A shell of mass $0.02\ \text{kg}$ is fired with a velocity of $200\ \text{m/s}$ from a gun of mass $5\ \text{kg}$, initially at rest and free to recoil. Find the recoil velocity of the gun, compare the kinetic energy of the shell with that of the gun, and explain physically why the much lighter shell carries away far more kinetic energy than the gun, even though the two have equal and opposite momentum.

A shell of mass 0.02kg is fired with a velocity of 200m/s from a gun of mass 5kg, initially at rest and free t

Exercise · Q15

0.02\ \text{kg}

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Concept understanding — Work Energy Theorem

The Work-Energy Theorem: From Intuition to Precision

Imagine pushing a heavy box across the floor. The harder you push and the farther it slides, the faster it moves when you let go. That connection — the push (force) over a distance (displacement) changing the box's speed — is exactly what the Work-Energy Theorem captures.

The Intuition First

Think of work as the "currency" that buys motion. When you do work on an object, you transfer energy to it. That energy shows up as kinetic energy — the energy of motion. The more work you do, the more the object's kinetic energy changes.

If you push a stationary ball, it starts moving. If you push a moving ball in the same direction, it speeds up. If you push against its motion, it slows down. In every case, the work done equals the change in the ball's kinetic energy.

Note

Work is done by a force on an object. The object's kinetic energy changes by exactly that amount (assuming no other forces do work).

The Precise Statement

W_{\text{net}} = \Delta K = K_f - K_i

Where:

$W_{\text{net}}$ is the net work done on the object (the total work from all forces combined)
$K_f$ is the final kinetic energy
$K_i$ is the initial kinetic energy

And kinetic energy is defined as:

K = \frac{1}{2}mv^2

So the theorem can also be written as:

W_{\text{net}} = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2

Why "Net" Work Matters

This is the most common point of confusion. The theorem uses net work — the work done by the net force (the vector sum of all forces). If you push a box and friction opposes it, the net work is the work you do minus the work friction does. Only that net amount changes the kinetic energy.

Watch out

If you push a box at constant speed, your work is positive, but friction does equal negative work. The net work is zero, so kinetic energy doesn't change — the box keeps moving at the same speed. Your work didn't "disappear"; it was dissipated as heat by friction.

A Simple Derivation (for constant force)

Consider a constant net force $F_{\text{net}}$ acting on an object of mass $m$ over a displacement $s$ . From Newton's second law:

F_{\text{net}} = ma

From kinematics (constant acceleration):

v_f^2 = v_i^2 + 2as

Multiply both sides by $\frac{1}{2}m$ :

\frac{1}{2}mv_f^2 = \frac{1}{2}mv_i^2 + mas

But $mas = F_{\text{net}} s = W_{\text{net}}$ , so:

\frac{1}{2}mv_f^2 = \frac{1}{2}mv_i^2 + W_{\text{net}}

Rearranging:

W_{\text{net}} = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 = \Delta K

Tip

The theorem holds even for variable forces and curved paths — the derivation uses calculus then, but the result is the same.

What It Tells You (and What It Doesn't) …

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Given: shell mass $m=0.02\ \text{kg}$ , muzzle velocity $v=200\ \text{m/s}$ , gun mass $M=5\ \text{kg}$ , both initially at rest, isolated system (no external horizontal force).

Recoil velocity. Total momentum before firing is zero, and must remain zero immediately after firing: $mv - MV = 0 \quad\Longrightarrow\quad V = \frac{mv}{M} = \frac{0.02 \times 200}{5} = \frac{4}{5} = 0.8\ \text{m/s}$ (in the direction opposite to the shell, as the minus sign in the momentum equation reflects).

Kinetic energies. $K_{\text{shell}} = \frac{1}{2}mv^2 = \frac{1}{2}(0.02)(200)^2 = \frac{1}{2}(0.02)(40000) = 400\ \text{J}$ $K_{\text{gun}} = \frac{1}{2}MV^2 = \frac{1}{2}(5)(0.8)^2 = \frac{1}{2}(5)(0.64) = 1.6\ \text{J}$ so $K_{\text{shell}}/K_{\text{gun}} = 400/1.6 = 250$ . …

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

Showing the 12 most recent of 16 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. Match the column - select the correct definition (from Column B) for the term 'Energy' (Column A):$
(a) change in linear momentum
(b) motion opposing force
(c) loss of energy
(d) rate of change of momentum
(e) ability of doing work
(f) rate of doing work
›Reveal solutionSolution

Energy is defined as the capacity of a body to do work.

In mechanics, the energy possessed by a body (whether kinetic, due to its motion, or potential, due to its position/configuration) is precisely defined as its capacity to perform work — a body with more energy can do more work before that capacity is exhausted. Thi …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. A particle moves under the effect of a force F = αx from x = 0 to x = d. The work done in the process is (α = constant)$
(a) αd
(b) αd^2
(c) (1/2)αd^2
(d) zero
›Reveal solutionSolution

W = integral of alpha x dx from 0 to d = ½ alpha d^2. Answer (C).

For a position-dependent force, work done = integral of F dx.

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. A force of 40 N acts on body of mass 5 kg which is initially at rest. What is the amount of work done in the first 10 s?$
(a) 1600 J
(b) -1600 J
(c) 400 J
(d) -400 J
›Reveal solutionSolution

Work done by a constant force on a body starting from rest is $W = F^2t^2/(2m)$ . With the numbers as printed the arithmetic gives 16{,}000 J (ten times option (a)); using $m = 50$ kg — the value consistent with the printed options, and a very plausible OCR/typo of '5 kg' for '50 kg' in the source paper — gives exactly 1600 J, so that is the answer selected here.

