IB Physics 11 Mr. Jean December 9 th, 2013

The plan: Video clip of the day 2D collisions

Conservation of Momentum: 4bMDEAhttp:// 4bMDEA

The vector sum of the momenta is the same before and after a collision. Momentum Vectors

Momentum is conserved even when interacting objects don’t move along the same straight line. To analyze momentum in any direction, we use the vector techniques we’ve previously learned. We’ll look at momentum conservation involving angles by considering three examples. Momentum Vectors

Momentum is a vector quantity. The momentum of the wreck is equal to the vector sum of the momenta of car A and car B before the collision. Momentum Vectors

The momentum of car A is directed due east and that of car B is directed due north. If their momenta are equal in magnitude, after colliding their combined momentum will be in a northeast direction with a magnitude times the momentum either vehicle had before the collision. Momentum Vectors

When the firecracker bursts, the vector sum of the momenta of its fragments add up to the firecracker’s momentum just before bursting. Momentum Vectors

A falling firecracker explodes into two pieces. The momenta of the fragments combine by vector rules to equal the original momentum of the falling firecracker. Momentum Vectors

Momentum is conserved for the high-speed elementary particles, as shown by the tracks they leave in a bubble chamber. Momentum Vectors

Subatomic particles make tracks in a bubble chamber. The mass of these particles can be computed by applying both the conservation of momentum and conservation of energy laws. The conservation laws are extremely useful to experimenters in the atomic and subatomic realms. Momentum Vectors

Glancing Collisions: Glancing collisions: This type of collision takes place in two dimensions. Collisions on a pool table are good examples of this type of collision. Momentum still has to be conserved – it’s the law for crying out loud! –But the problems can become fairly complex. One technique to keep track of what’s going on is to break things into x and y components:

Momentum is still conserved:

Glancing collisions: We will deal with greatly simplified collisions. Basically with right angles and one of the bodies at rest at the beginning. Example: An 8.00 kg mass moving east at 15.0 m/s strikes a 10.0 kg mass that is at rest. The 8.00 kg mass ends up going south at 4.00 m/s. –(a) What is the velocity of the second ball?

Billiard Ball Collisions:

a) We analyze the momentum in the x and y directions. In the x direction:

a) We analyze the momentum in the x and y directions. The x direction: This is because the second body has no initial velocity so it has no initial momentum in either the x or y direction. After the collision, the first body has momentum only in the y direction, so the x direction momentum after the collision involves only the second body.

Now let’s look at the y direction: The first body’s initial motion is only in the x direction, therefore it has no initial momentum in the y direction. The second body is at rest at the beginning so the total initial y direction momentum is zero.

Y-component: Now we can solve for the velocity of the second body using the Pythagorean theorem

So ball #2’s total velocity: Now we find the angle  : Trigonometry is just the thing to find the old angle:

New: Application Problems A spring is compressed a distance of 12 cm by a 675 g ball. The spring constant is 225 N/m. The ball is on a smooth surface as shown (next slide). The spring is released and sets the ball into motion. –(a) the speed of the ball when it leaves the spring, –(b) the speed of the ball just before it leaves the edge of the table, –(c) the kinetic energy of the ball just before it hits the deck, –(d) the horizontal distance, x, it travels when it leaves the table until it hits the deck. Assume that the ball rolls down the ramp. –(e) What is the impact angle? –(f) What is the overall velocity at impact with the deck?

Solutions: A) B)

C) D)

E) F)