2 MARK
A1. Define impulse. How is it related to momentum?
Impulse is the product of force and the time interval over which it acts.
Impulse causes a change in momentum:
[
\vec{J} = \Delta \vec{p}
]
A2. What is the mathematical expression for impulse, and what are its units?
[
\vec{J} = \vec{F},\Delta t
]
Units: Newton–second (N·s) or kg·m/s (both are equivalent).
A3. Explain how impulse can change the momentum of an object.
Impulse applies a force over time, which changes the object’s velocity.
Thus, impulse produces a change in momentum:
[
\vec{J} = \Delta \vec{p}
]
A4. State the impulse–momentum theorem.
The impulse acting on an object equals the change in its momentum:
[
\vec{F}\Delta t = \Delta \vec{p}
]
A5. How does increasing the time of impact affect the force during a collision?
Increasing the time reduces the force experienced because:
[
F = \frac{\Delta p}{\Delta t}
]
Larger time → smaller force for the same momentum change.
A6. What is the principle of conservation of linear momentum?
The total linear momentum of a system remains constant if no external force acts on it.
A7. Under what conditions is momentum conserved in a system?
Momentum is conserved when no external forces act on the system (i.e., system is isolated).
A8. How does an isolated system affect momentum conservation?
In an isolated system, external forces are absent, so internal forces cancel, and total momentum stays constant.
A9. Give an example where momentum is conserved during a collision.
Two ice skaters push off each other: the momentum one gains is equal and opposite to the momentum gained by the other.
A10. Explain the difference between elastic and inelastic collisions in terms of momentum conservation.
-
Elastic collision: Momentum and kinetic energy are conserved.
-
Inelastic collision: Momentum is conserved but kinetic energy is not (some energy becomes heat or sound).
A11. Define the law of conservation of momentum in your own words.
The total momentum of interacting objects remains unchanged unless acted on by an external force.
A12. Explain how momentum conservation applies during a collision between two objects.
Before and after the collision, the sum of momenta of both objects remains the same. Momentum lost by one object is gained by the other.
6 MARK
Below are step-by-step, 6-mark style solutions for B1–B6.
B1. Car braking: impulse and average force
Mass (m = 1500) kg
Initial velocity (u = 20) m/s
Final velocity (v = 0)
Time (t = 5) s
Impulse
[
J = m(v - u) = 1500(0 - 20) = -30,000\ \text{N·s}
]
(negative because it stops)
[
\boxed{J = -3.0\times 10^4\ \text{N·s}}
]
Average force
[
F_{\text{avg}} = \frac{J}{t} = \frac{-30000}{5} = -6000\ \text{N}
]
[
\boxed{F_{\text{avg}} = -6000\ \text{N}}
]
B2. Baseball hit: impulse and average force
Mass (m = 0.145) kg
Initial velocity (u = 40) m/s (toward batter)
Final velocity (v = -30) m/s (reversed direction)
Contact time (t = 0.002) s
Impulse
[
J = m(v-u) = 0.145(-30 - 40)
]
[
J = 0.145(-70) = -10.15\ \text{N·s}
]
[
\boxed{J = -10.15\ \text{N·s}}
]
Average force
[
F = \frac{J}{t} = \frac{-10.15}{0.002} = -5075\ \text{N}
]
[
\boxed{F_{\text{avg}} = -5.08\times10^3\ \text{N}}
]
B3. Hockey puck stopped: find initial velocity
Mass (m = 0.2) kg
Force (F = -10) N
Time (t = 0.1) s
Impulse:
[
J = F t = -10(0.1) = -1\ \text{N·s}
]
Impulse–momentum:
[
J = m(v-u)
]
Final velocity (v = 0):
[
-1 = 0.2(0 - u)
]
[
-1 = -0.2u
]
[
u = 5\ \text{m/s}
]
[
\boxed{u = 5\ \text{m/s}}
]
B4. Two skaters push apart
Momentum conserved:
[
m_1 v_1 + m_2 v_2 = 0
]
Given:
(m_1 = 50) kg, (v_1 = 3) m/s
(m_2 = 70) kg, (v_2 = ?)
[
50(3) + 70 v_2 = 0
]
[
150 + 70v_2 = 0
]
[
v_2 = -\frac{150}{70} = -2.14\ \text{m/s}
]
[
\boxed{v_2 = -2.14\ \text{m/s}}
]
(opposite direction)
B5. Inelastic collision (one rebounds)
Masses:
(m_1 = 10) kg
(m_2 = 15) kg
Velocities:
Before: (u_1 = 5) m/s, (u_2 = 0)
After: (v_1 = -1) m/s, (v_2 = ?)
