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Work, Energy, and Power ncert solutions physics class 11th physics ncert solutions class 11th Physics Class 11th Work, Energy and Power

6.25 Two inclined frictionless tracks, one gradual and the other steep meet at A from where two stones are allowed to slide down from rest, one on each track (Fig. 6.16). Will the stones reach the bottom at the same time? Will they reach there with the same speed? Explain. Given $θ_{1}$ = 30 $°$ , $θ_{2}$ = 60 $°$ , and h = 10 m, what are the speeds and times taken by the two stones ?

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Payal Gupta | Contributor-Level 10

5 months ago

6.25 From the law of conservation of energy,

the potential energy at the top = Kinetic energy at the bottom

mgh = (1/2)m $v_{1}^{2}$ ….(1)

and

mgh = (1/2)m $v_{2}^{2}$ ….(2)

$v_{1}$ = $v_{2}$ , Both the stones will reach with the same speed

For stone 1, the force acting on the stone 1 is given by , $F_{1}$ = m $* a_{1}$ = mg $\sin ? ?_{1}$

$a_{1}$ = g $\sin ? ?_{1}$

For stone 2, $a_{2}$ = g $\sin ? ?_{2}$

As $?_{2} > ?_{1}$ , $a_{2}$ > $a_{1}$

From v = u + at, we get t = v/a

Therefore $t_{2}$ < $t_{1}$ Stone 2 will reach faster than stone 1

From the law of conservation of energy

mgh = (1/2) mv²

v = $\sqrt{2 g h}$ = 14 m/s ( Given h = 10 m)

The time taken by two stones given as

$t_{1}$ &

...more

6.25 From the law of conservation of energy,

the potential energy at the top = Kinetic energy at the bottom

mgh = (1/2)m $v_{1}^{2}$ ….(1)

and

mgh = (1/2)m $v_{2}^{2}$ ….(2)

$v_{1}$ = $v_{2}$ , Both the stones will reach with the same speed

For stone 1, the force acting on the stone 1 is given by , $F_{1}$ = m $* a_{1}$ = mg $\sin ? ?_{1}$

$a_{1}$ = g $\sin ? ?_{1}$

For stone 2, $a_{2}$ = g $\sin ? ?_{2}$

As $?_{2} > ?_{1}$ , $a_{2}$ > $a_{1}$

From v = u + at, we get t = v/a

Therefore $t_{2}$ < $t_{1}$ Stone 2 will reach faster than stone 1

From the law of conservation of energy

mgh = (1/2) mv²

v = $\sqrt{2 g h}$ = 14 m/s ( Given h = 10 m)

The time taken by two stones given as

$t_{1}$ = v/ $a_{1}$ = v/g $\sin ? ?_{1}$ = 14/(9.81 $\sin ? 30 °)$ = 2.85 s

$t_{2}$ = v/ $a_{2}$ = v/g $\sin ? ?_{2}$ = 14/(9.81 $\sin ? 60 °)$ = 1.65 s

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An object is placed beyond the centre of curvature C of the given concave mirror. If the distance of the object is d₁ from C and the distance of the image formed is d₂ from C, the radius of curvature of this mirror is:

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$\Rightarrow (f + d_{1}) (f - d_{2}) = f^{2}$

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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) If spring is massless, the final state of the M₁ is state of rest.

(d) If the surface on which blocks are moving has friction, then collision cannot be elastic.

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This is a multiple choice answer as classified in NCERT Exemplar

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

a) So spring will continue compress until the blocks moves with the same velocity.

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A bullet of mass m fired at 30° to the horizontal leaves the barrel of the gun with a velocity v. The bullet hits a soft target at a height h above the ground while it is moving downward and emerges out with half the kinetic energy it had before hitting the target. Which of the following statements are correct in respect of bullet after it emerges out of the target?

(a) The velocity of the bullet will be reduced to half its initial value.

(b) The velocity of the bullet will be more than half of its earlier velocity.

(c) The bullet will continue to move along the same parabolic path.

(d) The bullet will move in a different parabolic path.

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This is a multiple choice answer as classified in NCERT Exemplar

(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

From eqn 1 and 2

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

So v₁ = v’/ $\sqrt[]{2}$ =v²(v’/2)

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hence after emerging from the target velocity of the bullet is more than half of its earlier velocity v’

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...more

This is a multiple choice answer as classified in NCERT Exemplar

(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

From eqn 1 and 2

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

So v₁ = v’/ $\sqrt[]{2}$ =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.

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A man, of mass m, standing at the bottom of the staircase, of height L climbs it and stands at its top.

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(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.

A cricket ball of mass 150 g moving with a speed of 126 km/h hits at the middle of the bat, held firmly at its position by the batsman. The ball moves straight back to the bowler after hitting the bat. Assuming that collision between ball and bat is completely elastic and the two remain in contact for 0.001s, the force that the batsman had to apply to hold the bat firmly at its place would be

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Payal Gupta

This is a multiple choice answer as classified in NCERT Exemplar

Time of contact =0.001s

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