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  1. 1. UNIT 4 TOPIC 1: FURTHER MECHANICS Part 1: MOMENTUM Prepared by: Pn Siti Fatimah Saipuddin INTEC
  2. 2. OBJECTIVES <ul><li>Able to express equation p = mv </li></ul><ul><li>Apply the principle of conservation of momentum, m 1 u 1 + m 2 u 2 = m 1 v 1 + m 2 v 2 </li></ul><ul><li>Special cases in collisions and energy, explosions </li></ul><ul><li>Apply the concept of impulse, Ft = mv-mu and force </li></ul>
  3. 3. LINEAR MOMENTUM <ul><li>The (linear) momentum of a body is defined by: </li></ul><ul><ul><li>Momentum = Mass x Velocity </li></ul></ul><ul><li>Exercise 2.1: </li></ul><ul><ul><li>A body A of mass 5 kg moves to the right with a velocity of 4 ms -1 . A body of mass 3 kg moves to the left with a velocity of 8 ms -1 . Calculate: </li></ul></ul><ul><ul><li>The momentum of A [ +20 kg ms -1 ] </li></ul></ul><ul><ul><li>The momentum of B [ -24 kg ms -1 ] </li></ul></ul><ul><ul><li>The total momentum of A and B [ -4 kg ms -1 ] </li></ul></ul><ul><li>  </li></ul>
  4. 4. CONSERVATION OF MOMENTUM <ul><li>Provided that no external forces are acting, it can be assumed that when collision happens between two bodies, the total momentum before collision is the same as that after collision. </li></ul><ul><li>This means that: </li></ul><ul><ul><li>m 1 u 1 + m 2 u 2 = m 1 v 1 + m 2 v 2 </li></ul></ul>
  5. 5. CONSERVATION OF MOMENTUM <ul><li>Exercise 2.2: </li></ul><ul><ul><li>A 2.0 kg object moving with a velocity of 8.0 ms -1 collides with a 4.0 kg object moving with a velocity of 5.0 ms -1 along the same line. If the two objects join together on impact, calculate their common velocity when they are initially moving </li></ul></ul><ul><ul><li>In the same direction [ 6.0 ms -1 ] </li></ul></ul><ul><ul><li>In opposite direction [ -0.67 ms -1 ] </li></ul></ul>
  6. 6. COLLISIONS AND ENERGY <ul><li>Momentum is conserved in a collision. Total energy is also conserved but the kinetic energy might not be conserved. It can be converted to other forms such as sound, work done during plastic deformation, etc. </li></ul><ul><li>In an elastic collision: </li></ul><ul><ul><li>Kinetic energy is conserved </li></ul></ul><ul><ul><li>Linear momentum is conserved </li></ul></ul><ul><ul><li>Energy is conserved </li></ul></ul><ul><li>In a non-elastic collision: </li></ul><ul><ul><li>Kinetic energy is not conserved </li></ul></ul><ul><ul><li>Linear momentum is conserved </li></ul></ul><ul><ul><li>Energy is conserved </li></ul></ul><ul><li>In a completely inelastic collision: </li></ul><ul><ul><li>The objects stick together on impact </li></ul></ul>
  7. 7. COLLISIONS AND ENERGY <ul><li>Exercise 2.3: </li></ul><ul><ul><li>Calculate the KE converted to other forms during the collisions in (a) and (b) of Exercise 2.2 </li></ul></ul><ul><ul><li>KE converted = [ 6J ] </li></ul></ul><ul><ul><li>KE converted = [ 113J ] </li></ul></ul><ul><li>  </li></ul><ul><li>Exercise 2.4: </li></ul><ul><li>  </li></ul><ul><ul><li>A 2.0 kg object moving with velocity 6.0 ms -1 collides with a stationary object of mass 1.0 kg. Assuming that the collision is perfectly elastic, calculate the velocity of each object after the collision. </li></ul></ul><ul><ul><li>[v 1 = 2.0 ms -1 and v 2 = 8.0 ms -1 ] </li></ul></ul>
  8. 8. COLLISIONS AND ENERGY <ul><li>Example: </li></ul><ul><li>Two particles S of mass 30g and T of mass 40g, both travel at the speed of 35 ms -1 in directions at right angles to each other. The two particles collide and stick together. Calculate their speed after the impact. [ 25.0 ms -1 ] </li></ul>T S
