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Physics: Motion and Mechanics - Momentum, Impulse, and Collisions

Grade 8IB

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' an object possesses. It depends on both the mass (mm) and the velocity (vv) of the object.

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The SI unit for momentum is kilograms-meters per second (kg⋅m/skg \cdot m/s).

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Conservation of Momentum: In an isolated system (where no external forces act), the total momentum before a collision is equal to the total momentum after the collision (pinitial=pfinalp_{initial} = p_{final}).

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Impulse (JJ) is defined as the change in momentum of an object. It occurs when a force (FF) is applied over a specific time interval (Δt\Delta t).

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The Impulse-Momentum Theorem states that FΔt=ΔpF \Delta t = \Delta p. This explains why increasing the time of impact (like using an airbag) reduces the force experienced during a collision.

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Elastic Collisions: Collisions where both momentum and kinetic energy are conserved.

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Inelastic Collisions: Collisions where momentum is conserved, but some kinetic energy is transformed into other forms (like heat or sound). Objects often stick together in perfectly inelastic collisions.

📐Formulae

p=m×vp = m \times v

Δp=m(v−u)\Delta p = m(v - u)

J=F×ΔtJ = F \times \Delta t

F×Δt=m(v−u)F \times \Delta t = m(v - u)

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

💡Examples

Problem 1:

A 0.5kg0.5 kg ball is rolling at a velocity of 4m/s4 m/s. What is its momentum?

Solution:

p=m×vp = m \times v p=0.5kg×4m/sp = 0.5 kg \times 4 m/s p=2kg⋅m/sp = 2 kg \cdot m/s

Explanation:

Momentum is calculated by multiplying the mass of the ball by its velocity.

Problem 2:

A car of mass 1000kg1000 kg traveling at 20m/s20 m/s crashes into a wall and comes to a complete stop in 0.5s0.5 s. Calculate the average force exerted on the car.

Solution:

m=1000kg,u=20m/s,v=0m/s,Δt=0.5sm = 1000 kg, u = 20 m/s, v = 0 m/s, \Delta t = 0.5 s Δp=m(v−u)=1000(0−20)=−20000kg⋅m/s\Delta p = m(v - u) = 1000(0 - 20) = -20000 kg \cdot m/s F=ΔpΔtF = \frac{\Delta p}{\Delta t} F=−200000.5=−40000NF = \frac{-20000}{0.5} = -40000 N

Explanation:

The change in momentum (impulse) is −20000kg⋅m/s-20000 kg \cdot m/s. Dividing this by the time of impact gives the force. The negative sign indicates the force acts in the opposite direction to the initial motion.

Problem 3:

A 2kg2 kg cart moving at 3m/s3 m/s collides with a stationary 1kg1 kg cart. If they stick together after the collision, what is their final velocity?

Solution:

m1u1+m2u2=(m1+m2)vm_1 u_1 + m_2 u_2 = (m_1 + m_2)v (2×3)+(1×0)=(2+1)v(2 \times 3) + (1 \times 0) = (2 + 1)v 6+0=3v6 + 0 = 3v v=63=2m/sv = \frac{6}{3} = 2 m/s

Explanation:

Using the Law of Conservation of Momentum, the total momentum before (6kg⋅m/s6 kg \cdot m/s) equals the total momentum after. Since the carts stick together, their combined mass is 3kg3 kg.