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System of Particles and Rotational Motion - Linear Momentum of a System of Particles

Grade 11CBSEPhysics

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

🔑Concepts

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The total linear momentum of a system of particles is defined as the vector sum of the individual linear momenta of all the particles in the system: P⃗=p⃗1+p⃗2+⋯+p⃗n\vec{P} = \vec{p}_1 + \vec{p}_2 + \dots + \vec{p}_n.

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The total linear momentum of the system is equivalent to the product of the total mass MM of the system and the velocity of its center of mass V⃗cm\vec{V}_{cm}: P⃗=MV⃗cm\vec{P} = M \vec{V}_{cm}.

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Newton's Second Law for a system of particles states that the rate of change of the total linear momentum is equal to the vector sum of all external forces acting on the system: F⃗ext=dP⃗dt\vec{F}_{ext} = \frac{d\vec{P}}{dt}.

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Internal forces between the particles of a system do not contribute to the change in the total linear momentum of the system because they occur in equal and opposite pairs (Newton's Third Law).

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Law of Conservation of Linear Momentum: If the net external force acting on a system is zero (F⃗ext=0\vec{F}_{ext} = 0), then the total linear momentum P⃗\vec{P} of the system remains constant or conserved.

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The conservation of momentum implies that if F⃗ext=0\vec{F}_{ext} = 0, the velocity of the center of mass V⃗cm\vec{V}_{cm} remains constant.

📐Formulae

P⃗=∑i=1nmiv⃗i\vec{P} = \sum_{i=1}^{n} m_i \vec{v}_i

P⃗=MV⃗cm\vec{P} = M \vec{V}_{cm}

F⃗ext=dP⃗dt\vec{F}_{ext} = \frac{d\vec{P}}{dt}

F⃗ext=MA⃗cm\vec{F}_{ext} = M \vec{A}_{cm}

If F⃗ext=0, then P⃗=constant\text{If } \vec{F}_{ext} = 0, \text{ then } \vec{P} = \text{constant}

💡Examples

Problem 1:

A radioactive nucleus of mass MM at rest decays into two fragments of masses m1m_1 and m2m_2 with velocities v⃗1\vec{v}_1 and v⃗2\vec{v}_2 respectively. If m1=2 kgm_1 = 2\text{ kg} and m2=4 kgm_2 = 4\text{ kg}, and the first fragment moves with a velocity of 10 m/s10\text{ m/s} along the +x+x axis, find the velocity of the second fragment.

Solution:

According to the Law of Conservation of Linear Momentum, since no external forces act on the nucleus during decay: P⃗initial=P⃗final\vec{P}_{initial} = \vec{P}_{final} As the nucleus is initially at rest, P⃗initial=0\vec{P}_{initial} = 0. Therefore: m1v⃗1+m2v⃗2=0m_1 \vec{v}_1 + m_2 \vec{v}_2 = 0 (2)(10i^)+(4)(v⃗2)=0(2)(10 \hat{i}) + (4)(\vec{v}_2) = 0 20i^+4v⃗2=020 \hat{i} + 4\vec{v}_2 = 0 4v⃗2=−20i^4\vec{v}_2 = -20 \hat{i} v⃗2=−5i^ m/s\vec{v}_2 = -5 \hat{i} \text{ m/s}

Explanation:

Because the initial momentum is zero and there are no external forces, the fragments must move in opposite directions such that their total momentum remains zero. The heavier fragment moves slower to balance the momentum of the lighter, faster fragment.

Problem 2:

A shell of mass 0.02 kg0.02\text{ kg} is fired from a gun of mass 10 kg10\text{ kg}. If the muzzle speed of the shell is 100 m/s100\text{ m/s}, calculate the recoil speed of the gun.

Solution:

Let m=0.02 kgm = 0.02\text{ kg}, v=100 m/sv = 100\text{ m/s} (velocity of shell), M=10 kgM = 10\text{ kg}, and VV be the recoil velocity of the gun. From conservation of momentum: mv+MV=0m v + M V = 0 (0.02)(100)+(10)(V)=0(0.02)(100) + (10)(V) = 0 2+10V=02 + 10V = 0 10V=−210V = -2 V=−0.2 m/sV = -0.2\text{ m/s} Recoil speed is ∣V∣=0.2 m/s|V| = 0.2\text{ m/s}.

Explanation:

The gun and shell system has no net external force in the horizontal direction. Thus, the forward momentum gained by the shell is exactly balanced by the backward momentum (recoil) of the gun.