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?Problem Solving Sample Problem A woman pushing a 3.5kg cart with a force of 380N along a horizontal surface. The cart moves at around 4.8m What is the final speed of the cart assuming that it came from rest? Given: m = 3.5kg F = 380N d = 4.8m Solution: Find Work: W = Fd = (380N)(4.8m) = 1,824] Find the final velocity using the equation: 2W 2(1,8241) 3.5kg = 32.28m/s 171 Practice Problem The Flash is seen running after a train with a mass of 450,000kg moving at 62m/s?. A) Find how much force is acted on the train. B) Calculate the amount of work he needs to put up with to catch upProblem Solving Sample Problem 1. A spring can be compressed by a force of 62 N by 0.25m, a) Find the force constant of the spring. B) How much force is needed to compress the spring by 0.5m? c) What is the elastic potential energy if it is compressed by 0.8m. a. Solution b. C. Given Given Given F = 62N x = 0.25m k = 248 N/m x = 0.5m k = 248N/m x = 0.8m Formula: Formula Formula F F = kx k= - EPE = -kx' F = (248 N/m) (0.5m) 1 248N 62N F = 124 N k = im )(0.8m) 0.25m EPE = 99 2 J k=248 N/m Practice Problem 1. A spring for the catapult was compressed by a 83 N force to 0.99m. a) Find the force constant of the spring. B) How much force is needed to compress the spring by 2.21m? c) What is the elastic potential energy if it is compressed by 0.2.5m