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Jamb Physics Key Points and Summaries on Newton's Laws of Motion for Jamb Candidates

Nov 20 2024 11:34 PM

Osason

Study Guide

Newton's Law of Motion | Jamb(UTME)

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Hi scholar, its no mistake you are here. propably you have been searching for information as per physics, I want to inform you that you are in the right place. You see, success is not a mirage if intentions are clothed with actions.
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We have the best interest of UTME candidate at heart that is why poscholars team has pooled out resources, exerted effort and invested time to ensure you are adequately prepared before you write the exam. Can you imagine an online platform where you can have access to key points and summaries in every topic in the Physics syllabus for Jamb UTME? Guess what! your imagination is now a reality.
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In this post, we have enumerated a good number of points from the topic Newton's laws of motion which was extracted from the Jamb syllabus. I would advice you pay attention to each of the point knowing and understanding them by heart. Happy learning.
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The table of content below will guide you on the related topics pertaining to "Newton's laws of motion" you can navigate to the one that captures your interest
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Table of Contents
  1. Jamb(utme) key points on Newtons' laws; inertia; mass; force; relationship between mass and accelaration; impulse and momentum
  2. Jamb(utme) key points on force-time graph; conservation of linear momentum and application
  3. Jamb(utme) Calculation problems and solutions involving impulse and momentum
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Jamb(utme) key points on Newtons' laws; inertia; mass; force; relationship between mass and accelaration; impulse and momentum

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Here are 60 points covering Newton's Laws of Motion, inertia, mass, force, the relationship between mass and acceleration, impulse, and momentum:
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Newton's Laws of Motion
  1. First Law (Law of Inertia): An object remains at rest or in uniform motion in a straight line unless acted upon by an external force.
  2. The First Law explains why objects don’t move unless a force is applied (e.g., a stationary chair stays at rest until pushed).
  3. Second Law (Law of Acceleration): The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass:
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    F=maF = ma
  4. The Second Law quantifies the relationship between force, mass, and acceleration.
  5. Third Law (Action-Reaction): For every action, there is an equal and opposite reaction.
  6. The Third Law explains phenomena like recoil in guns or the thrust in rockets.
  7. Newton’s Laws are the foundation of classical mechanics.
  8. These laws apply to objects in motion or at rest as long as relativistic speeds or quantum effects are negligible.
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Inertia
  1. Inertia is the property of matter that resists changes in motion or rest.
  2. Greater mass means greater inertia.
  3. A stationary truck is harder to move than a bicycle due to higher inertia.
  4. Inertia applies to rotational motion (rotational inertia depends on the mass distribution around an axis).
  5. Newton's First Law is also known as the Law of Inertia.
  6. Inertia explains why passengers lurch forward in a sudden stop or backward during acceleration.
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Mass
  1. Mass is the measure of the amount of matter in an object.
  2. The SI unit of mass is the kilogram kgkg.
  3. Mass is scalar and does not change with location (unlike weight).
  4. Mass is a measure of an object's resistance to acceleration.
  5. Weight is the force due to gravity acting on mass:
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    W=mgW = mg
  6. Mass is proportional to inertia; larger masses are harder to accelerate.
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Force
  1. Force is an interaction that changes the motion of an object.
  2. The SI unit of force is the newton NN.
  3. Force is a vector quantity with both magnitude and direction.
  4. A force can accelerate, decelerate, or change the direction of an object.
  5. The net force is the vector sum of all forces acting on an object.
  6. Balanced forces result in no acceleration, while unbalanced forces cause motion.
  7. Common types of forces include gravitational, frictional, and applied forces.
  8. The formula for force is:
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    F=maF = ma
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Relationship Between Mass and Acceleration
