PhysicsClass 11CBSE

When One Marble Hits Another and Stops Dead, Where Does Its Speed Go?

Compute average and instantaneous power as the dot product of force and velocity, tell elastic from inelastic collisions, derive final velocities in one-dimensional elastic collisions with special cases, and solve two-dimensional and perfectly inelastic collisions.

How fast is work done, and what survives a collision?

Two pumps can fill the same overhead tank, but the one that does it faster is more powerful. Power measures how quickly work is done.

In a collision, forces act for a moment and then vanish. Momentum always survives an isolated collision, but kinetic energy may not — and that one difference sorts collisions into types.

This part covers power, elastic and inelastic collisions, one-dimensional elastic collisions, and collisions in two dimensions.

How do you calculate average power, instantaneous power and power from force and velocity?

**Average power is work divided by time,

P_{a v g} = \frac{W}{t}

; instantaneous power is

P = \frac{d W}{d t}

, which for a force acting on a moving body equals the scalar product

P = F \cdot v

.

The unit is the watt**,

1

= 1

J/s;

1

horsepower

\approx 746

W.

Worked example 1 — a pump motor. A motor lifts

200

kg of water through

15

m in

30

s, with

g = 10

m/s

^{2}

W = m g h = 200 \times 10 \times 15 = 30000 J P = \frac{30000}{30} = 1000 W

Worked example 2 — a car. At a steady

25

m/s against

800

N of total resistance, the engine must supply

P = F v = 800 \times 25 = 20000 W = 20 kW

Worked example 3 — vector form.

F = 2 \hat{i} + 3 \hat{j}

N acts on a body with

v = 4 \hat{i} - \hat{j}

m/s:

P = (2) (4) + (3) (- 1) = 5 W

An everyday example. An electric water pump filling an overhead tank does the same work whether fast or slow, but a more powerful pump finishes sooner.

The substance. **

P = F v

explains low gear on a steep climb** — for the same engine power, lower speed gives a larger driving force.

How do you tell an elastic collision from an inelastic one?

In every isolated collision total momentum is conserved; in an elastic collision total kinetic energy is also conserved, in an inelastic collision some kinetic energy is lost, and in a perfectly inelastic collision the bodies stick together and the loss is greatest.

The coefficient of restitution

e = \frac{speed of separation}{speed of approach}

1

for elastic and

0

for perfectly inelastic collisions.

Worked example — test the kinetic energy. A

2

kg ball at

3

m/s hits a

1

kg ball at rest. Momentum before

= 6

kg m/s; kinetic energy before

= \frac{1}{2} (2) (9) = 9

J.

- Outcome A:

2

kg at

1

m/s,

1

kg at

4

m/s. Momentum

2 + 4 = 6

; KE

= 1 + 8 = 9

J — elastic
- Outcome B:

2

kg at

1.5

m/s,

1

kg at

3

m/s. Momentum

3 + 3 = 6

; KE

= 2.25 + 4.5 = 6.75

J — inelastic,

2.25

J lost

An everyday example. Carrom coins clicking together come close to elastic, while a lump of wet clay hitting the floor is perfectly inelastic.

The substance. Even in an elastic collision, kinetic energy dips during contact — it is briefly stored as deformation and then fully returned.

What are the final velocities in a one-dimensional elastic collision?

**Combining momentum and kinetic energy conservation for a body

m_{1}

u_{1}

striking

m_{2}

at rest gives

v_{1} = \frac{m _{1} - m _{2}}{m _{1} + m _{2}} u_{1}

and

v_{2} = \frac{2 m _{1}}{m _{1} + m _{2}} u_{1}

.

Derivation outline.** From

m_{1} (u_{1} - v_{1}) = m_{2} v_{2} m_{1} (u_{1}^{2} - v_{1}^{2}) = m_{2} v_{2}^{2}

dividing gives

u_{1} + v_{1} = v_{2}

— the relative speed of approach equals the relative speed of separation. Substituting back gives the formulas.

Special cases:

- Equal masses:

v_{1} = 0

v_{2} = u_{1}

— the velocities are exchanged
- Heavy target (

m_{2} ≫ m_{1}

v_{1} \approx - u_{1}

v_{2} \approx 0

— the light body bounces straight back

Worked example — heavy target. A

0.1

kg ball at

5

m/s hits a

10

kg block at rest.

v_{1} = \frac{0.1 - 10}{10.1} (5) \approx - 4.90 m/s v_{2} = \frac{0.2}{10.1} (5) \approx 0.099 m/s

An everyday example. In a game of marbles, one marble hitting an identical one head-on stops almost dead while the other rolls away.

The substance. The exchange of velocities needs both equal masses and a head-on elastic collision.

How do you solve two-dimensional collisions and find the energy lost in a perfectly inelastic collision?

