1conserve mometum bout d line of imapct
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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4 Answers
1hey m tryin bt if u wish u can luk at d solvd egs of this chap frm HCV same solvd eg mein hai
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]]