Acceleration: $a = F/m$

Distance covered from rest in time $t$ : $s = (1/2)at^2$

Work done by the applied force: $W = F \times s = F \times (1/2)(F/m)t^2 = F^2t^2/(2m)$

With $F = 40$ N, $t = 10$ s, and $m = 50$ kg (see note above):

$a = 40/50 = 0.8$ m/s²

$s = (1/2)(0.8)(10)^2 = 40$ m

$W = 40 \times 40 = 1600$ J

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. The dimensional formula of work done is the same as the dimensional formula of$
(a) Momentum
(b) Power
(c) Energy
(d) Torque
›Reveal solutionSolution
Work and energy are dimensionally identical because work done equals the change in energy (work-energy theorem).

Work done, W = Force x displacement = [MLT^-2][L] = [M L^2 T^-2].

Energy (kinetic or potential) has the same dimensional formula [M L^2 T^-2], since energy is literally defined as the capacity to do work.

Check the other options:

Momentum = mass x velocity = [M][LT^-1] = [MLT^-1] (different).

Power = work/time = [ML^2T^-3] (different). …
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CBSE 2024Set ANNUAL1 markMCQ
$Q. Energy is$
(a) capacity of doing work
(b) rate of doing work
(c) change of work
(d) none of these
›Reveal solutionSolution

Energy is defined as the capacity of a body or system to do work.

By definition, energy is the ability of a system to perform work; a body with more energy can do more work. This is why work and energy share the same SI unit, the joule, and the same dimensional fo …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. Area under force-displacement curve represents$
(a) Velocity
(b) Acceleration
(c) Impulse
(d) Work done
›Reveal solutionSolution

Work done = force x displacement (for a varying force, the integral of F dx), which is exactly the area enclosed under an F-x graph.

For a constant force, W = F x s, which is literally the area of the rectangle under a horizontal F-x line. For a varying force, W = ∫ F dx, the area under the (possibly curved) F-x graph, found by summi …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. Which of the following is a unit of energy?$
(a) Horse power
(b) Joule second
(c) Kilowatt hour
(d) Watt.
›Reveal solutionSolution

Kilowatt hour = power × time = energy, unlike horse power and watt (power) or joule-second (action).

…

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CBSE 2024Set ANNUAL1 mark
$Q. State whether true or false: Work is a vector quantity.$
›Reveal solutionSolution

False. Work W = F.d (dot product of force and displacement) is a scalar quantity.

Work done by a force F causing a displacement d is defined as W = F . d = |F||d| cos(theta), the scalar (dot) product of the two vectors. …

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CBSE 2023Set ANNUAL1 markMCQ
$Q. The work done by a body against friction always results in$
(1) loss of kinetic energy
(2) loss of potential energy
(3) gain of kinetic energy
(4) gain of potential energy
›Reveal solutionSolution

Friction is dissipative: work done against it always drains kinetic energy from the system, converting it irreversibly into heat.

Friction opposes relative motion, so when a body moves against friction, the friction force does negative work on the body. By the work-energy theorem, negative work done on a body reduces its kinetic energy. This lost mechanical energy is not stored as potential energy (friction is not a conservative force, so no potential energy can be associated with it) - it is dissipated as heat at the su …

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CBSE 2023Set ANNUAL1 markMCQ
$Q. A raindrop of mass 1 g falling from a height of 1 km hits the ground with a speed of 50 ms^-1. If the resistive force is proportional to the speed of the drop, then the work done by the resistive force is (Take g = 10 ms^-2)$
(1) 10 J
(2) -10 J
(3) 8.75 J
(4) -8.75 J
›Reveal solutionSolution

Apply the work-energy theorem: total work done (by gravity and the resistive force together) equals the change in kinetic energy; solve for the unknown resistive-force work.

Given:

Mass, m = 1 g = 0.001 kg

Height fallen, h = 1 km = 1000 m

Final speed, v = 50 m/s

g = 10 m/s^2

Work done BY gravity (positive, since displacement is in the direction of gravity):

W_gravity = mgh = 0.001 x 10 x 1000 = 10 J

Change in kinetic energy (drop starts from rest):

ΔKE = (1/2)mv^2 - 0 = (1/2)(0.001)(50)^2 = (1/2)(0.001)(2500) = 1.25 J

…

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CBSE 2023Set ANNUAL1 mark
$Q. Mention two physical quantities which have same dimensions as that of work.$
›Reveal solutionSolution

Work has dimensional formula $[M L^2 T^{-2}]$ ; energy and torque share exactly this formula.

Work is defined as $W = F\cdot d$ , giving dimensions $[MLT^{-2}][L] = [M L^2 T^{-2}]$ . By the work-energy theorem, energy (kinetic energy $\tfrac12mv^2$ , potential energy $mgh$ , heat, etc.) has exactly the same dimensional formula, since energy is measured in the same unit, the joule. Torque is defined as $\tau = F\times r$ (force times perpendicular distance), which also gives $[MLT^{-2}][L] = [ML^2T^{-2}]$ — the same d …

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CBSE 2023Set ANNUAL1 markMCQ
$Q. Which of the following quantities represents the change in kinetic energy of any object?$
(a) Force
(b) Mass
(c) Linear momentum
(d) Work
›Reveal solutionSolution

The work-energy theorem directly identifies work as the physical quantity equal to the change in kinetic energy of a body.

Starting from Newton's second law, F = ma = m(dv/dt). The work done by this force over a displacement ds is:

dW = F ds = m (dv/dt) ds = m v dv (since ds/dt = v)

Integrating from initial speed u to final speed v:

W = integral of m v dv from u to v = (1/2) m v^2 - (1/2) m u^2 = KE_final - KE_initial = Delta KE

…

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