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Physics I 95.141 LECTURE 17 11/8/10

Physics I 95.141 LECTURE 17 11/8/10. Outline/Administrative Notes . Outline Ballistic Pendulums 2D, 3D Collisions Center of Mass and translational motion. Notes HW Review Session on 11/17 shifted to 11/18. Last day to withdraw with a “W” is 11/12 (Friday). Ballistic Pendulum.

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Physics I 95.141 LECTURE 17 11/8/10

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  1. Physics I95.141LECTURE 1711/8/10

  2. Outline/Administrative Notes Outline Ballistic Pendulums 2D, 3D Collisions Center of Mass and translational motion • Notes • HW Review Session on 11/17 shifted to 11/18. • Last day to withdraw with a “W” is 11/12 (Friday)

  3. Ballistic Pendulum • A device used to measure the speed of a projectile. m h M M+m vo v1

  4. Ballistic Pendulum m M+m M vo v1

  5. Ballistic Pendulum M+m h M+m v1

  6. Exam Prep Problem • You construct a ballistic “pendulum” out of a rubber block (M=5kg) attached to a horizontal spring (k=300N/m). You wish to determine the muzzle velocity of a gun shooting a mass (m=30g). After the bullet is shot into the block, the spring is observed to have a maximum compression of 12cm. Assume the spring slides on a frictionless surface. • A) (10pts) What is the velocity of the block + bullet immediately after the bullet is embedded in the block? • B) (10pts) What is the velocity of the bullet right before it collides with the block? • C) (5pts) If you shoot a 15g mass with the same gun (same velocity), how far do you expect the spring to compress?

  7. Exam Prep Problem • k=300N/m, m=30g, M=5kg, Δxmax=12cm • A) (10pts) What is the velocity of the block + bullet immediately after the bullet is embedded in the block?

  8. Exam Prep Problem • k=300N/m, m=30g, M=5kg, Δxmax=12cm • B) (10pts) What is the velocity of the bullet right before it collides with the block?

  9. Exam Prep Problem • k=300N/m, m=30g, M=5kg, Δxmax=12cm • C) (5pts) If you shoot a 15g mass with the same gun (same velocity), how far do you expect the spring to compress?

  10. Collisions • In the previous lecture we discussed collisions in 1D, and the role of Energy in collisions. • Momentum always conserved! • If Kinetic Energy is conserved in a collision, then we call this an elastic collision, and we can write: • Which simplifies to: • If Kinetic Energy is not conserved, the collision is referred to as an inelastic collision. • If the two objects travel together after a collision, this is known as a perfectly inelastic collision.

  11. Collision Review • Imagine I shoot a 10g projectile at 450m/s towards a 10kg target at rest. • If the target is stainless steel, and the collision is elastic, what are the final speeds of the projectile and target?

  12. Collision Review • Imagine I shoot a 10g projectile at 450m/s towards a 10kg target at rest. • If the target is wood, and projectile embeds itself in the target, what are the final speeds of the projectile and target?

  13. Additional Dimensions • Up until this point, we have only considered collisions in one dimension. • In the real world, objects tend to exist (and move) in more than one dimension! • Conservation of momentum holds for collisions in 1, 2 and 3 dimensions!

  14. 2D Momentum Conservation • Imagine a projectile (mA) incident, along the x-axis, upon a target (mB) at rest. After the collision, the two objects go off at different angles • Momentum is a vector, in order for momentum to be conserved, all components (x,y,z) must be conserved.

  15. 2D Momentum Conservation • Imagine a projectile (mA) incident, along the x-axis, upon a target (mB) at rest. After the collision, the two objects go off at different angles

  16. Conservation of Momentum (2D) • Solving for conservation of momentum gives us 2 equations (one for x-momentum, one for y-momentum). • We can solve these if we have two unknowns • If the collision is elastic, then we can add a third equation (conservation of kinetic energy), and solve for 3 unknowns.

  17. Example problem • A cue ball travelling at 4m/s strikes a billiard ball at rest (of equal mass). After the collision the cue ball travels forward at an angle of +45º, and the billiard ball forward at -45º. What are the final speeds of the two balls?

