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a small particle of mass M/10 is moving horizontally at a height of 3R/2 from ground with velocity 10 m/s. a perfectly inelastic collission occurs at point P of sphere of mass m placed on smooth horizontal surface. the radius of sphere is R. (M=10kg andR=0.1m) all surfaces smooth.
1. speed of sphere just after collission is
a. approx 5 m/s b. approx. 15 m/s
c. approx 20 m/s
2. speed of sphere just after collission is
a. 27/43 b. 30/43 c. 35/43 d. 40/43
3. angular speed of the sphere just after the collision is
a. 0, b. 2 c. 2.5 d. 3
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UP 0 DOWN 0 0 4
4 Answers
1
3hi rashi ...............
""inelastic""" means the particle stops moving along the radial direction and its velocity is same along the tangential direction ..........................
[why because its completely inelastic and the common tanget is along radial dir.]
its R/2 ABOVE THE CENTER ............
draw a right angle triangle joining the point of collision center and drop a prependicular frm the point of collision to the horizontal radius [hope can visualise] ......
now u get the angle betw the horizontal and tangetial dir. as 60 ............
therefore vel. after collision is 10cos(60) = 5m/s ............
part 2 .....
m/10(5/2) + mv = m/10(5) .......
mv =m/10(5/2) ...
v = 1/4 .................. [not machin[2]] ......but ashish it dosnt stick to the shpere so
[M + M/10] is rong ..........
conserving ang. momentum
M/10(10)(R/2) = M/10(5/2)(R/2) + (2/5MR2)W .......
(1/10)(5/2)(1/2) = 2/5R(W) ..........
1/8 = 2/5RW .....
50/16 = W ..[AGAIN NOT MACHIN [2]]