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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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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 • A device used to measure the speed of a projectile. m h M M+m vo v1
Ballistic Pendulum m M+m M vo v1
Ballistic Pendulum M+m h M+m v1
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?
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?
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?
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?
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.
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?
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?
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!
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.
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
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.
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?
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?
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?
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!!
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.
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!
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.
Example F F F F
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.
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
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.
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.
Center of Mass (3D, n particles) • We can extend the previous CM calculations to n-particles lying anywhere in 3 dimensions.
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).
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!
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.
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?
Solid Objects (General) • If symmetry doesn’t work, we can solve for CM mathematically. • Divide mass into smaller sections dm.
Solid Objects (General) • If symmetry doesn’t work, we can solve for CM mathematically. • Divide mass into smaller sections dm.
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
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