2 = F \times x^3$ . Force has $[M L T^{-2}]$ , $x^3$ has $[L^3]$ , product is $[M L^4 T^{-2}]$ , giving $a+b+c = 1+4-2 = 3$ .

✓Final answer
$The correct option is (B) . ANSWER: B$
TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
$3\,\text{kg}$
›Reveal solutionSolution
The block moves at constant velocity down the incline, so friction exactly balances the component of gravity along the plane. Pulling it up the same distance requires work against both gravity and friction — the total work equals twice the gravitational potential energy gained, which is $30\,\text{J}$ .

The key insight is that "constant velocity" means no net force along the incline. For a block sliding down a rough incline at constant speed, the frictional force must exactly oppose and equal the component of gravity pulling it down. That tells us the magnitude of friction — and when we pull the block up, friction still acts opposite to motion (now down the plane), so the external force has to overcome both gravity and friction.

Let’s work through it step by step.

Set up the forces for downward motion. The block of mass $m = 3\,\text{kg}$ moves down with constant velocity, so the net force along the incline is zero. Let $\theta$ be the angle of the incline. The component of weight down the plane is $mg\sin\theta$ . Friction $f$ acts up the plane (opposing motion). Hence:

$f = mg\sin\theta.$

Work done by gravity when the block descends a height $h$ .

The block moves down along the incline through a vertical height $h = 50\,\text{cm} = 0.5\,\text{m}$ . The distance along the incline $s$ is related by $h = s\sin\theta$ , so $s = \frac{h}{\sin\theta}$ .

Gravity does work $mgh$ (positive because displacement is downward). But we don’t need that directly — we need the work to pull it up.

Work required to pull the block up the same incline through the same height.

When pulling upward at constant speed (again constant velocity, so no acceleration), the external force $F$ must balance both the component of gravity down the plane ( $mg\sin\theta$ ) and friction ( $f$ ), which now also acts down the plane (opposing upward motion). So:

$F = mg\sin\theta + f = mg\sin\theta + mg\sin\theta = 2mg\sin\theta.$

Work done by the external force. The displacement along the incline is $s = \frac{h}{\sin\theta}$ . Therefore:

$W = F \cdot s = (2mg\sin\theta) \cdot \frac{h}{\sin\theta} = 2mgh.$
Notice the $\sin\theta$ cancels — the result is independent of the incline angle.

Plug in the numbers. $m = 3\,\text{kg}$ , $g = 10\,\text{m/s}^2$ , $h = 0.5\,\text{m}$ :

$W = 2 \times 3 \times 10 \times 0.5 = 30\,\text{J}.$
Watch out

A common mistake is to think the work done is just $mgh = 15\,\text{J}$ , forgetting that friction also has to be overcome on the way up. Because friction equals $mg\sin\theta$ (from the constant-speed descent), the total force needed doubles, and so does the work.

Tip

The angle $\theta$ cancels out completely — you never need to find it. The work to pull the block up at constant speed on a rough incline (where it slides down at constant speed) is always $2mgh$ , regardless of the slope.

✓Final answer
$\boxed{30\,\text{J}}$
TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
$6\,\text{kg}$
›Reveal solutionSolution

Find $a = F/m = 2\,\text{ms}^{-2}$ , then apply $v^2 = u^2 + 2as$ to get $s = 5$ m.

Concept. Newton's second law gives the (uniform) acceleration produced by a constant force; the kinematic relation $v^2 = u^2 + 2as$ then connects the change in speed to the displacement, with no need to know the time.

Step 1 — acceleration.
$a = \frac{F}{m} = \frac{12\,\text{N}}{6\,\text{kg}} = 2\,\text{ms}^{-2}.$
Step 2 — displacement from the work–energy style kinematic equation.
$v^2 = u^2 + 2as \implies s = \frac{v^2 - u^2}{2a} = \frac{6^2 - 4^2}{2 \times 2} = \frac{36 - 16}{4} = 5\,\text{m}.$
Check (work–energy theorem). $\Delta KE = \tfrac{1}{2}(6)(36-16) = 60$ J; $W = Fs = 12 \times 5 = 60$ J. Consistent.

✓Final answer
$5\,\text{m}$

ANSWER: B
TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
$2~\mathrm{ms}^{-1}$
›Reveal solutionSolution
The magnetic force on a moving charge is given by $\vec{F} = q(\vec{v} \times \vec{B})$ , which is always perpendicular to both velocity and field. Here, the cross product yields a vector in the y–z plane, so the force lies in that plane — option (A).

The key concept is the Lorentz force law for a charge moving in a magnetic field:
$\vec{F} = q (\vec{v} \times \vec{B})$
The force is perpendicular to both the velocity and the magnetic field. That means the direction of $\vec{F}$ is given by the cross product, and its magnitude depends on the sine of the angle between $\vec{v}$ and $\vec{B}$ .