Conservation of momentum:
[
m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2
]
[
10(5) + 0 = 10(-1) + 15v_2
]
[
50 = -10 + 15v_2
]
[
60 = 15v_2
]
[
v_2 = 4\ \text{m/s}
]
[
\boxed{v_2 = 4\ \text{m/s}}
]
B6. Elastic collision (m₁ = 0.5 kg, m₂ = 1 kg)
Given:
(m_1 = 0.5) kg, (u_1 = 2) m/s
(m_2 = 1) kg, (u_2 = -1) m/s (opposite direction)
For elastic collisions:
[
v_1 = \frac{(m_1 - m_2)}{m_1 + m_2}u_1 + \frac{2m_2}{m_1 + m_2}u_2
]
[
v_2 = \frac{2m_1}{m_1 + m_2}u_1 + \frac{(m_2 - m_1)}{m_1 + m_2}u_2
]
Compute (v_1):
[
v_1 = \frac{0.5-1}{1.5}(2) + \frac{2(1)}{1.5}(-1)
]
[
v_1 = \frac{-0.5}{1.5}(2) + \frac{2}{1.5}(-1)
]
[
v_1 = -0.667 - 1.333 = -2.0\ \text{m/s}
]
Compute (v_2):
[
v_2 = \frac{2(0.5)}{1.5}(2) + \frac{1 - 0.5}{1.5}(-1)
]
[
v_2 = \frac{1}{1.5}(2) + \frac{0.5}{1.5}(-1)
]
[
v_2 = 1.333 - 0.333 = 1.0\ \text{m/s}
]
Final velocities
[
\boxed{v_1 = -2.0\ \text{m/s}}
]
[
\boxed{v_2 = 1.0\ \text{m/s}}
]
14 MARK
Below are well-structured, exam-ready answers for all C-type questions (C1–C9).
Each answer is written to fit 7-mark or 14-mark requirements as requested.
C1 (7 Marks)
Principle of Conservation of Momentum and its Application in Isolated & Non-isolated Systems
Definition:
The principle of conservation of linear momentum states that the total momentum of a system remains constant if no external force acts on it.
Mathematically:
[
\vec{p}{\text{initial}} = \vec{p}{\text{final}}
]
for isolated systems where
[
\sum \vec{F}_{\text{external}} = 0
]
Momentum in Isolated Systems
An isolated system is one where external forces are absent or cancel out.
Internal forces appear in action–reaction pairs and do not change total momentum.
Examples:
-
Two ice skaters pushing apart on frictionless ice.
-
Collisions between billiard balls.
-
Explosion of a firecracker (momentum of fragments adds up to original momentum).
In each case, total momentum before = total momentum after.
Momentum in Non-isolated Systems
A non-isolated system experiences a net external force.
In this case,
[
\frac{d\vec{p}}{dt} = \sum \vec{F}_{\text{external}}
]
so momentum is not conserved.
Examples:
-
A car braking (external friction force reduces momentum).
-
A falling object (gravity increases its momentum).
-
A rocket on Earth (external gravity + drag act).
Conclusion
Momentum is conserved only when external forces are zero.
Understanding isolated vs non-isolated systems is essential in analyzing collisions, explosions, propulsion, and many physical processes.
C2 (7 Marks)
Elastic vs Inelastic Collisions: Comparison Based on Momentum & Kinetic Energy
1. Momentum Conservation
-
Elastic collision:
Total momentum before = total momentum after
[
p_i = p_f
] -
Inelastic collision:
Momentum is still always conserved (if no external forces act).
[
p_i = p_f
]
2. Kinetic Energy
-
Elastic collision:
Total kinetic energy is conserved.
[
KE_i = KE_f
] -
Inelastic collision:
Kinetic energy is not conserved; some is converted into heat, sound, deformation.
[
KE_i > KE_f
]
3. Special Case—Perfectly Inelastic Collision
Objects stick together after impact and move with a common velocity.
4. Real-world Examples
-
Elastic: atomic/molecular collisions, billiard balls (approximately).
-
Inelastic: car crashes, clay ball hitting a wall, football impacts.
5. Summary Table
| Property | Elastic | Inelastic |
|---|---|---|
| Momentum | Conserved | Conserved |
| Kinetic Energy | Conserved | Lost |
| Deformation | Small | Large |
| Final motion | Separate bodies | May stick together |
C3 (7 Marks)
Role of Momentum Conservation in Collisions (Real-World Applications)
Momentum conservation provides the primary tool for analyzing collisions because forces during collisions are large but act for a very short time, making external forces negligible.
Why Momentum is Conserved
-
Internal forces during impact act as action–reaction pairs.
-
External forces (gravity, friction) are negligible over collision time.
Thus,
[
p_{\text{before}} = p_{\text{after}}
]
Real-World Applications
-
Car crashes
Accident investigators calculate speeds using conservation of momentum after a collision. -
Recoil of guns and cannons
When a bullet moves forward, the gun gains backward momentum. -
Spacecraft docking
Momentum conservation predicts resulting velocities of spacecraft when they connect. -
Sports
-
Hockey puck collisions
-
Billiards
-
Bat hitting a ball
-
-
Explosions
Momentum of fragments must sum to initial momentum.
Conclusion
Momentum conservation simplifies the analysis of complex collisions and is essential in engineering, sports physics, and forensic accident analysis.
C4 (7 Marks)
External Forces and Momentum Conservation; When Momentum is Not Conserved
Effect of External Forces
The total momentum of a system changes when an external force acts:
[
\frac{d\vec{p}}{dt} = \sum \vec{F}_{\text{ext}}
]
If
[
\sum \vec{F}_{\text{ext}} = 0
]
momentum is conserved.
