2. A 10 kg ball moving to the right with a speed of 6.0 m/s collides...

4.3 Ratings (830 Votes)

m1=10kg, m2=20kg, v1i=6m/s , v2i=-4m/s

a) By law of conservation of momentum,

m1v1i+m2v2i=m1v1f+m2v2f ------------(1)

since inelastic collision,

m1v1i+m2v2i=(m1+m2)vf

10*6+20*-4 = (10+20)vfÂ Â Â Â Â => vf = -0.67m/s

b) By eqn (1),

10*6+20*-4 =10*-7+20*v2fÂ Â Â Â Â Â => v2f = 2.5m/s â€¦â€¦â€¦â€¦â€¦â€¦directed to right

c) For perfectly elastic collision KE must be conserved.

1/2m1v1i^2 +1/2m2v2i^2 =1/2m1v1f^2+1/2m2v2f^2

1/2*10*6^2+1/2*20*-4^2 =1/2*10*7^2+1/2*20*2.5^2

340 Â¹ 307.5

Thus this is not perfectly elastic collision

d) F1 = m1(v1f-v1i)/Î”t = 10(-7-6)/0.03 = -4333.3 N â€¦â€¦â€¦â€¦â€¦..directed to left

F2= m2(v2f-v2i)/Î”t = 20(2.5-(-4))/0.03 = 4333.3 N â€¦â€¦â€¦â€¦â€¦..directed to right

e) By law of conservation of momentum,

m1v1ix+m2v2ix=m1v1fx+m2v2fx

10*6+20*-4 =10*-7+20*v2fxÂ Â Â Â Â => v2fx = 2.5m/s â€¦â€¦â€¦â€¦â€¦â€¦directed to right

f) By law of conservation of momentum,

m1v1iy+m2v2iy=m1v1fy+m2v2fy

10*0+20*0 =10*0+20*v2fxÂ Â Â Â Â => v2fx = 0m/s

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