Q: If the kinetic energy of a body is increased by 300%, its momentum will increase by

Kinetic energy \$k =\frac {P^2}{2m}\$where, P is momentum and m is mass. \$k_1=k, k_2=k+300\$ % of \$k=4k\$\$\Rightarrow \frac {P_2}{P_1}=\sqrt {\frac{k_2}{k_1}}=\sqrt{\frac{4k}{k}}=2\$\$\Rightarrow P_2=P_1 + 100\$% pf \$P_1\$Hence, momentum will increase by 100%.

Q: Certain force acting on a \$20\ kg\$ mass changes its velocity from \$5\ ms^{-1}\$ to \$2\ ms^{-1}\$. Calculate the work done by the force.

Given : Initial velocity \$u = 5\$ \$m/s\$ Final velocity \$v = 2\$ \$m/s\$The work done is equal to the change in the kinetic energy of an object.\$W = \dfrac{1}{2}m(v^2 - u^2)\$\$W = \dfrac{1}{2}\times 20 \times (2^2 - 5^2)\$\$\implies W = -210\$ \$J\$

Q: Find the components of \$\vec{a} = 2\hat{i} + 3\hat{j}\$ along the directions of vectors \$\hat{i} + \hat{j}\$ and \$\hat{i} - \hat{j}\$:

\$\textbf{Step 1: Calculation of component of}\,\, \vec{a}\,\, \text{along}\, \hat{i} + \hat{j} \$Let \$\vec{b} = \hat{i} + \hat{j} \$The unit vector \$\hat{b}\$\$\hat{b} = \dfrac{\hat{i} + \hat{j}}{| \hat{i} + \hat{j}|} = \dfrac{1}{\sqrt{2}} (\hat{i} + \hat{j}) \$ Component of vector \$\vec{a} \$ along \$\vec{b}\$ vector \$ = a \cos \theta \$ \$ = a \dfrac{\vec{a} . \vec{b}}{|ab|} = \dfrac{\vec{a} . \vec{b}}{b} = \vec{a} . \hat{b} \$ To get the vector form, \$(\vec{a} . \hat{b}) \hat{b} \$ \$ = \left [ (2 \hat{i} + 3 \hat{j}) . \dfrac{1}{\sqrt{2}} (\hat{i} + \hat{j}) \right ] \dfrac{1}{\sqrt{2}} (\hat{i} + \hat{j}) \$ \$= \left [\dfrac{1}{\sqrt{2}}(2 + 3) \right ] \dfrac{1}{\sqrt{2}} (\hat{i} + \hat{j}) \$ \$ = \dfrac{5}{2} (\hat{i} + \hat{j}) \$ \$\textbf{Step 2: Component of}\ \vec{a}\,\, \textbf{along}\,\, \hat{i} - \hat{j} \$Let \$\vec{c} = \hat{i} - \hat{j} \$ Unit vector, \$\hat{c} = \dfrac{1}{\sqrt{2}} (\hat{i} - \hat{j}) \$ Component of vector \$\vec{a}\$ along \$\vec{c} \$ vector \$ = (\vec{a} . \hat{c}) \hat{c} \$ \$ = \left [ (2 \hat{i} + 3 \hat{j}) . \dfrac{1}{\sqrt{2}} (\hat{i} - \hat{j}) \right ] \dfrac{1}{\sqrt{2}} (\hat{i} - \hat{j}) \$ \$ = \left [-\dfrac{1}{\sqrt{2}} \right ] \dfrac{1}{\sqrt{2}} (\hat{i} - \hat{j}) \$ \$ = \dfrac{-1}{2} \hat{i} + \dfrac{1}{2} \hat{j} \$

Q: The kinetic energy of an object of mass m, moving with a velocity of \$5 \:ms^{-1}\$ is 25 J. What will be its kinetic energy when its velocity is tripled?

