An electron and a proton are moving under the influence of mutual forces. In calculating the change in the kinetic energy of the system during motion, one ignores the magnetic force of one on another. This is because,

(a) The two magnetic forces are equal and opposite, so they produce no net effect.

(b) The magnetic forces do no work on each particle.

(c) The magnetic forces do equal and opposite (but non-zero) work on each particle.

(d) The magnetic forces are necessarily negligible.

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Payal Gupta | Contributor-Level 10

8 months ago

This is a multiple choice answer as classified in NCERT Exemplar

(b) When electron and proton are moving under influence of their mutual forces the magnetic forces will be perpendicular to their motion hence no work is done by these forces.

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Two blocks M₁ and M₂ having equal mass are free to move on a horizontal frictionless surface. M₂ is attached to a massless spring as shown in Fig. 6.10. Initially M₂ is at rest and M₁ is moving toward M₂ with speed v and collides head-on with M₂.

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(c) When M1 comes in contact with the spring. M1 is retarded by the spring force and M2 is accelerated by the spring force.

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(b) conserving energy between “O” ans ”A”

U_i+ K_i= U_f+ K_f

0+1/2mv²= mgh + 1/2mv’

$\frac{{(v')}^{2}}{2} = \frac{v^{2}}{2} = - g h$

(v’)²=v²-2gh = v^’= $\sqrt[]{v^{2} - 2 g h}$ ……….1

Let speed after emerging be v₁then

=1/2mv₁²=1/2[1/2mv’²]

1/2m(v₁)²=1/4m(v’)²=1/4m[v²-2gh]

V₁= $\sqrt[]{\frac{v^{2}}{2} - g h}$ ………….2

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$\frac{v'}{v_{1}} = \frac{\sqrt[]{v^{2} - 2 g h}}{\sqrt[]{v^{2} - 2 g h / \sqrt[]{2}}} = \sqrt[]{2}$

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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 =-