VA HW 10 PHYS 250 5pr x G A 2.50 kg particle has the x C consider A Rocket That

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VA HW 10 PHYS 250 5pr x G A 2.50 kg particle has the x C consider A Rocket That x f Facebook-Log In or so x C physics question Ichego x C ponses/submit?dep 3636499 03 greater than ess than the same as Additional Materials Seuliuri 9 -2 points HRW 109 Pa74 My Notes Ask Two 8 kg bodies, A and H, collide The velocities before the collision are V (13i 31 i) m/s and v (-15i q.ojy mfs. After the collision, v (-4.oi 18j) m/s. (d) Whal is e final velocily or B? m/s (b) What is the change in the total kinetic energy (including sig Additional Materials section 98 Submit Answer Save Progress 4. 2/2 points I Previous Answers HRW10 9 075 My Notes Ask A projectile proton with a speed of 390 m/s collides elastically with a target proton initially at rest. The twn protons then move along perpendicular paths, with the projectile path at 44 from the original di 648 PM a V a Type here to search 1/20/2017

Explanation / Answer

Using law of conservation of momentum

along X-axis

momentum before collision = momentum aftr collision

m1*u1x + m2*u2x = m1*v1x + m2*v2x

m1 = m2,hence cancels on both sides

13 - 15 = -4 + V2x

V2x = 2 m/sec

along y-axis

m1*u1y + m2*u2y = m1*v1y + m2*v2y

m1 = m2,hence cancels on both sides

31 + 9 = 18 + V2y

V2y = 22 m/sec

so v = (2 i + 22 j) m/sec

b) change in total kinetic energy is dK = Kf-Ki

initial kinetic energy is ki = 0.5*m1*u1^2 + 0.5*m2*u2^2 = (0.5*1.8*(13^2+31^2)) + (0.5*1.8*(15^2+9^2)) = 1294.4 J

Final kinetic energy Kf = (0.5*m1*v1^2)+(0.5*m2*v2^2) = (0.5*1.8*(4^2+18^2)) + (0.5*1.8*(2^2+22^2))

Kf = 745.2 J

dK = 745.2-1294.4 = -549.2 J

negative sign represents there is a loss in energy

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