Given, \$velocity \: of \: object, v = 5 \:ms^{-1}\$;\$kinetic \: energy \: of \: an \: object, \: E_K= 25 \: J\$\$\displaystyle E_K=\frac{1}{2}mv^2\$ or \$\displaystyle 25=\frac{1}{2}\times m \times (5)^2\$or \$\displaystyle m =\frac{50}{25} = 2 \: kg \$When the velocity triples,\$\displaystyle E_K=\frac{1}{2}mv^2=\frac{1}{2}\times 2 \times (15)^2= 225 \: J\$

Q: If the linear momentum of a body is increased by 50 %, the Kinetic energy will be increased by

If the initial momentum is \$p_0\$, the initial kinetic energy is \$K_0=\dfrac{p_0^2}{2m}\$When momentum is increased by 50 %, the momentum becomes \$p=p_0+(p_0\times \dfrac{50}{100})=(3/2)p_0\$ Now kinetic energy, \$K=\dfrac{p^2}{2m}=(9/4)\times (p_0^2/2m)=(9/4)K_0\$Thus the percentage increased in kinetic energy \$=\dfrac{K-K_0}{K_0}\times 100=[(9/4)-1]100=125\$ %

Q: Which one of the following statements is not true?

We know that force is equal to rate of change of momentum. \$\implies F = \dfrac{dp}{dt}\$\$\implies F \propto dp\$From the given equation we can easily observed thatlarger the change in momentum in a given time, larger will be the force.So , option (D) is incorrect statement.

Question 1

A ship A is moving Westwards with a speed of 10 km $h^{-1}$ and a ship B 100 km South of A is moving northwards with a speed of 10 km $h^{-1}$. The time after which the distance between them becomes shortest is

Accepted Answer

It is clear from the diagram that the shortest distance between the ship $A$ and $B$ is $PQ$.Here, $Sin\,45^0=\dfrac{PQ}{OQ}$$\implies PQ=100\times \dfrac{1}{\sqrt 2}=50\sqrt 2\,m$Also,$V_{AB}=\sqrt{V_A^2+V_B^2}=\sqrt{10^2+10^2}=10\sqrt 2\,km/h$So, the time taken to reach shortest path is$t=\dfrac{PQ}{V_{AB}}=\dfrac{50\sqrt2}{10\sqrt 2}=5\,h$

Question 2

If the kinetic energy of a body is increased by 300%, its momentum will increase by

Accepted Answer

Kinetic energy $k =\frac {P^2}{2m}$where, P is momentum and m is mass. $k_1=k, k_2=k+300$ % of $k=4k$$\Rightarrow \frac {P_2}{P_1}=\sqrt {\frac{k_2}{k_1}}=\sqrt{\frac{4k}{k}}=2$$\Rightarrow P_2=P_1 + 100$% pf $P_1$Hence, momentum will increase by 100%.

Question 3

The upper half of an inclined plane of inclination $\theta$ is perfectly smooth while the lower half rough. A block starting from rest at the top of the plane will again come to rest at the bottom if the coefficient of friction between the block and the lower half of the plane is given by.

Accepted Answer

Suppose length of plane is L. When block descends the plane, rise in kinetic energy $=$ fall in potential energy $=mg L\sin \theta i$Work done by friction $=\mu \times (reaction)\times $distance $0+\mu(mg \cos\theta )\times (L/2)$$=\mu(mg\cos \theta)\times (L/2)$Now, work done$=$ change in KE$mgL\sin \theta =\mu (mg \cos\theta )\times (L/2)$$\Rightarrow \tan\theta =\mu/2$$\therefore \mu =2\tan\theta$.

Question 4

Certain force acting on a $20\ kg$ mass changes its velocity from $5\ ms^{-1}$ to $2\ ms^{-1}$. Calculate the work done by the force.

Accepted Answer

Given :          Initial velocity     $u = 5$  $m/s$                      Final velocity     $v = 2$  $m/s$The work done is equal to the&#10;change in the kinetic energy of an object.$W = \dfrac{1}{2}m(v^2 - u^2)$$W = \dfrac{1}{2}\times 20 \times (2^2 - 5^2)$$\implies  W = -210$ $J$

Question 5

Find the components of $\vec{a} = 2\hat{i} + 3\hat{j}$ along the directions of vectors $\hat{i} + \hat{j}$ and $\hat{i} - \hat{j}$:

