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Work, Energy, and Power physics ncert solutions class 11th Physics Class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Six

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.

(e) The bullet will fall vertically downward after hitting the target.

(f) The internal energy of the particles of the target will increase.

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Payal Gupta | Contributor-Level 10

4 months ago

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

...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.

less

This is a multiple choice answer as classified in NCERT Exemplar(b) conserving energy between “O” ans ”A”<div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1745405520phpWT6vAD_480x360.jpeg" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1745405520phpWT6vAD.jpeg" alt="" width="375" height="119"></picture></div></div>Ui + Ki = Uf + Kf0+1/2mv2= mgh + 1/2mv’ <math> <mfrac> <mrow> <mrow> <msup> <mrow> <mrow> <mo>(</mo> <mi>v</mi> <mi>'</mi> <mo>)</mo> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mrow> <msup> <mrow> <mrow> <mi>v</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <mo>=</mo> <mo>-</mo> <mi>g</mi> <mi>h</mi> </math> (v’)2=v2-2gh = v’= <math> <mroot> <mrow> <mrow> <msup> <mrow> <mrow> <mi>v</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mo>-</mo> <mn>2</mn> <mi>g</mi> <mi>h</mi> </mrow> </mrow> <mrow></mrow> </mroot> </math> ……….1Let 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= <math> <mroot> <mrow> <mrow> <mfrac> <mrow> <mrow> <msup> <mrow> <mrow> <mi>v</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <mo>-</mo> <mi>g</mi> <mi>h</mi> </mrow> </mrow> <mrow></mrow> </mroot> </math> ………….2From eqn 1 and 2 <math> <mfrac> <mrow> <mrow> <mi>v</mi> <mi>'</mi> </mrow> </mrow> <mrow> <mrow> <msub> <mrow> <mrow> <mi>v</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> </mrow> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mrow> <mroot> <mrow> <mrow> <msup> <mrow> <mrow> <mi>v</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mo>-</mo> <mn>2</mn> <mi>g</mi> <mi>h</mi> </mrow> </mrow> <mrow></mrow> </mroot> </mrow> </mrow> <mrow> <mrow> <mroot> <mrow> <mrow> <msup> <mrow> <mrow> <mi>v</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mo>-</mo> <mn>2</mn> <mi>g</mi> <mi>h</mi> <mo>/</mo> <mroot> <mrow> <mrow> <mn>2</mn> <mi> </mi> </mrow> </mrow> <mrow></mrow> </mroot> </mrow> </mrow> <mrow></mrow> </mroot> </mrow> </mrow> </mfrac> <mo>=</mo> <mroot> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> <mrow></mrow> </mroot> </math> So v1 = v’/ <math> <mroot> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> <mrow></mrow> </mroot> </math> =v2(v’/2)v1>v’/2hence 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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This is a multiple choice answer as classified in NCERT Exemplar

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