krit.club logo

Space, Time and Motion - Forces and Momentum

Grade 12IBPhysics

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

🔑Concepts

•

Newton's Second Law in terms of momentum: The net force acting on an object is equal to the rate of change of its linear momentum, expressed as F=ΔpΔtF = \frac{\Delta p}{\Delta t}.

•

Linear Momentum: A vector quantity defined as the product of an object's mass mm and its velocity vv, given by p=mvp = mv. Its unit is kg m s−1\text{kg m s}^{-1} or N s\text{N s}.

•

Impulse: The change in momentum Δp\Delta p resulting from a force FF acting over a time interval Δt\Delta t. It is also represented as the area under a force-time (FF-tt) graph.

•

Law of Conservation of Linear Momentum: In an isolated system (where the external force Fext=0F_{ext} = 0), the total linear momentum remains constant over time.

•

Elastic Collisions: Collisions where both total linear momentum and total kinetic energy (EkE_k) are conserved.

•

Inelastic Collisions: Collisions where momentum is conserved, but kinetic energy is not. In a perfectly inelastic collision, the objects stick together and move with a common velocity.

•

Relation between Kinetic Energy and Momentum: Kinetic energy can be expressed in terms of momentum as Ek=p22mE_k = \frac{p^2}{2m}.

📐Formulae

p=mvp = mv

F=ΔpΔtF = \frac{\Delta p}{\Delta t}

J=FΔt=ΔpJ = F \Delta t = \Delta p

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

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

I=∫F dtI = \int F \, dt

💡Examples

Problem 1:

A tennis ball of mass 0.06 kg0.06\text{ kg} is moving horizontally at 25 m s−125\text{ m s}^{-1} towards a racket. It is hit by the racket and rebounds in the opposite direction at 30 m s−130\text{ m s}^{-1}. If the contact time is 0.05 s0.05\text{ s}, calculate the average force exerted by the racket.

Solution:

Taking the initial direction as positive: u=25 m s−1u = 25\text{ m s}^{-1} and v=−30 m s−1v = -30\text{ m s}^{-1}. Δp=m(v−u)=0.06(−30−25)=−3.3 kg m s−1\Delta p = m(v - u) = 0.06(-30 - 25) = -3.3\text{ kg m s}^{-1}. Average Force F=ΔpΔt=−3.30.05=−66 NF = \frac{\Delta p}{\Delta t} = \frac{-3.3}{0.05} = -66\text{ N}.

Explanation:

The change in momentum is calculated by considering the vector nature of velocity. The force is then found using Newton's second law.

Problem 2:

A car of mass 1500 kg1500\text{ kg} traveling at 10 m s−110\text{ m s}^{-1} collides with a stationary van of mass 2500 kg2500\text{ kg}. After the collision, the two vehicles stick together. Determine their common velocity.

Solution:

Using conservation of momentum: m1u1+m2u2=(m1+m2)vm_1 u_1 + m_2 u_2 = (m_1 + m_2)v. (1500)(10)+(2500)(0)=(1500+2500)v  ⟹  15000=4000v  ⟹  v=150004000=3.75 m s−1(1500)(10) + (2500)(0) = (1500 + 2500)v \implies 15000 = 4000v \implies v = \frac{15000}{4000} = 3.75\text{ m s}^{-1}.

Explanation:

In a perfectly inelastic collision, the total momentum before the crash equals the total momentum of the combined mass after the crash.

Forces and Momentum - Revision Notes & Key Formulas | IB Grade 12 Physics