Lab 04 - Lab Collisions_Tumani_Younge

PhET Simulation: Collisions Name________________________________________________Date _________ This lab activity consists of three Parts Elastic Collision in one Dimension, inelastic collision in one dimension and collision in two dimensions. To be familiar with simulation setting and controllers used to set the velocity, momentum, mass, kinetic energy using PhET simulation open the following link and play with it. https://phet.colorado.edu/sims/html/collision-lab/latest/collision-lab_en.html Objectives 1- Study collision in one dimension and collision in two dimensions. 2- Calculate the momentum and kinetic energy conservation in elastic and inelastic collisions. 3- Calculate the coefficient of restitution. Theory The following experiment explores the conservation of momentum and energy in a closed physical system. In this lab, we will see in practice how the conservation of momentum and total energy relate various parameters (masses, velocities) of the system independently of the nature of the interaction between the colliding bodies. Momentum: For a single object, momentum is defined as the product of the mass and velocity of an object. It is a vector quantity, possessing a magnitude and a direction. If m is an object's mass and v is its velocity (also a vector quantity), then the object's momentum is: P = m V ……… Eq. 1 Conservation of momentum: The conservation of momentum states that the total momentum of a system is constant if the net external force acting on the system is zero (in equation form p i = p f ). When collisions occur the forces between objects are internal forces and do not affect the total momentum of the system (it does affect the individual momentums of the objects however). Thus, the conservation of momentum can also be stated this way: The total momentum before a collision is equal to the total momentum after a collision (or p i = p f where p i is the momentum before the collision and p f is the momentum after the collision). For two objects:

p i = p 1i + p 2i & p f = p 1f + p 2f thus p i = p f turns into p 1i + p 2i = p 1f + p 2f and since p = m v then: m 1 V 1i + m 2 V 2i = m 1 V 1f + m 2 V 2f Elastic collision: In perfectly elastic collisions objects bounce of one another when they collide. In this type of collision both momentum and kinetic energy are conserved (or p i = p f and K i = K f ). The momentum is given by: P = m V The kinetic energy is given by K = 1 2 mv 2 Coefficient of restitution will be ⅇ = − v 1 f − v 2 f v 1 i − v 2 i Inelastic collision: A perfectly inelastic collision is one in which the objects stick together and move as a single unit after the collision. In this type of collision momentum is conserved, but kinetic energy is not conserved.

Table 1(b) Different mass m 1 = 1.5 kg and m 2 =2.5 kg; m 1 – moving to the right, m 2 – stationary V 1i (m/s) V 2i (m/s) V 1f (m/s) V 2f (m/s) P 1i (kgm/s) P 2i (kg.m/s) P 1f (kg.m/s) P 2f (kg.m/s) 3.00 .01 -.74 2.25 4.50 .03 -1.11 5.63 Table 1(c) Different mass m 1 = 1.5 kg and m 2 =2.5 kg; head on collision, object 2 is moving to the left. V 1i (m/s) V 2i (m/s) V 1f (m/s) V 2f (m/s) P 1i (kgm/s) P 2i (kg.m/s) P 1f (kg.m/s) P 2f (kg.m/s) P i (kg.m/s) P f (kg.m/s) K i (J) K i1 + K i2 K f (J) K f1 + K f2 4.95 4.95 P i (kg.m/s) P f (kg.m/s) K i (J) K i1 + K i2 K f (J) K f1 + K f2 4.53 4.52

2.03 -1.69 -2.62 1.10 3.05 -4.22 -3.93 2.75 P i = P 1 i + P 2 i , P f = P 1 f + P 2 f 1. Attach all formulas and calculations for one row of all tables and fill tables 1(a), 1(b) and 1(c). 2. Calculate the ratios K f /K i and P /P i for all tables . What do you observe? Explain. 3. Calculate the coefficient of restitution Part 2: Inelastic collision in one dimension Table 2(a) m 1 =2.0 kg and m 2 =3.0 kg; m 1 – moving to the right, m 2 – stationary V 1i (m/s) V 2i (m/s) V 1f (m/s) V 2f (m/s) P 1i (kgm/s) P 2i (kg.m/s) P 1f (kg.m/s) P 2f (kg.m/s) P i (kg.m/s) P f (kg.m/s) K i (J) K i1 + K i2 K f (J) K f1 + K f2 -1.17 -1.18 P i (kg.m/s) P f (kg.m/s) K i (J) K i1 + K i2 K f (J) K f1 + K f2

Table 3(a) Elastic m 1 =1.5 kg and m 2 =2.5 kg V 1xi (m/s) V 2xi (m/s) V 1xf (m/s) V 2xf (m/s) P 1xi (kg.m/s) P 2xi (kg.m/s) P 1xf (kg.m/s) P 2xf (kg.m/s) P xi (kg.m/s) P xf (kg.m/s) V 1yi (m/s) V 2yi (m/s) V 1yf (m/s) V 2yf (m/s) P 1yi (kg.m/s) P 2yi (kg.m/s) P 1yf (kg.m/s) P 2yf (kg.m/s) P yi (kg.m/s) P yf (kg.m/s) P i (kg.m/s) K i (J) P f (kg.m/s) K f (J)

Table 3(b) Inelastic m 1 =1.5 kg and m 2 =2.5 kg V 1xi (m/s) V 2xi (m/s) V 1xf (m/s) V 2xf (m/s) P 1xi (kg.m/s) P 2xi (kg.m/s) P 1xf (kg.m/s) P 2xf (kg.m/s) P xi (kg.m/s) P xf (kg.m/s) V 1yi (m/s) V 2yi (m/s) V 1yf (m/s) V 2yf (m/s) P 1yi (kg.m/s) P 2yi (kg.m/s) P 1yf (kg.m/s) P 2yf (kg.m/s) P yi (kg.m/s) P yf (kg.m/s) P i (kg.m/s) K i (J) P f (kg.m/s) K f (J) P i = √ p x ⅈ 2 + P y ⅈ 2 , P f = √ p fx 2 + P fy 2 , Ki= K i1 + K i2, , Kf= K if + K 2f Analysis part 3: 1.Attach all formulas and calculations for the first row and fill the data table 3a and 3b. 2.Calculate the ratios K f /K and P /P i for the two tables 3(a) and 3(b). What do you observe?