Here, the velocity is purely along the x-axis, while the magnetic field has components in all three directions. The cross product will tell us exactly which plane the force lies in.

Write down the vectors

Velocity: $\vec{v} = 2\hat{i} \ \mathrm{m/s}$

Magnetic field: $\vec{B} = 2\hat{i} + 2\hat{j} + 3\hat{k} \ \mathrm{T}$

Compute the cross product $\vec{v} \times \vec{B}$

$\vec{v} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & 0 \\ 2 & 2 & 3 \end{vmatrix}$
Expand:
$= \hat{i}(0 \cdot 3 - 0 \cdot 2) - \hat{j}(2 \cdot 3 - 0 \cdot 2) + \hat{k}(2 \cdot 2 - 0 \cdot 2)$ $= \hat{i}(0) - \hat{j}(6 - 0) + \hat{k}(4 - 0)$ $= -6\hat{j} + 4\hat{k}$

Interpret the result

The cross product has no $\hat{i}$ component — it lies entirely in the y–z plane. Since $\vec{F} = q(\vec{v} \times \vec{B})$ , the force vector is parallel to this cross product (scaled by $q$ ). So the force is also in the y–z plane.

Check the options

(A) In y–z plane → True

(B) Along –y axis → No, it has both –y and +z components

(C) Along +z axis → No, it has a –y component too

(D) Along –z axis → No, it has +z component

Watch out

A common mistake is to think the force must be along one axis. But unless the cross product simplifies to a single component, the force lies in a plane. Here, both y and z components are nonzero, so the force is not purely along any axis.

Tip

Notice that the x-component of $\vec{B}$ is parallel to $\vec{v}$ , so it contributes nothing to the cross product. Only the perpendicular parts of $\vec{B}$ (the y and z components) matter — and they produce a force in the y–z plane.

✓Final answer
$The correct option is (A) . ANSWER: A$
TG EAPCET 2023Set ap-2023-05-10-AN1 markMCQ
$2\,\text{kg}$
›Reveal solutionSolution
The block is at rest because static friction is strong enough to hold it. The frictional force equals the downhill component of weight, not the maximum possible friction. The answer is $10\,\text{N}$ .

The key idea here is that static friction is a self-adjusting force. It does not always act at its maximum value — it simply matches whatever force tries to slide the object, up to a limit. So you must first check whether the block would slide if friction were absent, then see if the available maximum static friction can prevent that sliding.

Let’s work through it.

Find the component of weight pulling the block down the incline. The weight is $mg = 2 \times 10 = 20\,\text{N}$ . The component parallel to the incline is

$mg\sin\theta = 20 \times \sin 30^\circ = 20 \times \frac12 = 10\,\text{N}.$
This is the force trying to slide the block downhill.

Find the maximum static friction available. The normal reaction on the incline is

$N = mg\cos\theta = 20 \times \cos 30^\circ = 20 \times \frac{\sqrt3}{2} = 10\sqrt3\,\text{N}.$
The maximum static friction is
$f_{\text{max}} = \mu_s N = 0.7 \times 10\sqrt3 = 7\sqrt3 \,\text{N}.$
Numerically, $7\sqrt3 \approx 12.12\,\text{N}$ .

Compare the downhill force with the maximum friction.

The downhill force is $10\,\text{N}$ , which is less than $7\sqrt3 \approx 12.12\,\text{N}$ .

So the block does not move. Static friction does not need to act at its maximum — it only needs to supply enough force to cancel the downhill pull.

Therefore, the actual frictional force equals the downhill component.

$f = mg\sin\theta = 10\,\text{N}.$
It acts up the incline, opposing the tendency to slide down.

Watch out

A common mistake is to calculate $f = \mu_s N = 7\sqrt3\,\text{N}$ and pick option (C). That would be the friction if the block were on the verge of sliding, but here the downhill force is smaller, so friction is less than the maximum.

Tip

Always ask: “Is the object actually moving or about to move?” If not, static friction is just the balancing force — never assume it’s at its limit.

✓Final answer
$\boxed{10\,\text{N}}$
TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
$3.3 \times 10^{-8}$
›Reveal solutionSolution
The force on a partially reflecting surface comes from both the absorbed and reflected parts of the light. Using momentum transfer per photon, the net force is $4.33 \times 10^{-8}$ N, matching option (B).

The key idea is that light carries momentum, and when it hits a surface, the change in momentum per second gives the force. For a perfectly absorbing surface, the force is $P/c$ ; for a perfectly reflecting surface, it's $2P/c$ . Here, the surface does both — 70% absorption and 30% reflection — so we need to combine the contributions.

