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Laws of Motion - Conservation of Momentum

Grade 11CBSEPhysics

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

🔑Concepts

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The Law of Conservation of Momentum states that if the net external force acting on a system is zero, the total linear momentum of the system remains constant. It is represented as F⃗ext=0  ⟹  P⃗=constant\vec{F}_{ext} = 0 \implies \vec{P} = \text{constant}.

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This law is a direct consequence of Newton's Second and Third Laws. According to Newton's Second Law, F⃗ext=dP⃗dt\vec{F}_{ext} = \frac{d\vec{P}}{dt}. If the external force is zero, the rate of change of momentum is zero, making momentum invariant.

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For an isolated system of two colliding bodies, the total momentum before collision is equal to the total momentum after collision: m1u⃗1+m2u⃗2=m1v⃗1+m2v⃗2m_1 \vec{u}_1 + m_2 \vec{u}_2 = m_1 \vec{v}_1 + m_2 \vec{v}_2.

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Internal forces within a system (e.g., forces during a collision) act in equal and opposite pairs and thus cancel each other out, having no effect on the total momentum of the system.

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Applications of this law include the recoil of a gun, the propulsion of rockets, and the explosion of a shell in mid-air where the center of mass continues its original parabolic path.

📐Formulae

P⃗=mv⃗\vec{P} = m \vec{v}

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

If F⃗ext=0, then ∑p⃗initial=∑p⃗final\text{If } \vec{F}_{ext} = 0, \text{ then } \sum \vec{p}_{initial} = \sum \vec{p}_{final}

m1u⃗1+m2u⃗2=m1v⃗1+m2v⃗2m_1 \vec{u}_1 + m_2 \vec{u}_2 = m_1 \vec{v}_1 + m_2 \vec{v}_2

V⃗recoil=−mMv⃗bullet\vec{V}_{recoil} = -\frac{m}{M} \vec{v}_{bullet}

💡Examples

Problem 1:

A bullet of mass 40 g40\text{ g} is fired from a gun of mass 8 kg8\text{ kg} with a muzzle velocity of 800 m/s800\text{ m/s}. What is the recoil velocity of the gun?

Solution:

Given: mass of bullet m=40 g=0.04 kgm = 40\text{ g} = 0.04\text{ kg}, velocity of bullet v=800 m/sv = 800\text{ m/s}, mass of gun M=8 kgM = 8\text{ kg}. Let VV be the recoil velocity. Since the initial momentum is zero, MV+mv=0MV + mv = 0. Therefore, V=−mvM=−0.04×8008=−4 m/sV = -\frac{mv}{M} = -\frac{0.04 \times 800}{8} = -4\text{ m/s}.

Explanation:

The conservation of momentum requires the gun to move in the opposite direction (indicated by the negative sign) to balance the forward momentum of the bullet, keeping the total system momentum at zero.

Problem 2:

A body of mass 2 kg2\text{ kg} moving with a velocity of 10 m/s10\text{ m/s} collides with another body of mass 3 kg3\text{ kg} moving at 5 m/s5\text{ m/s} in the same direction. If they stick together after collision, find their common velocity.

Solution:

Mass m1=2 kgm_1 = 2\text{ kg}, u1=10 m/su_1 = 10\text{ m/s}; Mass m2=3 kgm_2 = 3\text{ kg}, u2=5 m/su_2 = 5\text{ m/s}. Total initial momentum Pi=m1u1+m2u2=(2×10)+(3×5)=35 kg m/sP_i = m_1u_1 + m_2u_2 = (2 \times 10) + (3 \times 5) = 35\text{ kg m/s}. Since they stick together, let the common velocity be VV. Final momentum Pf=(m1+m2)V=(2+3)V=5VP_f = (m_1 + m_2)V = (2 + 3)V = 5V. By conservation of momentum, 5V=35  ⟹  V=7 m/s5V = 35 \implies V = 7\text{ m/s}.

Explanation:

In a perfectly inelastic collision where objects stick together, the total mass of the system becomes the sum of individual masses, and the final velocity is determined by the ratio of total initial momentum to total mass.

Conservation of Momentum - Revision Notes & Key Formulas | CBSE Class 11 Physics