Suppose that before collision, slider A (m_a = 200 g) is traveling at 3.6 m/s an

Question

Suppose that before collision, slider A (m_a = 200 g) is traveling at 3.6 m/s and slider B (m_b = 210g) is at rest. What is the total momentum hcfe collision1 (Find I he momentum of each slideT before the collision and then add them together) After the collision, presume that the sliders move m the same direction with v^j- 0.398 m/s and = 3.05 m/s. What is the total momentum following the collision? Find the momentum of each slider after the collision and combine them appropriately.) What is the total kinetic energy before collision? find the kinetic energy of each slider before the collision and add them together.) What is the total kinetic energy after collision? Is this an elastic collision? Why or why re^1 (Look in Chapter 7 of the tent to find what is meant by an elastic collision. What quantities are conserved in an elastic collision?)

Explanation / Answer

a)Pi= PAi + PBi = MAVAi + MBVBi = 0.200*3.6 + 0.210*0 = 0.72 kg*m/s

b)Pf= PAf + PBf = MAVAf + MBVBf= 0.200*0.398^2 + 0.210*3.05 = 0.7201 kg*m/s

c)KEi = KEAi + KEBi = 1/2MAVAi2 + 1/2MBVBi2 = 1/2* 0.200*3.6^2 + 1/2* 0.210*0^2= 1.296 J

d)KEf= KEAf + KEBf = 1/2MAVAf2 + 1/2MBVBf2 = 1/2* 0.200*0.398^2 + 1/2* 0.210*3.05^2= 0.9926 J

e) Since KEf not equal to KEi , collision is not elastic.

In elastic collision KE must conserved.

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