Think of it this way: each photon that gets absorbed transfers its full momentum to the surface. Each photon that gets reflected reverses its momentum, so the surface gets twice the momentum kick. The total force is just the sum of these two effects, weighted by the fractions of power absorbed and reflected.

Find the incident power and momentum flux. The incident beam carries power $P = 10$ W. The momentum per second (momentum flux) carried by the light is $P/c$ , since each photon of energy $E$ has momentum $E/c$ . Here $c = 3 \times 10^8$ m/s. So the incident momentum per second is

$\frac{P}{c} = \frac{10}{3 \times 10^8} = 3.33 \times 10^{-8} \text{ N}.$

Contribution from the absorbed part (70%). For absorption, the surface stops the light completely. The change in momentum per second for the absorbed portion is equal to the momentum it carried, since final momentum is zero. Force from absorption = (fraction absorbed) × (incident momentum per second)

$F_{\text{abs}} = 0.70 \times \frac{P}{c} = 0.70 \times 3.33 \times 10^{-8} = 2.33 \times 10^{-8} \text{ N}.$

Contribution from the reflected part (30%). For reflection, the light bounces back with equal speed but opposite direction. The change in momentum is twice the incident momentum (from $+p$ to $-p$ , so change = $2p$ ). Force from reflection = (fraction reflected) × $2 \times$ (incident momentum per second)

$F_{\text{ref}} = 0.30 \times 2 \times \frac{P}{c} = 0.60 \times 3.33 \times 10^{-8} = 2.00 \times 10^{-8} \text{ N}.$

Add the forces. Both forces act in the same direction (pushing the surface forward), so they add:

$F_{\text{total}} = F_{\text{abs}} + F_{\text{ref}} = (2.33 + 2.00) \times 10^{-8} = 4.33 \times 10^{-8} \text{ N}.$
Watch out

A common mistake is to treat the entire beam as either fully absorbed or fully reflected. Here, you must split the power into two parts and handle each with the correct momentum change factor (1 for absorption, 2 for reflection).

Tip

A quick formula for a surface that absorbs fraction $a$ and reflects fraction $r$ (with $a + r = 1$ ) is:

$F = \frac{P}{c} (a + 2r) = \frac{P}{c} (1 + r).$

Since $a = 1 - r$ , this simplifies to $F = \frac{P}{c}(1 + r)$ . Here $r = 0.30$ , so $F = \frac{10}{3\times10^8} \times 1.30 = 4.33 \times 10^{-8}$ N.

✓Final answer
$4.33 \times 10^{-8}$
TG EAPCET 2022Set eng-2022-07-19-AN1 markMCQ
$\vec{F}$
›Reveal solutionSolution
The key is to first find the mass of block A from the given force and acceleration, then treat both blocks as a system under the new force to find the common acceleration, and finally apply Newton’s second law to block B alone to get the force on it. The force on block B is $10~\mathrm{N}$ .

Concept and intuition

We have two separate situations. In the first, a known force on block A alone gives us its acceleration — so we can find the mass of A. In the second, the same block A is pushed against block B, and a different force is applied to the combination. Since friction is neglected, the two blocks move together as one system. The force that block B experiences is simply the net force needed to accelerate block B at the system’s common acceleration. This is a classic “two-block pushed together” problem: find the system acceleration, then isolate one block.

Find the mass of block A From the first scenario:

$F = m_A a \quad\Rightarrow\quad 10 = m_A \times 20$
So
$m_A = \frac{10}{20} = 0.5~\mathrm{kg}$

Treat A and B as a single system under the new force The total mass is

$m_{\text{total}} = m_A + m_B = 0.5 + 1.5 = 2.0~\mathrm{kg}$
The applied force is $F' = 20~\mathrm{N}$ . The common acceleration of the system (no friction) is
$a = \frac{F'}{m_{\text{total}}} = \frac{20}{2.0} = 10~\mathrm{m/s^2}$

Find the force on block B Block B is accelerated only by the contact force from block A (call it $F_{AB}$ ). Applying Newton’s second law to block B alone:

$F_{AB} = m_B \cdot a = 1.5 \times 10 = 15~\mathrm{N}$
This is the force that block A exerts on block B — which is the force “on block B” asked in the problem.

Watch out

A common mistake is to think the force on B equals the applied force $F'$ (20 N) or to use the first acceleration (20 m/s²) in the second part. Both are wrong because the mass and the applied force have changed.

Tip

Notice that the force on B (15 N) is less than the applied force (20 N) because part of the applied force goes into accelerating block A itself. The ratio of forces on the blocks is exactly the ratio of their masses.

✓Final answer
$The correct option is (A) . ANSWER: A$
TG EAPCET 2021Set ap-2021-08-09-AN1 markMCQ
$S = 2500, \; P = 6.250$
›Reveal solutionSolution

Force = rate of momentum transfer gives $S = 500\ \text{m/s}$ ; the power delivered to the bullets is $KE_{\text{bullet}} \times \text{firing rate} = 3.125\ \text{kW}$ . Answer (C).

