Introduction to Physics: test on velocity, energy, kinetic friction | PHYS 1211, Exams of Physics

Download Introduction to Physics: test on velocity, energy, kinetic friction | PHYS 1211 and more Exams Physics in PDF only on Docsity! PHYSICS 1211 EXAM III NAME [(E 7 ***Y ou may use g ~10 m/s’ if you wish****** 1. A 1 kg ball which has a velocity of 10 m/sec collides with two other balls, masses 2 kg and 3 kg. I will call the direction of the incident velocity the +x direction. After they have all finished their collisions the 1 kg ball is moving with speed 5 m/sec but in the opposite direction in which it was originally moving, the 2 kg ball is moving with speed u in the -y direction, and the 3 kg ball is moving in a direction 37° above the +x with a speed v. after a) Determine u and v. Pye 10 eq ™/s P= 0 hyz 75 + 30 4037 Pag Bar Sm 37 2a ee 2--5+2.4v =10 _ 346.20 «067 2K =)d b) Was energy conserved, lost, or gained in this collision? If lost or gained, by how much? E,= arlke 2505 3 is Jit — Ey = bles? +hBx 6.0 tLe 2x S605 = [62,74 - BG= $2.75 geomsh ‘ote: you must do this problem using work and energy concepts. No credit will be given i, you use F=ma and equations of motion for uniform acceleration!} 2. A 20 kg box is sliding on a horizontal floor with a speed of 10 m/sec. The coefficient of kinetic friction between the box and the floor is 4, = 0.2 a) How far will the box slide before coming to rest? + aq Fy 4ON O= [000-405 —> $=2s > KE x2Ox 0" = (000 T we 0 W427 705 b) How much work was done by the friction? W.z- Aoxrs = —/000 J £ c) Suppose that you want the box to stop after sliding 10 m. You decide to achieve this by applying a constant horizontal force F to the box. Tams is F? (Be sure to specify its Mizeetier da w= —-(4otF) 1/0 =-/000 = wes Fx/0 C= CON Coppel dination of velraity ) d) What was the total amount of work (in part c)) done by you and the friction? - (000 T