Work Energy & Power

Ans: 18

Sol: Here P = 1 J/s , m = 2 kg

$\displaystyle P \times t = W = \Delta K$

$\displaystyle P \times t = \frac{1}{2} m v^2$

$\displaystyle 1 \times t = \frac{1}{2} (2) v^2$

$\displaystyle t = v^2 $

$\displaystyle v = \sqrt{t} $

$\displaystyle \frac{ds}{dt} = \sqrt{t} $

$\displaystyle \int_{0}^{s}ds = \int_{0}^{9} \sqrt{t} dt $

$\displaystyle s = [\frac{t^{3/2}}{3/2}]_{0}^{9}$

$\displaystyle s = \frac{2}{3} \times 27 = 18 m$

Ans: 10

Sol: Applying conservation of momentum along x-axis

$\displaystyle m u = m v_1 cos\theta_1 + 10 m v_2 cos\theta_2 $

$\displaystyle u = v_1 cos\theta_1 + 10 v_2 cos\theta_2 $ …(i)

Applying conservation of momentum along y-axis

$\displaystyle 0 = m v_1 sin\theta_1 – 10m v_2 sin\theta_2 $ ….(ii)

According to question ,

$\displaystyle \frac{1}{2} m v_1^2 = \frac{1}{2} (\frac{1}{2} m u^2) $

$\displaystyle v_1 = \frac{u}{\sqrt{2}}$

As collision is Elastic .

$\displaystyle \frac{1}{2}(10 m)v_2^2 = \frac{1}{2} (\frac{1}{2} m u^2) $

$\displaystyle v_2 = \frac{u}{\sqrt{20}} = \frac{u}{2 \sqrt{5}}$

From (ii)

$\displaystyle m v_1 sin\theta_1 = 10m v_2 sin\theta_2 $

$\displaystyle (\frac{u}{\sqrt{2}}) sin\theta_1 = 10 (\frac{u}{2 \sqrt{5}}) sin\theta_2 $

$\displaystyle sin\theta_1 = \sqrt{10} sin\theta_2 $

n = 10

Ans: 10

Sol: Applying Conservation of Energy ,

Total Mechanical Energy at A = Total Mechanical Energy at P

$\large m g h_1 = \frac{1}{2}m v^2 + m g h_2$

$\large \frac{1}{2}m v^2 = m g h_1 – m g h_2 $

$\large K.E = m g (h_1 – h_2) $

$\large K.E = 1 \times 10 (2 – 1) $

K.E = 10 J

Ans: (c)

Sol: After connecting the pipe the change in energy is equal to the loss in Potential Energy

$\large \Delta U = U_i -U_f $

$\large \Delta U = [(S x_1 d)\frac{x_1}{2} + (S x_2 d)\frac{x_2}{2}]g – [2 \times S (\frac{x_1 + x_2}{2} d )(\frac{x_1 + x_2}{4})]g $

$\large \Delta U = \frac{S d g}{4}[\frac{x_1^2}{2} + \frac{x_2^2}{2}-\frac{(x_1+ x_2)^2}{4}] $

$\large \Delta U = \frac{S d g}{4} [2 x_1^2 + 2x_2^2 -x_1^2 -x_2^2 -2x_1 x_2] $

$\large \Delta U = \frac{1}{4}g d S (x_2 -x_1)^2 $

Ans: (d)

Sol: Initial K.E of the block , $\large K.E_i = \frac{1}{2}m v^2$

After first collision ,

$\large m v = (m+m)v_1$

$\large v_1 = \frac{v}{2}$

After 2nd collision with mass 2m ,

$\large 2m v_1 = (2m + 2m)v_2 $

$\large 2m (\frac{v}{2}) = (2m + 2m)v_2 $

$\large v_2 = \frac{v}{4}$

After 3rd collision with mass 4m ,

$\large 4m v_2 = (4m + 4m)v_3 $

$\large 2m (\frac{v}{4}) = (2m + 2m)v_3 $

$\large v_3 = \frac{v}{8}$

After final collision with mass 8m ,

$\large 8m v_3 = (8m + 8m)v_4 $

$\large 8m (\frac{v}{8}) = (8m + 8m)v_4 $

$\large v_4 = \frac{v}{16}$

Final Kinetic Energy $\large K.E_f = \frac{1}{2}(16m)v_4^2 $

$\large K.E_f = \frac{1}{2}(16m)(\frac{v}{16})^2 $

$\large K.E_f = \frac{1}{16}(\frac{1}{2}m v^2 ) $

Percentage Change in K.E is $\large = \frac{K_i – K_f}{K_i}\times 100 $

$\large = (1- \frac{K_f}{K_i})\times 100 $

$\large = (1-\frac{1}{16})\times 100 $

= 94 %