Given: firing rate $n = 5\ \text{s}^{-1}$ , bullet mass $m = 5\ \text{g} = 0.005\ \text{kg}$ , holding force $F = 12.5\ \text{N}$ .

Speed. The average force needed to hold the gun equals the momentum delivered to the bullets each second:
$F = n\,m\,S \implies 12.5 = 5 \times 0.005 \times S = 0.025\,S \implies S = \frac{12.5}{0.025} = 500\ \text{m/s}.$
Power. Kinetic energy of one bullet:
$KE = \tfrac{1}{2} m S^2 = \tfrac{1}{2}(0.005)(500)^2 = 625\ \text{J}.$
Energy is imparted to bullets at the firing rate, so the power delivered is
$P = KE \times n = 625 \times 5 = 3125\ \text{W} = 3.125\ \text{kW}.$
Thus $S = 500\ \text{m/s}$ and $P = 3.125\ \text{kW}$ .

✓Final answer
$S = 500\ \text{m/s}$
TG EAPCET 2021Set ap-2021-08-09-AN1 markMCQ
$x = 0$
›Reveal solutionSolution
Work done by a variable force is the area under the force–displacement graph. Summing the triangle and rectangle gives $20\ \mathrm{J}$ , option (D).

The work done by a variable force equals the area under the $F$ -vs- $x$ curve: when the force is not constant you integrate $F\,dx$ , which geometrically is the area between the curve and the $x$ -axis. The graph here is piecewise linear, so we split it into simple shapes.

From $x=0$ to $2$ m the force rises linearly from $0$ to $5$ N — a triangle:

$W_1 = \tfrac12 \times \text{base} \times \text{height} = \tfrac12 \times 2 \times 5 = 5\ \mathrm{J}.$

From $x=2$ to $5$ m the force is constant at $5$ N — a rectangle:

$W_2 = \text{base} \times \text{height} = 3 \times 5 = 15\ \mathrm{J}.$

Total work:

$W_\text{total} = W_1 + W_2 = 5 + 15 = 20\ \mathrm{J}.$
Tip

Always read the force value on the flat part of the graph carefully — the area (and hence the work) scales directly with it.

✓Final answer
$The correct option is (D) . ANSWER: D$
TG EAPCET 2021Set ap-2021-08-09-FN1 markMCQ
$15\,\mathrm{kg}$
›Reveal solutionSolution
The key idea is to use the impulse–momentum theorem: the net impulse equals the change in momentum. The force is constant, so $F \Delta t = m(v_f - v_i)$ . With $v_i = +15\ \mathrm{m/s}$ , $v_f = -5\ \mathrm{m/s}$ , $m=15\ \mathrm{kg}$ , and $\Delta t = 5\ \mathrm{s}$ , we get $F = -60\ \mathrm{N}$ , which corresponds to option (D).

Concept and intuition:

When a constant force acts over a time interval, it changes the object’s momentum. The impulse–momentum theorem says:
$\text{Impulse} = \text{Force} \times \text{time} = \text{change in momentum}$
Here the object reverses direction, so the change in velocity is large — and the force must be opposite to the original motion. We don’t need to know acceleration or distance; just the start and end velocities, the mass, and the time.

Step-by-step reasoning:

Choose a sign convention.

Let the initial direction of motion be positive.

So initial velocity: $v_i = +15\ \mathrm{m/s}$ .

The final velocity is $5\ \mathrm{m/s}$ in the opposite direction, so $v_f = -5\ \mathrm{m/s}$ .

Write the impulse–momentum equation.

For a constant force $F$ acting for time $\Delta t$ :

$F \Delta t = m v_f - m v_i = m (v_f - v_i)$

Substitute the known values. $m = 15\ \mathrm{kg}$ , $\Delta t = 5\ \mathrm{s}$ , $v_i = +15$ , $v_f = -5$ :

$F \times 5 = 15 \times [(-5) - (+15)]$

Simplify the velocity change.

$v_f - v_i = -5 - 15 = -20\ \mathrm{m/s}$
So:
$5F = 15 \times (-20) = -300$

Solve for $F$ .

$F = \frac{-300}{5} = -60\ \mathrm{N}$
The negative sign means the force acts opposite to the initial direction.

Watch out

A common mistake is to forget the sign of the final velocity. If you treat both speeds as positive, you get $F = 30\ \mathrm{N}$ , which isn’t even an option. Always set a direction as positive and stick to it.

Tip

Notice that the magnitude of the force is $60\ \mathrm{N}$ , but the direction matters. The problem asks for the force acting on the object, so the sign is part of the answer.

✓Final answer
$The correct option is (D) . ANSWER: D$

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