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Motion, Forces and Energy - Momentum

Grade 11IGCSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Momentum (pp) is a vector quantity defined as the product of an object's mass and its velocity: p=mvp = mv. Its unit is kg⋅m/skg \cdot m/s.

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The Principle of Conservation of Momentum states that in a closed system (where no external forces act), the total momentum before a collision or explosion is equal to the total momentum after: ∑pbefore=∑pafter\sum p_{before} = \sum p_{after}.

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Impulse is defined as the change in momentum, calculated as the product of the resultant force and the time for which it acts: Impulse=FΔt=ΔpImpulse = F \Delta t = \Delta p.

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Newton's Second Law can be expressed in terms of momentum: the resultant force acting on an object is equal to the rate of change of momentum: F=ΔptF = \frac{\Delta p}{t}.

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In an elastic collision, both momentum and kinetic energy (12mv2\frac{1}{2}mv^2) are conserved. In an inelastic collision, momentum is conserved but kinetic energy is not (some is converted to heat or sound).

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Safety features like crumple zones and seatbelts increase the time taken (Δt\Delta t) for the change in momentum to occur, thereby reducing the impulsive force (FF) acting on the passengers.

📐Formulae

p=mvp = mv

m1u1+m2u2=m1v1+m2v2m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2

F=m(v−u)tF = \frac{m(v - u)}{t}

Impulse=FΔt=Δp=mv−muImpulse = F \Delta t = \Delta p = m v - m u

💡Examples

Problem 1:

A car of mass 1200 kg1200 \text{ kg} traveling at 20 m/s20 \text{ m/s} collides with a stationary van of mass 1800 kg1800 \text{ kg}. After the collision, the two vehicles stick together. Calculate their common velocity vv.

Solution:

Total momentum before = (1200×20)+(1800×0)=24000 kg m/s(1200 \times 20) + (1800 \times 0) = 24000 \text{ kg m/s}. Total mass after = 1200+1800=3000 kg1200 + 1800 = 3000 \text{ kg}. Using conservation of momentum: 24000=3000×v24000 = 3000 \times v. Therefore, v=240003000=8 m/sv = \frac{24000}{3000} = 8 \text{ m/s}.

Explanation:

Since the vehicles stick together, they move with a shared velocity. We apply the conservation of momentum law where the sum of initial momenta equals the sum of final momenta.

Problem 2:

A tennis ball of mass 0.06 kg0.06 \text{ kg} is hit by a racket. Its velocity changes from 15 m/s15 \text{ m/s} in one direction to 25 m/s25 \text{ m/s} in the opposite direction. If the racket is in contact with the ball for 0.02 s0.02 \text{ s}, calculate the average force exerted on the ball.

Solution:

Take the initial direction as positive. u=15 m/su = 15 \text{ m/s}, v=−25 m/sv = -25 \text{ m/s}. Change in momentum Δp=m(v−u)=0.06×(−25−15)=0.06×(−40)=−2.4 kg m/s\Delta p = m(v - u) = 0.06 \times (-25 - 15) = 0.06 \times (-40) = -2.4 \text{ kg m/s}. Force F=Δpt=−2.40.02=−120 NF = \frac{\Delta p}{t} = \frac{-2.4}{0.02} = -120 \text{ N}.

Explanation:

Force is the rate of change of momentum. The negative sign indicates that the force acts in the opposite direction to the initial motion. The magnitude of the force is 120 N120 \text{ N}.

Momentum - Revision Notes & Key Formulas | IGCSE Grade 11 Physics