A curved surface is shown in Fig. The portion BCD is free of friction. There are three spherical balls of identical radii and masses. Balls are released from rest one by one from A which is at a slightly greater height than C With the surface AB, ball 1 has large enough friction to cause rolling down without slipping; ball 2 has a small friction and ball 3 has a negligible friction.
(a) For which balls is total mechanical energy conserved?
(b) Which ball(s) can reach D?
(c) For balls which do not reach D, which of the balls can reach back A?
This is a long answer type question as classified in NCERT Exemplar
(a) As ball 1 is rolling down without slipping there is no dissipation of energy hence, total mechanical energy is conserved. Ball 3 is having negligible friction hence, there is no loss of energy.
(b) Ball 1 acquires rotational energy, ball 2 loses energy by friction. They cannot cross at C. Ball 3 can cross over.
(c) Ball 1, 2 turn back before reaching C. Because of loss of energy, ball 2 cannot reach back to A. Ball 1 has a rotational motion in “wrong” sense when it reaches B. It cannot roll back to A, because of kinetic friction.
<p><span data-teams="true">This is a long answer type question as classified in NCERT Exemplar</span></p><p><strong> (a)</strong> As ball 1 is rolling down without slipping there is no dissipation of energy hence, total mechanical energy is conserved. Ball 3 is having negligible friction hence, there is no loss of energy.</p><p><strong> (b)</strong> Ball 1 acquires rotational energy, ball 2 loses energy by friction. They cannot cross at C. Ball 3 can cross over.</p><p><strong> (c) </strong>Ball 1, 2 turn back before reaching C. Because of loss of energy, ball 2 cannot reach back to A. Ball 1 has a rotational motion in “wrong” sense when it reaches B. It cannot roll back to A, because of kinetic friction.</p>
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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