  1. Acceleration is inversely proportional to mass for a constant force:
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    a=Fma = \frac{F}{m}
  2. Objects with greater mass require more force to achieve the same acceleration as lighter objects.
  3. For example, pushing a small car is easier than pushing a truck due to the difference in mass.
  4. Mass affects acceleration, but force determines the magnitude of motion.
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Impulse
  1. Impulse is the product of force and the time over which it acts: J=FΔtJ = F \Delta t
  2. The SI unit of impulse is NsN·s (newton-second).
  3. Impulse is a vector quantity.
  4. Impulse equals the change in momentum:
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    J=ΔpJ = \Delta p
  5. Longer impact times result in smaller forces (e.g., airbags).
  6. Shorter impact times create higher forces (e.g., a hammer hitting a nail).
  7. Impulse explains why following through in sports increases momentum.
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Momentum
  1. Momentum is the product of an object's mass and velocity:
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    p=mvp = mv
  2. Momentum is a vector quantity with the same direction as velocity.
  3. The SI unit of momentum is kgm/skg·m/s.
  4. An object at rest has zero momentum.
  5. Momentum is conserved in isolated systems (Law of Conservation of Momentum).
  6. In collisions, total momentum before impact equals total momentum after, assuming no external forces.
  7. Elastic collisions conserve both momentum and kinetic energy.
  8. Inelastic collisions conserve momentum but not kinetic energy.
  9. Momentum depends equally on mass and velocity; doubling either doubles the momentum.
  10. Momentum conservation explains rocket propulsion and the recoil of guns.
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Impulse-Momentum Theorem
  1. Impulse causes a change in momentum: J=ΔpJ = \Delta p
  2. This theorem is used to calculate the effects of forces in collisions and impacts.
  3. For example, the momentum change in a bouncing ball depends on the impulse exerted by the surface.
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Applications in Everyday Life
  1. Seatbelts reduce force by increasing the time over which momentum changes during car crashes.
  2. In sports, follow-through helps impart greater momentum to a ball.
  3. Airbags extend impact time, reducing the force on passengers.
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Engineering
  1. Crumple zones in cars reduce momentum transfer to passengers during crashes.
  2. Rocket engines generate thrust by expelling gases to conserve momentum.
  3. Turbines use momentum changes in fluids to generate mechanical work.
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Physics and Astronomy
  1. In space, astronauts move by pushing off surfaces, demonstrating momentum conservation.
  2. In particle physics, momentum conservation helps analyze collisions in particle accelerators.
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Jamb(utme) key points on force-time graph; conservation of linear momentum and application

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Here are 50 easy-to-understand points covering the force-time graph, conservation of linear momentum, and its applications:
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Force-Time Graph**
  1. A force-time graph shows how the force acting on an object varies over time.
  2. The x-axis represents time (t)(t) and the y-axis represents force (F)(F).
  3. The area under a force-time graph represents impulse.
  4. Impulse is the product of force and the time over which it acts.
  5. The formula for impulse is:
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    J=F(t)dtJ = \int F(t) \, dt
  6. Impulse is equal to the change in momentum of the object: J=ΔpJ = \Delta p
  7. If the force is constant, the area under the graph is a rectangle (J=FΔt)(J = F \cdot \Delta t).
  8. If the force varies, the area can be calculated using geometry (triangles, trapezoids) or integration.
  9. A spike in the graph indicates a sudden large force acting over a very short time.
  10. The slope of the force-time graph is not directly meaningful but indicates how quickly the force changes.
  11. In collisions, force-time graphs often show a sharp rise and fall, representing impact forces.
  12. Longer impact times (wider graph) reduce peak forces, minimizing damage (e.g., airbags in cars).
  13. A flat line on the graph indicates a constant force.
  14. A zero line indicates no force acting on the object.
  15. Negative areas below the x-axis represent forces acting in the opposite direction.
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Conservation of Linear Momentum
  1. Momentum (p)(p) is the product of mass and velocity: p=mvp = mv
  2. The Law of Conservation of Linear Momentum states that in an isolated system (no external forces), the total momentum remains constant.
  3. Mathematically: pinitial=pfinalp_{initial} = p_{final}
  4. The principle applies to collisions, explosions, and any interaction between objects.
  5. Internal forces, like those during a collision, do not affect the system's total momentum.
  6. Momentum conservation is valid for both elastic and inelastic collisions.
  7. In elastic collisions, both momentum and kinetic energy are conserved.
  8. In inelastic collisions, only momentum is conserved, while kinetic energy is not.
  9. Momentum conservation also applies in two-dimensional motion, where components in each direction are conserved separately.
  10. In explosions, the total momentum before and after the event remains zero if the system was initially at rest.
  11. Momentum conservation explains how objects recoil after releasing energy (e.g., firing a gun).
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Applications of Conservation of Momentum
Collisions
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27. In car crashes, momentum conservation helps calculate post-collision velocities of vehicles. 28. Sports like billiards and bowling rely on momentum transfer for effective gameplay. 29. In head-on collisions, the combined momentum of two colliding objects is distributed after the impact. 30. In elastic collisions, objects bounce off each other while conserving momentum and energy.