Situations Where Momentum Is NOT Conserved
-
Car braking
External friction force reduces momentum. -
Falling object
Earth’s gravity accelerates object → momentum increases. -
Rocket near Earth
Gravity and air resistance change momentum. -
Cart pushed by an external force
Momentum increases due to applied push. -
Object sliding with friction
External friction decreases its momentum.
Conclusion
Momentum is conserved only in isolated systems.
In non-isolated systems, external forces change the total momentum, making the principle inapplicable.
C5 (14 Marks)
Elastic Collisions: Conservation Laws & Derivation of Final Velocities (1-D)
1. Conservation of Momentum
[
m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2
\tag{1}
]
2. Conservation of Kinetic Energy
[
\frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2
\frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2
\tag{2}
]
3. Derivation Using Both Equations
From kinetic energy conservation:
[
m_1(u_1 - v_1)(u_1 + v_1) = m_2(v_2 - u_2)(v_2 + u_2)
]
Using momentum equation (1):
[
m_1(u_1 - v_1) = m_2(v_2 - u_2)
]
Divide the two:
[
u_1 + v_1 = v_2 + u_2
\tag{3}
]
4. Solve Equations (1) and (3)
Equation (3):
[
v_2 = u_1 + v_1 - u_2
]
Plug into (1):
[
m_1u_1 + m_2u_2 = m_1v_1 + m_2(u_1 + v_1 - u_2)
]
Solving gives:
✔ Final velocities in 1-D elastic collision
[
\boxed{
v_1 = \frac{(m_1 - m_2)}{m_1+m_2}u_1 + \frac{2m_2}{m_1+m_2}u_2
}
]
[
\boxed{
v_2 = \frac{2m_1}{m_1+m_2}u_1 + \frac{(m_2 - m_1)}{m_1+m_2}u_2
}
]
Significance
-
Elastic collisions conserve both momentum AND kinetic energy.
-
These equations describe molecular collisions, billiard balls, and many physics problems.
C6 (7 Marks)
Center of Mass: Concept & Importance
The center of mass (CM) of a system is the point where the entire mass of the system can be considered concentrated.
For discrete objects:
[
x_{cm} = \frac{\sum m_i x_i}{\sum m_i}
]
Importance
-
Predicting Motion
The CM follows Newton’s laws, even if internal motions are complex. -
System Simplification
A complicated system (e.g., rotating objects, many-body systems) can be treated as a single particle located at the CM. -
Understanding Collisions
Total momentum relates to the velocity of the CM. -
Rocket motion
CM shifts as fuel burns. -
Human motion
Used in sports biomechanics.
C7 (7 Marks)
Motion of the Center of Mass in System Analysis (Especially When No External Forces Act)
Key Idea
If
[
\sum F_{\text{external}} = 0
]
then
[
\vec{v}_{cm} = \text{constant}
]
The system moves as if all mass were concentrated at the center of mass.
Applications
-
Free-floating astronauts
Their CM moves uniformly even if they move limbs internally. -
Explosion of a firework
Fragments fly in all directions but the CM continues along the original path. -
Two skaters pushing apart
Skaters move away, but CM stays fixed (no external force). -
Projectile motion
For a thrown object that breaks apart midair, the CM follows a smooth parabolic path.
Conclusion
Center of mass motion simplifies analysis of complex systems and is essential for understanding multi-body dynamics.
C8 (7 Marks)
Rocket Propulsion and Momentum Conservation (Derivation of Rocket Equation)
Rocket propulsion is based on momentum conservation in an isolated system (rocket + exhaust gases).
Derivation
At time (t):
Rocket mass = (m), velocity = (v).
After ejecting a small mass (dm) with relative exhaust speed (u):
Momentum before:
[
P_i = mv
]
Momentum after:
[
P_f = (m-dm)(v+dv) + (dm)(v - u)
]
Applying conservation (P_i = P_f):
[
mv = (m-dm)(v+dv) + dm(v-u)
]
Simplifying gives:
[
m dv = u,dm
]
Since mass decreases, take (dm < 0):
[
\boxed{v = u \ln\left(\frac{m_0}{m}\right)}
]
This is the Tsiolkovsky rocket equation.
Meaning
A rocket accelerates because gases are ejected backward, giving the rocket forward momentum.
C9 (7 Marks)
Rocket Propulsion and Newton’s Third Law
Rocket propulsion can be fully explained by action–reaction forces.
Action
The rocket expels gas backward at high speed.
Reaction
The gases exert an equal and opposite force on the rocket, pushing it forward.
Physics Explanation
-
No air or wall is needed for a rocket to accelerate.
-
The force arises from internal pressure pushing gas out and pushing the rocket forward.
Real-World Evidence
-
Rockets work in space (no external medium required).
-
Fire extinguishers push students on trolleys (recoil effect).
-
Balloon rockets demonstrate backward gas ejection → forward motion.
Conclusion
Rocket propulsion is a direct consequence of Newton's 3rd law and momentum conservation, forming the basis of all satellite launches and space travel.