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Use the formula that provided here plus don’t copy from chegg. Explain me step by step would be helpful please 12,13,15,16 12. A 4-kg ball moving to the right at 12 m/s collides elastically with a 2-kg ball moving to the left at 6 m/s. What are the speeds of the 4-kg and the 2-kg balls after the collision? answer is 4 m/s and 10 m

Use the formula that provided here plus don’t copy from chegg. Explain me step by step would be helpful please 12,13,15,16 12. A 4-kg ball moving to the right at 12 m/s collides elastically with a 2-kg ball moving to the left at 6 m/s. What are the speeds of the 4-kg and the 2-kg balls after the collision? answer is 4 m/s and 10 m

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Use the formula that provided here plus don’t copy from chegg. Explain me step by step would be helpful please 12,13,15,16 12. A 4-kg ball moving to the right at 12 m/s collides elastically with a 2-kg ball moving to the left at 6 m/s. What are the speeds of the 4-kg and the 2-kg balls after the collision? answer is 4 m/s and 10 m/s 13. A 2-kg block is placed at the top of a 5-m high, frictionless ramp and is released from rest. At the bottom of the ramp the 2-kg block collides with a 1-kg block that is initially at rest. The blocks stick together. What is the velocity of the blocks after the collision? answer is 6.6m/s 15. A 3-kg box is moving at 6.0 m/s atop a hill that is 5.0 m above a flat frictionless surface. It descends the hill (also frictionless) and at the bottom of the hill contacts an ideal spring, which compresses 1.53 m before stopping. What is the spring constant k ? answer is 172n/m 16. A 2-kg box is released from rest from the top of a 5-m-high frictionless ramp. At the bottom of the ramp the box has an elastic collision with an 8-kg box that is initially at rest. The 2-kg box bounces off the 8-kg box and goes back up the ramp. To what height does the 2-kg block return up the ramp before coming to rest? answer is 1.8m
As a student at NJIT I will conduct myself in a professional manner and Will comply wIth the proVislons 8f the NJIT Academic Honor Code. I also understand that I must subscribe to the following pledge : On my honor, I pledge that I have not violated the provisions of the NJIT Academic Honor Code. The exam is closed book and closed notes. v? Circular motion: a, = R Pavr = At Weight: Fg= mg,; g = 9.8 m/s?; Fnet = ma; 1 Kinetic energy: K = m v; Potential energy: Ug = mgy E = Er E = U+K 2 1 rev = 2n rad; v2 = o'r; Rotational motion: v = o r; a, = r a = a r; 0 = 0o + at; 1 0 = 0o t+ . 20a = m? - 0o? T=r x F; T= rFsin%; K= 2 mR 1 Icy= -mR 2 Στ- Ια; Ipoint mass = mr? Idisk = 2 + out 1 1 Irod(end) = 3 Ihoop = mR? Irod(center) = -mL? Iball = mR2 Ishell = mR? 12 I= Icom + MD² work: W=t0; 2 W = 2 K= dW P = W Pavr = At 2 dt Rolling: 1 -m Vcom ? Vcom = Ro K= T= f,R Fsmax = H,Fn Incline: Fg= mgsin0 Fg mgcos® Angular momentum: Lpoint mass =m rxy L= mrvsin 0; L=m (r,Vy - ryV3)k L=Io LI = L I 01 =I 202 m,X1+m2X2 X com = miyı+m2y2 m, +m2 Y com = m1 +m2
Transcribed Image Text:As a student at NJIT I will conduct myself in a professional manner and Will comply wIth the proVislons 8f the NJIT Academic Honor Code. I also understand that I must subscribe to the following pledge : On my honor, I pledge that I have not violated the provisions of the NJIT Academic Honor Code. The exam is closed book and closed notes. v? Circular motion: a, = R Pavr = At Weight: Fg= mg,; g = 9.8 m/s?; Fnet = ma; 1 Kinetic energy: K = m v; Potential energy: Ug = mgy E = Er E = U+K 2 1 rev = 2n rad; v2 = o'r; Rotational motion: v = o r; a, = r a = a r; 0 = 0o + at; 1 0 = 0o t+ . 20a = m? - 0o? T=r x F; T= rFsin%; K= 2 mR 1 Icy= -mR 2 Στ- Ια; Ipoint mass = mr? Idisk = 2 + out 1 1 Irod(end) = 3 Ihoop = mR? Irod(center) = -mL? Iball = mR2 Ishell = mR? 12 I= Icom + MD² work: W=t0; 2 W = 2 K= dW P = W Pavr = At 2 dt Rolling: 1 -m Vcom ? Vcom = Ro K= T= f,R Fsmax = H,Fn Incline: Fg= mgsin0 Fg mgcos® Angular momentum: Lpoint mass =m rxy L= mrvsin 0; L=m (r,Vy - ryV3)k L=Io LI = L I 01 =I 202 m,X1+m2X2 X com = miyı+m2y2 m, +m2 Y com = m1 +m2
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