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Recoil and Explosions
  1. A gun recoils when fired due to conservation of momentum.
  2. Rockets work based on momentum conservation: expelling gases downward creates upward thrust.
  3. In fireworks, the fragments of an explosion spread out in different directions while conserving total momentum.
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Astronomy and Space
  1. Satellites use momentum exchange during docking and separation from spacecraft.
  2. In space, astronauts move by pushing off objects, relying on momentum conservation.
  3. Collisions between celestial bodies, like asteroids, are analyzed using momentum conservation.
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Vehicle Design
  1. Momentum conservation principles are applied in designing crumple zones to reduce crash impact forces.
  2. Airbags increase the time over which momentum is transferred, reducing peak forces on passengers.
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Engineering and Physic
  1. Momentum conservation is used to design systems like jet engines and turbines.
  2. The study of particle collisions in physics relies on momentum conservation to understand subatomic interactions.
  3. Pendulums and Newton’s cradle demonstrate momentum conservation in motion.
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Everyday Scenarios
  1. When a person jumps out of a stationary boat, the boat moves backward due to momentum conservation.
  2. Momentum transfer allows a skateboarder to propel forward by pushing against the ground.
  3. A child sliding on ice throws an object backward to move forward.
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Industrial Applications
  1. Hydraulic presses and hammers use momentum principles to deliver large forces during impact.
  2. Conveyor belts rely on momentum transfer to move goods efficiently.
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Traffic Safety
  1. Analyzing traffic accidents involves calculating the pre- and post-collision momentum to reconstruct events.
  2. Seatbelts help spread the force over time, reducing the momentum change experienced by passengers.
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Sports
  1. Catchers in baseball reduce force by pulling their hands back, increasing the time of momentum transfer.
  2. In soccer, players impart momentum to the ball by kicking it with varying force and direction.
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Jamb(utme) Calculation problems and solutions involving impulse and momentum

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Here are 30 calculation questions on impulse and momentum with their respective solutions:
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1. Calculating Momentum
Question: A 5 kg object is moving with a velocity of 10m/s10m/s. What is its momentum?
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Solution: p=mv=5×10=50kgm/sp = mv = 5 \times 10 = 50kg·m/s
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2. Momentum After Velocity Change
Question: A 3 kg object changes velocity from 4m/s4m/s to 10m/s10m/s. Find the change in momentum.
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Solution: Δp=m(vfinalvinitial)=3×(104)=18kgm/s\Delta p = m(v_{final} - v_{initial}) = 3 \times (10 - 4) = 18kg·m/s
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3. Impulse from Force and Time
Question: A constant force of 20N20N acts on an object for 3s3s. What is the impulse?
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Solution: J=Ft=20×3=60NsJ = F \cdot t = 20 \times 3 = 60N·s
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4. Impulse from Momentum Change
Question: A ball’s velocity changes from 6m/s6m/s to 2m/s2m/s. If its mass is 2kg2kg, find the impulse.
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Solution: J=Δp=m(vfinalvinitial)=2×(26)=8NsJ = \Delta p = m(v_{final} - v_{initial}) = 2 \times (2 - 6) = -8N·s
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5. Force from Impulse
Question: If an impulse of 50Ns50N·s acts over 5s5s, what is the force?
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Solution: F=Jt=505=10NF = \frac{J}{t} = \frac{50}{5} = 10N
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6. Velocity After Impulse
Question: A 4kg4kg object at rest experiences an impulse of 24Ns24N·s. Find its final velocity.
Solution: J=Δp=mvfinal    vfinal=Jm=244=6m/sJ = \Delta p = m \cdot v_{final} \implies v_{final} = \frac{J}{m} = \frac{24}{4} = 6m/s
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7. Impulse with Direction Change
Question: A 1kg1kg ball moving at 5m/s5m/s rebounds at 3m/s3 m/s. What is the impulse?
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Solution: J=m(vfinalvinitial)=1×(35)=8NsJ = m(v_{final} - v_{initial}) = 1 \times (-3 - 5) = -8N·s
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8. Force Over Time
Question: A 10kg10kg object accelerates from 0m/s0m/s to 15m/s15m/s in 5s5s. Find the average force.
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Solution: J=Δp=mΔv=10×15=150Ns,F=Jt=1505=30NJ = \Delta p = m \cdot \Delta v = 10 \times 15 = 150N·s, \quad F = \frac{J}{t} = \frac{150}{5} = 30N
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9. Time for a Given Force
Question: A 25N25N force is applied to change the momentum of an object by 75Ns75N·s. How long does the force act?
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Solution: t=JF=7525=3st = \frac{J}{F} = \frac{75}{25} = 3s
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10. Velocity from Momentum
Question: If the momentum of a 10kg10kg object is 120kgm/s120 kg·m/s, find its velocity.
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Solution: v=pm=12010=12m/sv = \frac{p}{m} = \frac{120}{10} = 12m/s
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11. Final Momentum
Question: A 2kg2kg object is moving at 3m/s3m/s. A force of 6N6N acts for 4s4s. Find the final momentum.
