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Motion, Forces and Energy - Momentum

Grade 12IGCSEPhysics

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

πŸ”‘Concepts

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Momentum is a vector quantity defined as the product of an object's mass (mm) and its velocity (vv). It is measured in kilogram metres per second (kgβ‹…m/skg \cdot m/s) or Newton-seconds (Nβ‹…sN \cdot s).

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The Principle of Conservation of Momentum states that in an isolated system (where no external forces act), the total momentum before an interaction is equal to the total momentum after the interaction: βˆ‘pinitial=βˆ‘pfinal\sum p_{initial} = \sum p_{final}.

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Impulse is defined as the product of the resultant force (FF) acting on an object and the time interval (Ξ”t\Delta t) for which it acts. It is equivalent to the change in momentum: Ξ”p=FΞ”t\Delta p = F\Delta t.

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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 its momentum.

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In an elastic collision, both total momentum and total kinetic energy (EkE_k) are conserved. In an inelastic collision, momentum is conserved, but kinetic energy is not (some energy is dissipated as heat or sound).

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If two objects of mass m1m_1 and m2m_2 collide and stick together (perfectly inelastic), they move with a common final velocity vv such that m1u1+m2u2=(m1+m2)vm_1u_1 + m_2u_2 = (m_1 + m_2)v.

πŸ“Formulae

p=mvp = mv

F=Ξ”pΞ”tF = \frac{\Delta p}{\Delta t}

Impulse=FΞ”t=m(vβˆ’u)Impulse = F \Delta t = m(v - u)

m1u1+m2u2=m1v1+m2v2m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2

Ek=p22mE_k = \frac{p^2}{2m}

πŸ’‘Examples

Problem 1:

A car of mass 1200 kg1200\,kg traveling at 15 m/s15\,m/s collides with a stationary van of mass 2000 kg2000\,kg. After the collision, they stick together. Calculate their common velocity vv immediately after the impact.

Solution:

Using the conservation of momentum: m1u1+m2u2=(m1+m2)vm_1u_1 + m_2u_2 = (m_1 + m_2)v (1200Γ—15)+(2000Γ—0)=(1200+2000)v(1200 \times 15) + (2000 \times 0) = (1200 + 2000)v 18000=3200v18000 = 3200v v=180003200=5.625 m/sv = \frac{18000}{3200} = 5.625\,m/s

Explanation:

Since there are no external forces, the total momentum before the collision (only the car has momentum) must equal the total momentum after the collision (both masses combined).

Problem 2:

A tennis ball of mass 0.06 kg0.06\,kg hits a wall at 30 m/s30\,m/s and rebounds in the opposite direction at 20 m/s20\,m/s. If the contact time is 0.01 s0.01\,s, calculate the average force exerted by the wall on the ball.

Solution:

First, calculate the change in momentum (taking the initial direction as positive): Ξ”p=m(vβˆ’u)\Delta p = m(v - u) Ξ”p=0.06Γ—(βˆ’20βˆ’30)=0.06Γ—(βˆ’50)=βˆ’3.0 kgβ‹…m/s\Delta p = 0.06 \times (-20 - 30) = 0.06 \times (-50) = -3.0\,kg \cdot m/s Now, find the force: F=Ξ”pΞ”t=βˆ’3.00.01=βˆ’300 NF = \frac{\Delta p}{\Delta t} = \frac{-3.0}{0.01} = -300\,N

Explanation:

Velocity is a vector, so the rebound velocity must be negative if the approach velocity is positive. The force is negative because it acts in the opposite direction to the initial motion.

Momentum - Revision Notes & Key Formulas | IGCSE Grade 12 Physics