Collisions in 1- and 2-D Momentum and Energy Conservation Physics 109 Experiment Number 7 2017
Outline Brief summary of Binary Star Experiment Some thoughts on conservation principles Description of the Collision Experiment Collisions in one dimension Perfectly Elastic Partially Elastic Totally Inelastic Collision in two dimensions Elastic collision of two pucks Impulse Other problems based on collisions
Energies from Binary Star Expt. Energy Plot Energy, joules 0.07 0.06 0.05 0.04 0.03 0.02 0.01 0 0 5 Total 10 15 time, 0.1 sec K1 K2 20 V
Energies with Linear Fit Energy Plot Energy, joules 0.07 0.06 0.05 0.04 0.03 0.02 0.01 0 5 Total 10 15 time, 0.1 sec K1 K2 20 V
More about Momentum Re-write the second law, F = ma = m V/ t Since m does not change, we can bring it inside the Delta symbol: F = mv/ t = change in momentum, P = mv with time. So, another expression of the second law is as follows: F = P/ t. Remember that P is a vector quantity! (which is why I made it bold)
So we change focus.. Before we were dealing with how to describe motion.. Now we are facing issues related to the conservation of energy and momentum. The Binary Stars experiment introduces these concepts But, collisions really brings conservation issues to the forefront!
Example of Elastic Collisions Helium atoms at room temperature (slowed 2 x 109 times)
Perfectly Elastic Collisions Momentum and Energy are Conserved in the Collision. Equal masses, with one initially stationary After collision, the first mass stops, and the second mass moves with the same velocity as the first mass.
Partially Elastic Collisions In this case, Q (energy lost to heat etc.) is not zero. The assumption of momentum conservation is tested. The masses of the two cars are not identical so we need to measure both final velocities
Totally Inelastic Collisions In this case, the two masses stick together upon colliding. The final velocity is given by the ratio of the first mass to the sum of the two masses, times the initial velocity
Elastic Collisions in Two Dimensions x A B y A B Puck A Puck Air Table B
Two Dimensional Elastic Collision This case is like the one-dimensional case The two masses now move in two dimensions. Assume conservation of momentum Sum of the x components of the initial velocities is equal to the sum of the x components of the final velocities. Same for y velocity components.
A truck hits a stationary car and they stick together... What type of collision is this? A. Elastic B. Partially Elastic C. Inelastic D. None of the above
What is conserved in the prior collision? A. B. C. D. Energy Momentum Both Energy and Momentum Nothing
Impulse Collisions generally occur over a short time period. We are interested in determining the force involved in a collision. This force is called an IMPULSIVE FORCE Impulsive force = Momentum Change/time
Impulse a quantity related to Momentum Impulse is equal to the change in momentum. It is given the symbol J, and thus J = P This also equal to the net force multiplied by the time interval of the event. J = P = Fnet t
Baseball and Bat example Baseball of mass 0.15 kg, approaches plate at -32 m/s. Bat reverses direction to +38 m/s. Total momentum change is 10.5 kg.m/s The encounter time is 0.5 x 10-3s So the impulse force is 21 kn!
How this event would look:
How we analyze the event:
Center of Mass Definition one dimension: Xcm = Sum of MiXi/Sum of Mi Two dimensions: Xcm as above, Ycm = Sum of MiYi/Sum of Mi Book example: Mass = 2kg X= -2m 2kg 4kg 0m +3m cm Xcm = ((-2)(2) + (0)(2) + (4)(3))/8 = 1m
Center of Mass (cont) You will encounter this in many other contexts. Symmetrical bodies are easy More complicated bodies can be divided into pieces, the center of mass determined for each piece, then added up Calculus will simplify this even more...
Problems Problems from the text Parachutist Car and wall Train cars Sled and children A collision Perfectly elastic collision Totally inelastic collision Truck and car problems A croquet ball problem
A 65 kg parachutist hits the ground at 35 Km/hr and stops in 140 ms. How does the impulsive force compare with his weight? A. Less than. C. Greater than. B. Equal D. Cannot be determined.