  18. Example Problem II • Now imagine a collision between two masses (mA=1kg and mB=2kg) travelling at vA=2m/s and vB= -2m/s along the x-axis. If mA bounces back at an angle of -30º, what are the final velocities of each ball?

  19. Example Problem II • Now imagine a collision between two masses (mA=1kg and mB=2kg) travelling at vA=2m/s and vB= -2m/s on the x-axis. If mA bounces back at an angle of -30º, what are the final velocities of each ball, assuming the collision is elastic?

  20. Simplification of Elastic Collisions • In 1D, we showed that the conservation of Kinetic Energy can be written as: • This does not hold for more than one dimension!!

  21. Problem Solving: Collisions • Choose your system. If complicated (ballistic pendulum, for example), divide into parts • Consider external forces. Choose a time interval where they are minimal! • Draw a diagram of pre- and post- collision situations • Choose a coordinate system • Apply momentum conservation (divide into component form). • Consider energy. If elastic, write conservation of energy equations. • Solve for unknowns. • Check solutions.

  22. Center of Mass • Conservation of momentum is powerful for collisions and analyzing translational motion of an object. • Up until this point in the course, we have chosen objects which can be approximated as a point particle of a certain mass undergoing translational motion. • But we know that real objects don’t just move translationally, they can rotate or vibrate (general motion)  not all points on the object follow the same path. • Point masses don’t rotate or vibrate!

  23. Center of Mass • We need to find an addition way to describe motion of non-point mass objects. • It turns out that on every object, there is one point which moves in the same path a particle would move if subjected to the same net Force. • This point is known as the center of mass (CM). • The net motion of an object can then be described by the translational motion of the CM, plus the rotational, vibrational, and other types of motion around the CM.

  24. Example F F F F

  25. Center of Mass • If you apply a force to an non-point object, its center of mass will move as if the Force was applied to a point mass at the center of mass!! • This doesn’t tell us about the vibrational or rotation motion of the rest of the object.

  26. Center of Mass (2 particles, 1D) • How do we find the center of mass? • First consider a system made up of two point masses, both on the x-axis. xB mA xA x=0 xB x-axis

  27. Center of Mass (n particles, 1D) • If, instead of two, we have n particles on the x-axis, then we can apply a similar formula to find the xCM.

  28. Center of Mass (2D, 2 particles) • For two particles lying in the x-y plane, we can find the center of mass (now a point in the xy plane) by individually solving for the xCM and yCM.

  29. Center of Mass (3D, n particles) • We can extend the previous CM calculations to n-particles lying anywhere in 3 dimensions.

  30. Example • Suppose we have 3 point masses (mA=1kg, mB=3kg and mC=2kg), at three different points: A=(0,0,0), B=(2,4,-6) and C=(3,-3,6).

  31. Solid Objects • We can easily find the CM for a collection of point masses, but most everyday items aren’t made up of 2 or 3 point masses. What about solid objects? • Imagine a solid object made out of an infinite number of point masses. The easiest trick we can use is that of symmetry!

  32. CM and Translational Motion • The translational motion of the CM of an object is directly related to the net Force acting on the object. • The sum of all the Forces acting on the system is equal to the total mass of the system times the acceleration of its center of mass. • The center of mass of a system of particles (or objects) with total mass M moves like a single particle of mass M acted upon by the same net external force.

  33. Example • A 60kg person stands on the right most edge of a uniform board of mass 30kg and length 6m, lying on a frictionless surface. She then walks to the other end of the board. How far does the board move?

  34. Solid Objects (General) • If symmetry doesn’t work, we can solve for CM mathematically. • Divide mass into smaller sections dm.

  35. Solid Objects (General) • If symmetry doesn’t work, we can solve for CM mathematically. • Divide mass into smaller sections dm.

  36. Example: Rod of varying density • Imagine we have a circular rod (r=0.1m) with a mass density given by ρ=2x kg/m3. x L=2m

  37. Example: Rod of varying density • Imagine we have a circular rod (r=0.1m) with a mass density given by ρ=2x kg/m3. x L=2m

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