A water reservoir tank of capacity $250\;m^3$ is situated at a height of $20 \; m$ from the water level. What will be the power of an electric motor to be used to fill the tank in $3$ hours? Efficiency of motor is $70\;\%$

Given

Volume $(v) = 250\; m^3$

Height $(h) = 20 \; m$

Time $(t) = 3 \; hrs = 10800 \; sec$

Efficiency of motor $(\eta) = 70 \%$

Power of the electric motor $(P_{in}) = ?$

We have, $\eta = $$\frac{P_{in}}{P_{out}}$$*100\%$ .......... (i)

 Firstly we have to find ,

$P_{out} = $$\frac{Work\;done}{time}$ ......... (ii)

But,

Work done to fill the tank $W = mgh = \rho * v * g * h = 1000 * 250 * 10 * 20 = 5 * 10^7\;J $

 [Since, density of water in SI system is $1000\;kg/m^3$]

Then from equation (ii), we get

$P_{out} = $$\frac{5 \;* \;10^7}{10800}$ $ = 4.63 * 10^3\; Watt$

Finally from equation (i), we get

$P_{in} = $$\frac{P_{out}}{\eta}$$ = $$\frac{4.63\; * \;10^3}{0.7}$$ = 6614\;Watt$

∴ Power of the electric motor $(P_{in}) = 6614\;Watt$

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A stationary mass explodes into two parts of mass $4 \; kg$ and $40 \; kg$. The initial kinetic energy of larger mass is $10 \; J$. What is the initial kinetic energy of the smaller mass and its velocity.

Given,

Mass ($M_1) = 4\;kg$

Mass ($M_2) = 40\;kg$

Initial kinetic energy of larger mass $(E_2) = 10\;J$

Initial kinetic energy of the smaller mass $(E_1) = \;?$

Velocity of the smaller part $(V_1) = \;?$

We have,

$\frac{E_1}{E_2} = \frac{M_2}{M_1}$

$E_1 = $$\frac{E_2\;*\;M_2}{M_1}$$ = \frac{10 \;*\; 40}{4}$$ = 100\;J$

Now to calculate the velocity of the smaller part,

$E_1 = \frac{1}{2}\;M_1\;V_1^2$

$V_1^2 = \frac{2\;*\;E_1}{M_1}$

⇒ $V_1 = \sqrt{\frac{2\;*\;E_1}{M_1}} = \sqrt{\frac{2\;*\;100}{4}} = 7.07\;m/s  $

∴ Initial kinetic energy of the smaller mass $(E_1) = \;100\;J$

∴ Velocity of the smaller part $(V_1) = \;7.07\;m/s$

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An explosive of mass M placed at a point explodes into one-third and two-third parts. If the initial kinetic energy of the smaller part is $1000 \; J$. What will be the initial kinetic energy of the bigger part?

Given,

Total mass = $M$

Mass of smaller fragment ($m_1 = $$\frac{1}{3}$$M$)

Mass of bigger fragment ($m_2 = $$\frac{2}{3}$$M$)

Kinetic Energy of smaller fragment ($E_1 = 1000\;J$)

Kinetic Energy of bigger fragment ($E_2 = \;?$)

We have,

$\frac{E_2}{E_1} = \frac{m_1}{m_2}$

$E_2 = $$\frac{m_1}{m_2}$$\;*\;E_1 = $$\frac{\frac{1}{3}M}{\frac{2}{3}M}$$\;*\;1000 = 500\;J$

∴ Initial kinetic energy of bigger fragment is $500\;J$

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A ball of mass $4 \; kg$ moving with a velocity $10 \; m/s$ collides with another body of mass $16 \; kg$ moving with $4 \; m/s$ from the opposite direction and then coalesces into a single body. Compute the loss of energy on impact.

