Problem Solving Sample Problem A 0.23kg case was dropped accidentally on the floor breaking into three pieces. The first piece has a mass of 0.12kg found in the +y-direction with a speed of 1.5m/s and the second one flew to the x-direction with a speed of 1.3m/s and a mass of 0.03kg. Find a) the mass of the third piece and b) the horizontal and vertical component of its velocity. Get the momentum of each piece in the given. p = mv pi = (0.12kg) (1.5m/s) = 0.18 kg.m/s along the + y-axis p2 = (0.03kg) (1.3m/s) = 0.039 kg.m/s along the + x-axis Given: Piece Mass (kg) Horizontal Vertical component of component of Momentum Momentum (kg.m/s) (kg.m/s) 0.12 0 0.18 2 0.03 0.039 O a) Mass of the third piece m3 =0.23kg - 0.12kg - 0.03kg = 0.08kg b) using the conservation of momentum, we can get the horizontal and vertical component of momentum. The plate fell to the ground which means it has no horizontal momentum before the interaction with the floor. Conservation of horizontal component 0 = 0.039kg.m/s +P3x P3x = -0.039 kg.m/s -0.039 kg . m/s Pax = 2= -0.49m/s 0. 08kg Conservation of vertical component 0 =0.18kg.m/s + pay pay = 0.18 kg.m/s -0. 18kg . m/s V3) = = -2.25m/s 0. 08kg The values suggest that the piece travelled to the third quadrant of the cartesian plane. Practice Problem An explosion occurred breaking an object into three parts. The mass of the object is lkg. the first piece with a mass of 0.36kg traveled to the -y-direction at a speed of 2.5m/s while the second piece moved at a speed of 4m/s with a mass of 0.5kg to the + x-direction. Find a) the mass of the third piece and b) the horizontal and vertical component of its velocity.Problem Solving Sample Problem (Inelastic collision) Two objects A with a mass of 3.5kg and B with a mass of 5.5kg are approaching each other with velocities 4.5m/s and 7.2m/s respectively. Find the a) velocity after the impact assuming that this is a perfectly inelastic collision and b) the KE lost during the collision. Given: mA = 3.5kg ms = 5.5kg VA = 4.5m/s VB = -7.2m/s a) Find the velocity. mv, + m2v2 = (m, + mz)V2 (3.5kg)(4.5m/s) + (5.5kg)(-7.2m/s) = (3.5kg + 5.5kg)v 15.75kg . m/s + (-39.6kg . m/s) = (3.5kg + 5.5kg)v -23.85kg . m/s = (99)v -2. 65m/s = v the negative sign means that the objects are moving to the left or -x-direction. DJ LOST KE KE before = ma(vA) , ma(vs) (3.5kg)(4.5m/s) (5.5kg)(-7.2m/s) 70.88 285.12 2 = 35.44 + 142.56 = 178/ KE after = ma + is(v) 3.5kg + 5.5kg (-2.65m/s) 63.2 2 2 = =31.61 KE lost = 1781 - 31.6J = 146.4J Sample Problem (Elastic collision) Problem above is changed into an elastic collision. A) how much kinetic energy is lost? B) how fast each mass is moving after, colliding? Given: mA = 3.5kg ms = 5.5kg VA = 4 5m/s vn = -7.2m/s al no KE is lost because it is an elastic collision b) velocity of each object (3.5kg) (4.5 ) + (5.6kg) (-7.2-") = 3.5kgvaz + 5.6kgvg: 15.75kg. m/s + (-40.32kg. m/s) = 35kgve + 5.6kgvps (-24.57ky.m/$) = 3 5kgv.2 + 5.6kgVN You need to find the coefficient of restitution in an elastic collision is 1, This is needed to get the values of vaz 4.5m/s -(-7.2m/s) 10.7 -142 + Una = 11.7m/s +ing = 11.7m/8 + VAZ Substitute the answer from the I to the equation of law of conservation of momentum we can get the answers for the two velocities (-24.57kg.m/s) = 3.5kgv2 + 5.6kg(11.7m/s + VA2) (-24.57kg.m/s) = 3.5kgvaz + 5.6kg(11.7m/s+ VA2) (-24.57kg.") = 3.5kgvx2 + 65.52kg. m/s + 5.6VAZ (-91.44kg. " ) = 8.8VAZ -10.39kg. m/s = VAZ VB2 = 11. 7m "+ (-10.39) = 1.31m/s The values show that object a moved to the left and object b moved to the right Practice Problem 1. Two objects A with a mass of 4.7kg and B with a mass of 3.8g are approaching each other with velocities 5.5m/s and 9.8m/s respectively. Find the a) velocity after the impact assuming that this is a perfectly inelastic collision and b) the KE lost during the collision. 2. Problem above is changed into an elastic collision. A) how much kinetic energy is lost? B) how fast each mass is moving after colliding?Problem Solving Sample Problem A school service with a mass of 2,835kg moving at a speed of 5.3m/s. a) What is the momentum of the bus if there are 30 students with an average mass of 130kg? b) What is its momentum if 12 students got off the bus but continues to move with the same speed and direction? c) What is the impulse of the bus it it has been moving for 30 mins with 30 student and a force of 350N? Given: m of bus = 2835kg m of each student = 130kg v = 5.3m/s 1 = 30min F =350N Solution: Formula: m = m of bus + (# of students) (mass of each student) = 2835kg + 30students (130kg) = 6,735kg a) Find the momentum with 30 students p = mv = (6735kg) (5.3m/s) = 35,695.5kg . m/s b) Momentum with 30 - 12 = 18 students Formula: m = m of bus + (# of students) (mass of each student) = 2835kg + 18 students (130kg) = 5175kg p = mv = (5175kg) (5.3m/s) = 27,427.5kg . m/s cJ Impulse for 30 students in 30 min or 1800s 1 = Ft = (350N) (1800s) = 6.3x105N - s Practice Problem An armored service van with a mass of 3,500kg moving at a speed of 10.3m/s. a) What is the momentum of the van if there are 4 passengers with an average mass of 125kg and 5 sacks with an average mass of 15kg? b) What is its momentum if the 5 sacks are dropped along with 1 passenger but continues to move with the same speed and direction