Transcript Lecture 17
Physics I 95.141 LECTURE 17 11/8/10 95.141, F2010, Lecture 17 Department of Physics and Applied Physics Outline/Administrative Notes • Outline – Ballistic Pendulums – 2D, 3D Collisions – Center of Mass and translational motion 95.141, F2010, Lecture 17 Department of Physics and Applied Physics • Notes – HW Review Session on 11/17 shifted to 11/18. – Last day to withdraw with a “W” is 11/12 (Friday) Ballistic Pendulum • A device used to measure the speed of a projectile. m M vo 95.141, F2010, Lecture 17 Department of Physics and Applied Physics M+m v1 h Ballistic Pendulum m M vo 95.141, F2010, Lecture 17 Department of Physics and Applied Physics M+m v1 Ballistic Pendulum M+m h M+m v1 95.141, F2010, Lecture 17 Department of Physics and Applied Physics 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? 95.141, F2010, Lecture 17 Department of Physics and Applied Physics 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? 95.141, F2010, Lecture 17 Department of Physics and Applied Physics 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? 95.141, F2010, Lecture 17 Department of Physics and Applied Physics 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? 95.141, F2010, Lecture 17 Department of Physics and Applied Physics Collisions • In the previous lecture we discussed collisions in and the role of Energy in collisions. – Momentum always conserved! psystem psystem 1D, – If Kinetic Energy is conserved in a collision, then we call this an elastic collision, and we can write: m Av A mB v B m AvA mB vB – Which simplifies to: 1 1 1 1 m Av 2A m B v B2 m Av A2 m B v B2 2 2 2 2 v A vB vB vA – 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. 95.141, F2010, Lecture 17 Department of Physics and Applied Physics 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? 95.141, F2010, Lecture 17 Department of Physics and Applied Physics 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? 95.141, F2010, Lecture 17 Department of Physics and Applied Physics 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! 95.141, F2010, Lecture 17 Department of Physics and Applied Physics 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 A , B • Momentum is a vector, in order for momentum to be conserved, all components (x,y,z) must be conserved. 95.141, F2010, Lecture 17 Department of Physics and Applied Physics 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 A , B 95.141, F2010, Lecture 17 Department of Physics and Applied Physics Conservation of Momentum (2D) • Solving for conservation of momentum gives us 2 equations (one for x-momentum, one for ymomentum). • 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. 95.141, F2010, Lecture 17 Department of Physics and Applied Physics 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? 95.141, F2010, Lecture 17 Department of Physics and Applied Physics 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? 95.141, F2010, Lecture 17 Department of Physics and Applied Physics 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? mAv A mBvB mAvA cos A mBvB cos B vA 3.26 m s vB 0.82 m s 95.141, F2010, Lecture 17 Department of Physics and Applied Physics 0 m AvA sin A mB vB sin B B 63.8 Simplification of Elastic Collisions • In 1D, we showed that the conservation of Kinetic Energy can be written as: v A vB vB vA • This does not hold for more than one dimension!! 95.141, F2010, Lecture 17 Department of Physics and Applied Physics Problem Solving: Collisions 1. Choose your system. If complicated (ballistic pendulum, for example), divide into parts 2. Consider external forces. Choose a time interval where they are minimal! 3. Draw a diagram of pre- and post- collision situations 4. Choose a coordinate system 5. Apply momentum conservation (divide into component form). 6. Consider energy. If elastic, write conservation of energy equations. 7. Solve for unknowns. 8. Check solutions. 95.141, F2010, Lecture 17 Department of Physics and Applied Physics 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! 95.141, F2010, Lecture 17 Department of Physics and Applied Physics 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. 95.141, F2010, Lecture 17 Department of Physics and Applied Physics Example F F F F 95.141, F2010, Lecture 17 Department of Physics and Applied Physics 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. 95.141, F2010, Lecture 17 Department of Physics and Applied Physics 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. x=0 mA xB xA xB 95.141, F2010, Lecture 17 Department of Physics and Applied Physics x-axis 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. 95.141, F2010, Lecture 17 Department of Physics and Applied Physics 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. 95.141, F2010, Lecture 17 Department of Physics and Applied Physics Center of Mass (3D, n particles) • We can extend the previous CM calculations to nparticles lying anywhere in 3 dimensions. 95.141, F2010, Lecture 17 Department of Physics and Applied Physics 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). 95.141, F2010, Lecture 17 Department of Physics and Applied Physics 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! 95.141, F2010, Lecture 17 Department of Physics and Applied Physics CM and Translational Motion • The translational motion of the CM of an object is directly related to the net Force acting on the object. MaCM Fext • 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. 95.141, F2010, Lecture 17 Department of Physics and Applied Physics 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? 95.141, F2010, Lecture 17 Department of Physics and Applied Physics Solid Objects (General) • If symmetry doesn’t work, we can solve for CM mathematically. – Divide mass into smaller sections dm. dm r 95.141, F2010, Lecture 17 Department of Physics and Applied Physics Solid Objects (General) • If symmetry doesn’t work, we can solve for CM mathematically. – Divide mass into smaller sections dm. xCM xCM 1 M xdm 1 M x dm i i i yCM 1 M 95.141, F2010, Lecture 17 Department of Physics and Applied Physics ydm zCM 1 zdm M 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 95.141, F2010, Lecture 17 Department of Physics and Applied Physics 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 95.141, F2010, Lecture 17 Department of Physics and Applied Physics