## A body of mass 2 kg is driven by an engine delivering a constant Power of 1 J/s . The body starts from rest …..

Q: A body of mass 2 kg is driven by an engine delivering a constant Power of 1 J/s . The body starts from rest and moves in a straight line . After 9 sec , the body has moved a distance (in m) …..

Click to See Solution :
Ans: 18
Sol: Here P = 1 J/s , m = 2 kg

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

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

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

$\displaystyle t = v^2$

$\displaystyle v = \sqrt{t}$

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

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

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

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

## A particle of mass m is moving along x-axis with initial velocity u i . It collides elastically with a particle of mass 10 m at rest ….

Q: A particle of mass m is moving along x-axis with initial velocity $u \hat{i}$. It collides elastically with a particle of mass 10 m at rest and then moves with half its initial kinetic energy (as shown) If $\large sin\theta_1 = \sqrt{n} sin\theta_2$ . Then value of n is …..

Click to See Solution :
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

## A particle (m =1 kg) slides down a frictionless track (AOC) starting from rest at a point A (height 2 m) . …

Q: A particle (m =1 kg) slides down a frictionless track (AOC) starting from rest at a point A (height 2 m) . After reaching C , the particle continues to move freely in air as a projectile . When it reaches its highest point P (height 1 m) , the kinetic energy of particle in Joule is …. (take g = 10 m/s2)

Click to See Solution :
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

## Two identical cylindrical vessels are kept on the ground and each contain the same liquid of density d …..

Q: Two identical cylindrical vessels are kept on the ground and each contain the same liquid of density d . The area of the base of both vessel are S but height of liquid in one vessel is x1 and in other x2 . When both cylinders are connected through a pipe of negligible volume very close to the bottom , the liquid flows from one vessel to other until it comes to equilibrium at a new height . The change in energy in the process is

(a) $\large \frac{3}{4}g d S (x_2 -x_1)^2$

(b) $\large g d S (x_2 -x_1)^2$

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

(d) $\large g d S (x_2 + x_1)^2$

Click to See Solution :
Ans: (c)

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

$\large \Delta U = U_i -U_f$

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

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

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

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

## Blocks of mass m ,2m , 4m and 8m are arranged in a line on a frictional floor .Another block of mass m , moving with speed v along the same line ….

Q: Blocks of mass m , 2m , 4m and 8m are arranged in a line on a frictional floor . Another block of mass m , moving with speed v along the same line as shown collides with mass m in perfectly inelastic manner . All the subsequent collisions are perfectly inelastic . By the time the last block of mass 8m starts moving the total energy loss is p% of the original energy . The value of p is close to

(a) 87

(b) 37

(c) 77

(d) 94

Click to See Solution :
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 %