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Space, Time and Motion - Momentum and Impulse

Grade 12IBPhysics

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

🔑Concepts

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Linear momentum is defined as the product of an object's mass and its velocity: p⃗=mv⃗\vec{p} = m\vec{v}. It is a vector quantity, meaning it has both magnitude and direction, measured in kg⋅m⋅s−1kg \cdot m \cdot s^{-1}.

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Newton's Second Law in terms of momentum states that the net resultant force acting on an object is equal to the rate of change of its linear momentum: Fnet=ΔpΔtF_{net} = \frac{\Delta p}{\Delta t}.

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Impulse (JJ) is the product of the average force and the time interval over which it acts: J=FΔtJ = F \Delta t. It is equivalent to the change in momentum: J=Δp=m(v−u)J = \Delta p = m(v - u).

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The area under a Force-time (FF-tt) graph represents the impulse delivered to an object, which corresponds to the total change in its momentum.

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The Law of Conservation of Linear Momentum states that in an isolated system (where no external forces act), the total linear momentum remains constant: ∑pinitial=∑pfinal\sum p_{initial} = \sum p_{final}.

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In an elastic collision, both total momentum and total kinetic energy (EkE_k) are conserved. In an inelastic collision, only total momentum is conserved, while some kinetic energy is dissipated as heat, sound, or used in deformation.

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The relationship between kinetic energy (EkE_k) and momentum (pp) is given by 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}

Impulse=∫F dt\text{Impulse} = \int F \, dt

💡Examples

Problem 1:

A tennis ball of mass 0.060 kg0.060\text{ kg} is moving horizontally at 40 m s−140\text{ m } s^{-1} when it is struck by a racket. The ball rebounds in the opposite direction at 30 m s−130\text{ m } s^{-1}. If the ball is in contact with the racket for 0.005 s0.005\text{ s}, calculate the average force exerted by the racket on the ball.

Solution:

Taking the initial direction as positive: u=40 m s−1u = 40\text{ m } s^{-1} v=−30 m s−1v = -30\text{ m } s^{-1} m=0.060 kgm = 0.060\text{ kg} Δt=0.005 s\Delta t = 0.005\text{ s}

Change in momentum: Δp=m(v−u)=0.060×(−30−40)=−4.2 kg m⋅s−1\Delta p = m(v - u) = 0.060 \times (-30 - 40) = -4.2\text{ kg } m \cdot s^{-1}

Average force: F=ΔpΔt=−4.20.005=−840 NF = \frac{\Delta p}{\Delta t} = \frac{-4.2}{0.005} = -840\text{ N}

Explanation:

The negative sign indicates that the force acts in the direction opposite to the initial velocity. The impulse delivered is 4.2 N s4.2\text{ N } s.

Problem 2:

A block of mass 2.0 kg2.0\text{ kg} moving at 5.0 m s−15.0\text{ m } s^{-1} collides with a stationary block of mass 3.0 kg3.0\text{ kg}. The two blocks stick together after the collision. Calculate the final velocity of the combined mass.

Solution:

Using conservation of momentum: m1u1+m2u2=(m1+m2)vm_1 u_1 + m_2 u_2 = (m_1 + m_2)v (2.0×5.0)+(3.0×0)=(2.0+3.0)v(2.0 \times 5.0) + (3.0 \times 0) = (2.0 + 3.0)v 10=5.0v10 = 5.0v v=2.0 m s−1v = 2.0\text{ m } s^{-1}

Explanation:

Since the blocks stick together, it is a perfectly inelastic collision. The total momentum before the collision (10 kg m⋅s−110\text{ kg } m \cdot s^{-1}) must equal the total momentum after the collision.