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Solution: J=Ft=6×4=24Ns,Δp=J,pfinal=mvinitial+J=23+24=30kgm/sJ = F \cdot t = 6 \times 4 = 24N·s, \quad \Delta p = J, \quad p_{final} = m \cdot v_{initial} + J = 2 \cdot 3 + 24 = 30kg·m/s
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12. Momentum Conservation
Question: Two carts collide. One is 3kg3kg moving at 5m/s5m/s, and the other is 2kg2 kg moving at 3m/s-3m/s. What is their total momentum?
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Solution: ptotal=(35)+(23)=156=9kgm/sp_{total} = (3 \cdot 5) + (2 \cdot -3) = 15 - 6 = 9kg·m/s
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13. Final Velocity (Elastic Collision)
Question: Two balls collide elastically. A 1kg1kg ball moving at 2m/s2m/s hits a stationary 3kg3kg ball. Find the final velocity of the 3kg3kg ball.
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Solution: vfinal=2m1vinitialm1+m2=2×1×21+3=1m/sv_{final} = \frac{2m_1 v_{initial}}{m_1 + m_2} = \frac{2 \times 1 \times 2}{1 + 3} = 1m/s
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14. Force Acting Over Time
Question: A 10N10N force acts on a 2kg2kg object for 2s2s. What is the change in velocity?
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Solution: J=Ft=10×2=20Ns,Δv=Jm=202=10m/sJ = F \cdot t = 10 \times 2 = 20N·s, \quad \Delta v = \frac{J}{m} = \frac{20}{2} = 10m/s
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15. Finding Impulse
Question: An object of mass 5kg5kg initially moving at 3m/s3m/s is stopped by a force. What is the impulse applied?
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Solution: J=Δp=m(vinalvinitial)=5×(03)=15NsJ = \Delta p = m(v_{inal} - v_{initial}) = 5 \times (0 - 3) = -15N·s
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16. Momentum After Collision
Question: A 6kg6kg object moving at 4m/s4m/s collides with a stationary 3kg3kg object. After the collision, they stick together. What is their velocity?
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Solution:
  • Total momentum before collision: pinitial=(64)+(30)=24kgm/sp_{initial} = (6 \cdot 4) + (3 \cdot 0) = 24kg·m/s
  • Total mass after collision: mtotal=6+3=9kgm_{\text{total}} = 6 + 3 = 9kg
  • Final velocity: vfinal=pinitialmtotal=249=2.67m/sv_{final} = \frac{p_{initial}}{m_{total}} = \frac{24}{9} = 2.67m/s
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17. Momentum in Two Dimensions
Question: A 2kg2kg ball moving east at 3m/s3m/s collides with a 4kg4kg ball moving north at 2m/s2m/s. What is the total momentum?
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Solution:
  • Momentum components: px=23=6kgm/s,py=42=8kgm/sp_x = 2 \cdot 3 = 6kg·m/s, \quad p_y = 4 \cdot 2 = 8kg·m/s
  • Total momentum magnitude: ptotal=px2+py2=62+82=36+64=10kgm/sp_{total} = \sqrt{p_x^2 + p_y^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = 10kg·m/s
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18. Impulse to Stop
Question: A 10kg10kg object moving at 15m/s15m/s is stopped by a force. What is the impulse required?
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Solution: J=Δp=m(vfinalvinitial)=10(015)=150NsJ = \Delta p = m(v_{final} - v_{initial}) = 10 \cdot (0 - 15) = -150N·s
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19. Time to Stop an Object
Question: A 5kg5kg object moving at 20m/s20 m/s is stopped by a force of 25N25N. How long does it take to stop?
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Solution:
  • Impulse required: J=Δp=mv=520=100NsJ = \Delta p = m \cdot v = 5 \cdot 20 = 100N·s
  • Time: t=JF=10025=4st = \frac{J}{F} = \frac{100}{25} = 4s
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20. Force During Collision
Question: A 0.5kg0.5kg ball moving at 10m/s10m/s stops in 0.2s0.2s during a collision. What is the average force?
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Solution:
  • Impulse: J=Δp=mv=0.510=5NsJ = \Delta p = m \cdot v = 0.5 \cdot 10 = 5N·s
  • Force: F=Jt=50.2=25NF = \frac{J}{t} = \frac{5}{0.2} = 25N
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21. Velocity After Force Application
Question: A 3kg3kg object at rest experiences a force of 9N9N for 5s5s. What is its final velocity?