  9. 9. EXPLOSIONS <ul><li>An object explodes as a result of some internal forces. As the result, the total momentum of the separate parts will be the same as that of the original body, which is normally zero. </li></ul><ul><li>Exercise 2.4: </li></ul><ul><li>  </li></ul><ul><ul><li>Figure below shows two trolleys A and B initially at rest, separated by a compressed spring. The spring is now released and the 3.0 kg trolley moves with a velocity of 1.0 ms -1 to the right. Calculate: </li></ul></ul><ul><ul><li>The velocity of the 2.0 kg trolley [-1.5 ms -1 ] </li></ul></ul><ul><ul><li>The total KE of the trolleys [3.75J] </li></ul></ul>
  10. 10. IMPULSE AND FORCE <ul><li>If a force, F acts on a body of mass, m for a time, t so that the velocity of the body changes from u to v , then: </li></ul><ul><li>F = (rate of change of momentum) = (mv-mu) </li></ul><ul><li> t </li></ul><ul><li>Ft = mv – mu = impulse </li></ul><ul><li>Exercise 2.5: </li></ul><ul><ul><li>A stationary golf ball is hit with a club which exerts an average force of 80 N over a time of 0.025 s. Calculate: </li></ul></ul><ul><ul><li>The change in the momentum [2.00 kg ms -1 ] </li></ul></ul><ul><ul><li>The velocity acquired by the ball if it has a mass of 0.020 kg [ 100 ms -1 ] </li></ul></ul>
  11. 11. IMPULSE AND FORCE <ul><li>Exercise 2.6: </li></ul><ul><ul><li>Figure below shows how the force acting on a body varies with time. The increase in momentum of the body, measured in Ns as the result of this force acting for four seconds is _________ </li></ul></ul><ul><li>[24 Ns] </li></ul>
  12. 12. FRICTION <ul><li>Friction is the force that opposes the relative motion between two solid surfaces which are in contact. </li></ul><ul><li>The frictional force before relative motion between the surfaces occurs is known as static friction. </li></ul><ul><li>Limiting static friction, F s = μ s R </li></ul><ul><ul><li>The limiting static friction, Fs between two surfaces just before relative motion occurs. </li></ul></ul><ul><ul><li>Independent of the surface area of contact </li></ul></ul><ul><li>Kinetic friction, F k = μ k R </li></ul><ul><ul><li>The friction between two surfaces when there is relative motion between the surfaces </li></ul></ul><ul><ul><li>Independent of the surface area of contact and the relative speed between the surfaces </li></ul></ul><ul><li>The value of coefficient: </li></ul><ul><ul><li>μ k < μ s </li></ul></ul>
  13. 13. SUMMARY <ul><li>Momentum, p = Mass, m x Velocity, v </li></ul><ul><li>Principle of conservation of momentum states that: </li></ul><ul><li>m 1 u 1 + m 2 u 2 = m 1 v 1 + m 2 v 2 </li></ul><ul><li>In an elastic collision: </li></ul><ul><ul><li>Kinetic energy, Linear momentum, and Energy are conserved </li></ul></ul><ul><li>In a non-elastic collision: </li></ul><ul><ul><li>Kinetic energy is not conserved </li></ul></ul><ul><ul><li>Linear momentum and Energy are conserved </li></ul></ul><ul><li>Ft = mv – mu = impulse </li></ul><ul><li>Limiting static friction, F s = μ s R </li></ul><ul><li>Kinetic friction, F k = μ k R </li></ul><ul><li>μ k < μ s </li></ul>
  16. 16. EXAMPLE: <ul><li>A 1500kg car traveling east with the speed of 25 ms -1 collides at an intersection with a 2500kg van traveling north at a speed of 20ms -1 . find the direction and the magnitude of the velocity of the wreckage after the collision, assuming that the collision undergoes perfectly inelastic collision </li></ul>θ = 53.1 ° v f = 15.6 ms -1