**In two dimensions, momentum is conserved separately along

x

and along

y

; when bodies stick together, the kinetic energy lost is the difference between the total before and after.

Worked example 1 — sticking at a crossing.** A

2

kg trolley at

5

m/s east collides with a

3

kg trolley at

4

m/s north, and they move off together.

p_{x} = 2 \times 5 = 10 p_{y} = 3 \times 4 = 12 v = \frac{1 0 ^{2} + 1 2 ^{2}}{5} = \frac{244}{5} \approx 3.12 m/s

Direction:

tan^{- 1} \frac{12}{10} \approx 50.2°

north of east.

K_{b e f or e} = 25 + 24 = 49 J K_{a f t er} = \frac{1}{2} (5) \frac{244}{25} = 24.4 J lost = 24.6 J

Worked example 2 — one dimension. A

2

kg body at

6

m/s sticks to a

1

kg body at rest.

v = \frac{12}{3} = 4 m/s K : 36 J \to 24 J lost = 12 J

The general result is

Δ K = \frac{m _{1} m _{2} ( u _{1} - u _{2} ) ^{2}}{2 ( m _{1} + m _{2} )} = \frac{2 \times 1 \times 36}{6} = 12

J.

An everyday example. Two vehicles colliding at a road crossing and moving off tangled together follow exactly this vector momentum rule.

The substance. **For a target at rest, the fraction of kinetic energy lost is

\frac{m _{2}}{m _{1} + m _{2}}

** — here one third.

Exam tip

What earns full marks on power and collisions?

Always write the momentum equation first, then decide from the question whether kinetic energy is conserved.

- Power:

P = \frac{W}{t}

P = F \cdot v

- Elastic: momentum and KE conserved;

e = 1

- Perfectly inelastic: common final velocity;

e = 0

- 1D elastic, target at rest:

v_{1} = \frac{m _{1} - m _{2}}{m _{1} + m _{2}} u_{1}

v_{2} = \frac{2 m _{1}}{m _{1} + m _{2}} u_{1}

- 2D: separate

x

and

y

momentum equations

The trap. Assuming kinetic energy is conserved in every collision. Only momentum is guaranteed.

Did you know

Why does a heavy ball hitting a light one send it off at nearly double speed?

Put

m_{1} ≫ m_{2}

into the elastic formula:

v_{2} = \frac{2 m _{1}}{m _{1} + m _{2}} u_{1} \approx 2 u_{1}

.

Suppose a

10

kg ball at

2

m/s strikes a

0.1

kg ball at rest:

v_{2} = \frac{2 \times 10}{10.1} \times 2 \approx 3.96 m/s v_{1} = \frac{9.9}{10.1} \times 2 \approx 1.96 m/s

The heavy ball barely slows, yet the light ball flies off at almost twice its speed. Seen from the heavy ball, the light one approaches at

2

m/s and bounces back at

2

m/s — add the heavy ball's own

2

m/s and you get

4

Exam relevance

How are power and collisions tested in JEE Main and NEET?

Power and collisions are regular Work, Energy and Power topics in both JEE Main and NEET, and JEE Advanced extends collisions to rotating bodies and the centre-of-mass frame.

What gets asked. Power of pumps and engines,

P = F v

for vehicles, velocities after one-dimensional elastic collisions, coefficient of restitution with bouncing balls, energy lost when bodies stick, and two-dimensional collisions with momentum components. Collisions link directly to centre of mass in the next chapter.

Question types. Numericals, ratio-based questions and assertion-reason statements on energy conservation.

The trap that costs marks. Adding momenta in two dimensions as plain numbers instead of as components.

Key takeaways

What must you be able to do from this part?

- Power: pump gives

1000

W; car needs

20

kW;

F \cdot v = 5

W
- Collision types: momentum always conserved; KE only in elastic collisions
- 1D elastic:

2

kg at

3

m/s on

1

kg gives

1

m/s and

4

m/s; equal masses exchange velocities
- 2D sticking:

3.12

m/s at

50.2°

with

24.6

J lost
- 1D sticking:

12

J lost, one third of the initial KE

A

3

kg ball at

4

m/s collides elastically head-on with a

1

kg ball at rest. Find both final velocities and check that kinetic energy is conserved.

How fast is work done, and what survives a collision?

How do you calculate average power, instantaneous power and power from force and velocity?

How do you tell an elastic collision from an inelastic one?

What are the final velocities in a one-dimensional elastic collision?

How do you solve two-dimensional collisions and find the energy lost in a perfectly inelastic collision?

What earns full marks on power and collisions?

Why does a heavy ball hitting a light one send it off at nearly double speed?

How are power and collisions tested in JEE Main and NEET?

What must you be able to do from this part?

Related study notes

Ready to put this into practice?