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Solution:
  • Impulse: J=Ft=95=45NsJ = F \cdot t = 9 \cdot 5 = 45N·s $
  • Final velocity: vfinal=Jm=453=15m/sv_{final} = \frac{J}{m} = \frac{45}{3} = 15m/s $
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22. Elastic Collision
Question: A 1kg1kg ball moving at 4m/s4m/s collides elastically with a 2kg2kg ball at rest. Find the velocity of the 1kg1kg ball after the collision.
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Solution:
  • Velocity formula for elastic collision: v1=(m1m2)v1+2m2v2m1+m2v_1' = \frac{(m_1 - m_2)v_1 + 2m_2v_2}{m_1 + m_2} Substituting values: v1=(12)4+2201+2=43=1.33m/sv_1' = \frac{(1 - 2)4 + 2 \cdot 2 \cdot 0}{1 + 2} = \frac{-4}{3} = -1.33m/s
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23. Recoil Velocity
Question: A 50kg50kg person jumps off a 200kg200kg boat with a velocity of 3m/s3m/s. What is the boat’s recoil velocity?
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Solution:
  • Momentum conservation: m1v1+m2v2=0m_1v_1 + m_2v_2 = 0 Solving for v2:v_2: v2=m1v1m2=503200=0.75m/sv_2 = -\frac{m_1v_1}{m_2} = -\frac{50 \cdot 3}{200} = -0.75m/s
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24. Velocity of a System After Explosion
Question: A 6kg6kg object explodes into two parts: 4kg4kg moving at 5m/s5m/s and 2kg2kg moving at vm/svm/s. Find vv.
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Solution:
  • Momentum conservation: 60=45+2v6 \cdot 0 = 4 \cdot 5 + 2 \cdot v Solving for vv: v=452=10m/sv = -\frac{4 \cdot 5}{2} = -10 \, \text{m/s}
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25. Kinetic Energy Loss in Inelastic Collision
Question: A 2kg2kg object moving at 6m/s6m/s collides inelastically with a 4kg4kg object at rest. Find the kinetic energy lost.
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Solution:
  • Initial kinetic energy: KEinitial=12mv2=12262=36JKE_{initial} = \frac{1}{2} m v^2 = \frac{1}{2} \cdot 2 \cdot 6^2 = 36J
  • Final velocity: vfinal=ptotalmtotal=262+4=2m/sv_{final} = \frac{p_{total}}{m_{total}} = \frac{2 \cdot 6}{2 + 4} = 2m/s
  • Final kinetic energy: KEfinal=12mtotalvfinal2=12622=12JKE_{final} = \frac{1}{2} m_{total} v_{final}^2 = \frac{1}{2} \cdot 6 \cdot 2^2 = 12J
  • Energy lost: KElost=3612=24JKE_{lost} = 36 - 12 = 24J
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26. Impulse in a Bouncing Ball
Question: A 1kg1kg ball hits the ground at 5m/s5m/s and rebounds at 4m/s4m/s. What is the impulse?
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Solution: J=m(vfinalvinitial)=1(4(5))=9NsJ = m(v_{final} - v_{initial}) = 1 \cdot (4 - (-5)) = 9N·s
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27. Impulse During a Collision
Question: A 2kg2kg object moving at $ 3m/s collides and stops in
0.1s0.1s. What is the force during the collision?
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Solution: J=Δp=2(03)=6Ns,F=Jt=60.1=60NJ = \Delta p = 2 \cdot (0 - 3) = -6N·s, \quad F = \frac{J}{t} = \frac{-6}{0.1} = -60N
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28. Explosion Momentum Conservation
Question: A 10kg10kg stationary object explodes into two fragments: 6kg6kg moving at 2m/s2m/s and 4kg4kg moving at vv. Find vv.
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Solution:
  • Momentum conservation: 100=(62)+(4v)10 \cdot 0 = (6 \cdot 2) + (4 \cdot v) Solving for ( v ): v=624=3m/sv = -\frac{6 \cdot 2}{4} = -3m/s
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29. Impulse from Average Force
Question: A 5N5N force acts for 4s4s on an object. Find the impulse.
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Solution: J=Ft=54=20NsJ = F \cdot t = 5 \cdot 4 = 20N·s
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30. Final Velocity After Recoil
Question: A 20kg20kg object recoils at 2m/s-2m/s. What is its momentum?
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Solution: p=mv=20(2)=40kgm/sp = m \cdot v = 20 \cdot (-2) = -40kg·m/s
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If you are a prospective Jambite and you think this post is resourceful enough, I enjoin you to express your view in the comment box below. I wish you success ahead. Remember to also give your feedback on how you think we can keep improving our articles and posts.
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