A bob of mass m suspended by a light string of length L is whirled into a vertical circle as shown in Fig. 6.11. What will be the trajectory of the particle if the string is cut at
This is a short answer type question as classified in NCERT Exemplar
At B the velocity of B is vertically downward, therefore when string is cut at B then it fall downwards.
At C velocity along horizontally right, so when we cut it at C then it will move to right. But under the action of gravity its path becomes parabola.
At X when we cut it moves tangentially in forward direction. So under of the action of gravity it also follows parabola path.
<p><span data-teams="true">This is a short answer type question as classified in NCERT Exemplar</span></p><p>At B the velocity of B is vertically downward, therefore when string is cut at B then it fall downwards.</p><p>At C velocity along horizontally right, so when we cut it at C then it will move to right. But under the action of gravity its path becomes parabola.</p><p>At X when we cut it moves tangentially in forward direction. So under of the action of gravity it also follows parabola path.</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1745464799phpcgpvLc_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1745464799phpcgpvLc.jpeg" alt="" width="182" height="159"></picture></div></div>
This is a multiple choice answer as classified in NCERT Exemplar
(b) conserving energy between “O” ans ”A”
Ui + Ki = Uf + Kf
0+1/2mv2= mgh + 1/2mv’
(v’)2=v2-2gh = v’= ……….1
Let speed after emerging be v1 then
=1/2mv12=1/2[1/2mv’2]
1/2m(v1)2=1/4m(v’)2=1/4m[v2-2gh]
V1=………….2
From eqn 1 and 2
So v1 = v’/=v2(v’/2)
v1>v’/2
hence after emerging from the target velocity of the bullet is more than half of its earlier velocity v’
(d) as the velocity of the bullet changes to v’ which is less than v’ hence , path
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This is a multiple choice answer as classified in NCERT Exemplar
(b) conserving energy between “O” ans ”A”
Ui + Ki = Uf + Kf
0+1/2mv2= mgh + 1/2mv’
(v’)2=v2-2gh = v’= ……….1
Let speed after emerging be v1 then
=1/2mv12=1/2[1/2mv’2]
1/2m(v1)2=1/4m(v’)2=1/4m[v2-2gh]
V1=………….2
From eqn 1 and 2
So v1 = v’/=v2(v’/2)
v1>v’/2
hence after emerging from the target velocity of the bullet is more than half of its earlier velocity v’
(d) as the velocity of the bullet changes to v’ which is less than v’ hence , path, followed will change and the bullet reaches at point B instead of A
(f) as the bullet is passing through the target the loss in energy of the bullet is transferred to particles of the target . therefore their internal energy increases.
This is a multiple choice answer as classified in NCERT Exemplar
(b, d) When a man of mass m climbs up the staircases of height L, work done by the gravitational force on the man is mgl work done by internal muscular forces will be mgL as the change in kinetic is almost zero.
Hence total work done =-mgL + mgL=0
As the point of application of the contact forces does not move hence work done by reaction forces will be zero.
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