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Forces and Motion - Momentum and Collisions

Grade 9IB

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

🔑Concepts

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Momentum (pp) is a vector quantity that describes the quantity of motion of an object. It is the product of mass (mm) and velocity (vv). Units: kg⋅m/skg \cdot m/s.

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The Law 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.

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Impulse is defined as the change in momentum (Δp\Delta p). It is also equal to the product of the average force (FF) and the time interval (Δt\Delta t) over which the force is applied.

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In an Elastic Collision, both momentum and kinetic energy (Ek=12mv2E_k = \frac{1}{2} m v^2) are conserved. These are typically observed between subatomic particles or ideal gas molecules.

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In an Inelastic Collision, momentum is conserved but kinetic energy is not. Some kinetic energy is transformed into other forms like heat or sound. If the objects stick together, it is a perfectly inelastic collision.

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

📐Formulae

p=mvp = m v

Δp=m(v−u)\Delta p = m(v - u)

F=ΔpΔtF = \frac{\Delta p}{\Delta t}

FΔt=ΔpF \Delta t = \Delta p

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

Ek=12mv2E_k = \frac{1}{2} m v^2

💡Examples

Problem 1:

A car of mass 1500 kg1500\text{ kg} is traveling at 20 m/s20\text{ m/s} when the driver applies the brakes, bringing the car to rest in 4.0 s4.0\text{ s}. Calculate the average braking force.

Solution:

m=1500 kg,u=20 m/s,v=0 m/s,Δt=4.0 sm = 1500\text{ kg}, u = 20\text{ m/s}, v = 0\text{ m/s}, \Delta t = 4.0\text{ s} Δp=m(v−u)=1500(0−20)=−30000 kg⋅m/s\Delta p = m(v - u) = 1500(0 - 20) = -30000\text{ kg} \cdot m/s F=ΔpΔt=−300004.0=−7500 NF = \frac{\Delta p}{\Delta t} = \frac{-30000}{4.0} = -7500\text{ N}

Explanation:

First, calculate the change in momentum (Δp\Delta p) by subtracting the initial momentum from the final momentum. Then, use the force formula F=ΔpΔtF = \frac{\Delta p}{\Delta t}. The negative sign indicates the force acts in the opposite direction to the car's initial motion (deceleration).

Problem 2:

A 2.0 kg2.0\text{ kg} ball moving at 6.0 m/s6.0\text{ m/s} collides with a stationary 4.0 kg4.0\text{ kg} ball. After the collision, the 2.0 kg2.0\text{ kg} ball rebounds at 2.0 m/s2.0\text{ m/s} in the opposite direction. Calculate the velocity of the 4.0 kg4.0\text{ kg} ball after the collision.

Solution:

m1=2.0 kg,u1=6.0 m/s,v1=−2.0 m/sm_1 = 2.0\text{ kg}, u_1 = 6.0\text{ m/s}, v_1 = -2.0\text{ m/s} m2=4.0 kg,u2=0 m/s,v2=?m_2 = 4.0\text{ kg}, u_2 = 0\text{ m/s}, v_2 = ? m1u1+m2u2=m1v1+m2v2m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 (2.0)(6.0)+(4.0)(0)=(2.0)(−2.0)+(4.0)v2(2.0)(6.0) + (4.0)(0) = (2.0)(-2.0) + (4.0)v_2 12.0=−4.0+4.0v212.0 = -4.0 + 4.0 v_2 16.0=4.0v2  ⟹  v2=4.0 m/s16.0 = 4.0 v_2 \implies v_2 = 4.0\text{ m/s}

Explanation:

Apply the Law of Conservation of Momentum. Note that velocity is a vector, so the rebound velocity of the first ball is negative (−2.0 m/s-2.0\text{ m/s}). Solve the linear equation for v2v_2.