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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

6.24 A bullet of mass 0.012 kg and horizontal speed 70 m s–1 strikes a block of wood of mass 0.4 kg and instantly comes to rest with respect to the block. The block is suspended from the ceiling by means of thin wires. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block.

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Payal Gupta | Contributor-Level 10

5 months ago

6.24 Mass of the bullet, $m_{b}$ = 0.012 kg

Initial speed of the bullet, u = 70 m/s

Mass of the wooden block , $m_{w}$ = 0.4 kg

Initial speed of the wooden block = 0

Let's assume, final speed of the bullet = v

Applying the law of conservation of momentum

$m_{b} * u + m_{w} * 0 = (m_{b} + m_{w}) v$

Hence v = ( $m_{b} * u) / (m_{b} + m_{w})$ = 2.04 m/s

Let h be the height by which the block rise. Applying law of conservation of energy

Potential energy of the combined bullet + block = Kinetic energy of the combination

$(m_{b} + m_{w}) * g * h =$ (1/2) $* (m_{b} + m_{w}) * v^{2}$

h = $v^{2}$ /2g = 0.212 m

The heat produced = Initial kinetic energy of the bullet – final kinetic energy of the combination

= (1/2) $m_{b}$ $u^{2}$ - (1/2) $* (m_{b} + m_{w}) * v^{2}$

= (1/2

...more

6.24 Mass of the bullet, $m_{b}$ = 0.012 kg

Initial speed of the bullet, u = 70 m/s

Mass of the wooden block , $m_{w}$ = 0.4 kg

Initial speed of the wooden block = 0

Let's assume, final speed of the bullet = v

Applying the law of conservation of momentum

$m_{b} * u + m_{w} * 0 = (m_{b} + m_{w}) v$

Hence v = ( $m_{b} * u) / (m_{b} + m_{w})$ = 2.04 m/s

Let h be the height by which the block rise. Applying law of conservation of energy

Potential energy of the combined bullet + block = Kinetic energy of the combination

$(m_{b} + m_{w}) * g * h =$ (1/2) $* (m_{b} + m_{w}) * v^{2}$

h = $v^{2}$ /2g = 0.212 m

The heat produced = Initial kinetic energy of the bullet – final kinetic energy of the combination

= (1/2) $m_{b}$ $u^{2}$ - (1/2) $* (m_{b} + m_{w}) * v^{2}$

= (1/2) $*$ 0.012 X $70^{2}$ - (1/2) $* (0.012 + 0.4) * {2.04}^{2}$ J

= 29.4 – 0.857 J

= 28.54 J

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This is a multiple choice answer as classified in NCERT Exemplar

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

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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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This is a multiple choice answer as classified in NCERT Exemplar

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