Given the speed, mass of the car and a maximum for the impulsive force, how do we calculate the minimum time for the car to stop? t? A. ½ mv2/force C. mv/force V M B. mgx/force D. None of the above.
What is the speed of the cars after they couple? 7 Mph, 56T. 2.6 Mph, 31 T? A. More than 7 Mph B. Less than 2.6 Mph C. Between 2.6 and 7 Mph D. Cannot be determined
What was lost in the collision? 7 Mph, 56T. 2.6 Mph, 31 T? A. Momentum B. Kinetic Energy C. Potential Energy D. Heat
A child and sled of 33 kg mass is sliding horizontally at 10 m/s. Another child jumps on with negligible speed and speed drops to 6.4 m/s. What principle do we use to solve this problem? X kg 33kg 10 m/s 33kg 6.4 m/s A. Conservation of energy B. Conservative forces. C. Conservation of momentum D. Conservation of mass
Solution to sled question Using conservation of momentum: 33(10) = (33+x)(6.4) 6.4x = 118.8 x = 18.56 kg
An object, A, strikes another, B, which is initially at rest. The collision reverses the direction of object A. Which statement is correct? A. B. C. D. Object A is more massive. Object B is more massive. Impossible to determine relative masses. None of the above.
In a perfectly elastic collision: A. B. C. D. Momentum is conserved. Energy is conserved. Both of the above. Neither of the above.
In a totally inelastic collision: A. Momentum is conserved. B. Energy is conserved. C. Both of the above. D. Neither of the above
A truck of mass 9M moves with a speed of v = 15km/hr and collides with a parked car of mass M. The collision is elastic. What happens? A. The truck stops and the car is propelled backwards. B. Both vehicles continue in the same direction at the same speed. C. Both vehicles continue in the same direction at different speeds. D. None of the above.
Solution to previous question Momentum conserved: 9M(15) = 9M(V1f) + M(V2f) Energy conserved: 1/2(9M)(15^2)=1/2(9M)(V1f)^2 + 1/2(M) (V2f)^2 Working thru the math gives: V1f = 12 km/hr and V2f = 27 km/hr
Reverse the collision a car of mass M collides with a parked truck of mass 9M, at a speed of 15 km/hr. Now what happens? A. The car stops, and the truck moves off in the same direction as the car initial dir. B. Both vehicles move on in the same direction, but the car is slower. C. The car rebounds and moves backward, and the truck moves more slowly in the same direction as the car initial dir. D. None of the above.
Solution to previous question Conservation of Momentum: 15M = M(V1f) + 9M(V2f) Conservation of Energy: 1/2M(15)^2 = 1/2M(V1f)^2 + 1/2(9M)(V2f)^2 Working through the math: V2f = 3 km/hr and V1f = -12 km/hr
You and a cat You (10Mcat) and a cat (M) are moving with equal momentum. What is the relationship between your kinetic energies? A. They are equal B. You have higher KE C. The cat has higher KE D. You cannot tell
You and a cat solution Your mass is 10 times the cat mass. This means that the cat has to run 10 times faster to have the same momentum (you) 10mv = (cat) m(10v) = P Your KE is ½ x 10 x v^2 = 5v^2 The cat KE is ½ x 1 x (10v)^2 = 50v^2 Clearly the cat has the higher KE.
A moving croquet ball strikes a stationary one of equal mass. The incident ball goes off at 30 deg. to its original direction. In what direction does the other ball move? θ 30o A. 0 deg. B. 30 deg. C. 60 deg. D. 75 deg.
Discussion of the previous problem: Momentum: v1 (initial) = v1(final) + v2 (final) Energy: (v1i)2 = (v1f)2 + (v2f)2 After several steps (see pages 136-7 in the text) 2v1fv2f cos (θ + 30) = 0 So, cos (θ + 30) = 0, and θ = 60o (We will test this using the air table and pucks. It probably will not work. Why?)