Given,

Mass of first ball $(m_1) = 4\;kg$

Initial velocity of first ball $(u_1) = 10\;m/s$

Mass of second ball $(m_2) = 16\;kg$

Initial velocity of second ball $(u_2) = 4\;m/s$

Loss of energy on impact $(∇E) = \;?$

Since ball are moving in opposite direction.

Then, from the principle of conservation of linear momentum,

$m_1\;.\;u_1 \;+\; m_2\;.\;(-\;u_2) = (m_1\; +\; m_2)\;v$               [$v$ is the common velocity]

⇒ $v = $$\frac{m_1\;u_1\; -\; m_2\;u_2}{m_1\; + \;m_2} = \frac{4\;*\;10 \;- \;16\;*\;4}{4\; +\; 16}$ $\;\;\; =\;-\;1.2\;m/s$

[i.e. is in the direction of $m_2$]

Kinetic Energy before collision $E_1 = $$\frac{1}{2}$$\;m_1\;u_1^2\; + \;$$\frac{1}{2}$$\;m_2\;u_2^2$

$= $$\frac{1}{2}$$\;*\;4\;*\;10^2 \;+ \;$$\frac{1}{2}$$\;*\;16\;*\;4^2 = 328\;J $

Kinetic Energy after collision $E_f = $$\frac{1}{2}$$\;(m_1\;+\;m_2)\;v^2$

$= $$\frac{1}{2}$$\;(4\;+\;16)\;*\;(1.2)^2 = 14.4\;J$

Loss of energy on impact (∇E) = Kinetic Energy before collision - Kinetic Energy after collision.

= 328 - 14.4 = 313.6\;J

∴ Loss of energy is 313.6 J.

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A $0.15 \; kg$ glider is moving to the right on a friction less horizontal air track with a speed of $0.8 \; m/s$. It has a ahead on collision with a $0.3 \; kg$ glider that is moving to the left with a speed of $2.2 \; m/s$. Find the final velocity (magnitude and direction) of each glider if the collision is elastic.

Given,

Mass of glider $(M_1) = 0.15\;kg$

Mass of glider $M_2 = 0.3\;kg$

Initial velocity $u_1 = 0.8\;m/s$               [moving to the right]

Initial velocity $u_2 = -\;2.2\;m/s$            [moving to the left]

Final velocity $v_1 = ?$

Final velocity $v_2 = ?$

From the principle of conservation of linear momentum, we have

$M_1\;u_1 + M_2\;u_2 = M_1\;v_1 + M_2\;v_2$

or, $0.15 * 0.8 + 0.3 * (-2.2) = 1.15\;v_1 + 0.3\;v_2$

or, $-0.54 = 0.15\;v_1 + 0.3\;v_2$ .......... (i)

From the question, we know that it is elastic collision.

So, for the elastic collision,

$u_1 + v_1 = u_2 + v_2$

or, $0.8 + v_1 = -\;2.2 + v_2$

or, $v_1 - v_2 = -\;3$ .......... (ii)

Solving equation (i) and (ii), we get

∴ $v_2 = -\;0.2\;m/s$

∴ $v_1 = -\;3.2\;m/s$

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A typical car weights about $1200 \; N$. If the coefficient of rolling friction is $\mu_r = 0.015$. What horizontal force is needed to make the car move with constant speed of $72 \; km/hr$ on a level road? Also calculate the power developed by the engine to maintain this speed.

Given,

Weight of the car $(W) =1200 \; N$

Coefficient of friction $(\mu) = 0.015$

Velocity $(v) = 72\;km/hr = 20\;m/s$

Horizontal force $(F) = ?$

Power developed by the engine $(P) = ?$

Now we have,

$F = F_k + ma$    [Here, $F_k = \mu_r\;.\;mg$ = frictional force]

If the speed is uniform, then a = 0,

$F = \mu_r\;.\;mg = 0.015 * 1200 = 18\;N$

Therefore, horizontal force needed $(F) = 18\;N$

Also, Power developed by the engine $(P) = F * v$

$= 18 * 20 = 360\;N$