Accepted Answer

$\textbf{Step 1: Calculation of component of}\,\, \vec{a}\,\, \text{along}\, \hat{i} + \hat{j}  $Let $\vec{b} = \hat{i} + \hat{j} $The unit vector $\hat{b}$$\hat{b} = \dfrac{\hat{i} + \hat{j}}{| \hat{i} + \hat{j}|} = \dfrac{1}{\sqrt{2}} (\hat{i} + \hat{j}) $ Component of vector $\vec{a} $ along $\vec{b}$ vector $ = a \cos \theta $ $ = a \dfrac{\vec{a} . \vec{b}}{|ab|} = \dfrac{\vec{a} . \vec{b}}{b} = \vec{a} . \hat{b} $ To get the vector form,   $(\vec{a} . \hat{b}) \hat{b} $ $ = \left [ (2 \hat{i} + 3 \hat{j}) . \dfrac{1}{\sqrt{2}} (\hat{i} + \hat{j}) \right ] \dfrac{1}{\sqrt{2}} (\hat{i} + \hat{j}) $   $= \left [\dfrac{1}{\sqrt{2}}(2 + 3) \right ] \dfrac{1}{\sqrt{2}} (\hat{i} + \hat{j}) $ $ = \dfrac{5}{2} (\hat{i} + \hat{j}) $ $\textbf{Step 2: Component of}\ \vec{a}\,\, \textbf{along}\,\, \hat{i} - \hat{j} $Let $\vec{c} = \hat{i} - \hat{j} $ Unit vector, $\hat{c} = \dfrac{1}{\sqrt{2}} (\hat{i} - \hat{j}) $ Component of vector $\vec{a}$ along $\vec{c} $ vector $ = (\vec{a} . \hat{c}) \hat{c} $ $ = \left [ (2 \hat{i} + 3 \hat{j}) . \dfrac{1}{\sqrt{2}} (\hat{i} - \hat{j}) \right ] \dfrac{1}{\sqrt{2}} (\hat{i} - \hat{j}) $     $ = \left [-\dfrac{1}{\sqrt{2}} \right ] \dfrac{1}{\sqrt{2}} (\hat{i} - \hat{j}) $ $ = \dfrac{-1}{2} \hat{i} + \dfrac{1}{2} \hat{j} $

Question 6

The kinetic energy of an object of mass m, moving with a velocity of $5 \:ms^{-1}$ is 25 J. What will be its kinetic energy when its velocity is tripled?

Accepted Answer

Given, $velocity \: of \: object, v = 5 \:ms^{-1}$;$kinetic \: energy \: of \: an \: object, \: E_K=  25 \: J$$\displaystyle E_K=\frac{1}{2}mv^2$ or $\displaystyle 25=\frac{1}{2}\times m \times  (5)^2$or $\displaystyle m =\frac{50}{25} = 2 \: kg $When the velocity triples,$\displaystyle E_K=\frac{1}{2}mv^2=\frac{1}{2}\times 2 \times (15)^2=  225 \: J$

Question 7

If the linear momentum of a body is increased by 50 %, the Kinetic energy will be increased by

Accepted Answer

If the initial momentum is $p_0$, the  initial kinetic energy is $K_0=\dfrac{p_0^2}{2m}$When momentum is increased by 50 %, the momentum becomes $p=p_0+(p_0\times \dfrac{50}{100})=(3/2)p_0$ Now kinetic energy, $K=\dfrac{p^2}{2m}=(9/4)\times (p_0^2/2m)=(9/4)K_0$Thus the percentage increased in kinetic energy $=\dfrac{K-K_0}{K_0}\times 100=[(9/4)-1]100=125$ %

Question 8

State work-energy theorem. Prove it for a variable force.

Accepted Answer

Work energy theorem -It states that the work done by the net force acting on a body is equal to the changed produced in kinetic energy of the body. Let F be the variable force $\therefore$ Work done by this variable force , $W=\int_{x_i}^{x_f} F\cdot dx$where $x_i$ is the initial position and $x_f$ is final position.Also Kinetic energy of an object , $K=\dfrac 12 mv^2$ $\implies \dfrac{dK}{dt}=mv \dfrac{dv}{dt}$$\implies \dfrac{dK}{dt}= ma\dfrac{dx}{dt}$$\implies \dfrac{dK}{dt}F\dfrac{dx}{dt}$$\implies dK=Fdx$$\implies \int _{K_i}^{K_f}dK=\int_{x_i}^{x_f} F\cdot dx$$\implies \Delta K= W$Where $\Delta K $ is the change in kinetic energy

Question 9

Which one of the following statements is not true?

Accepted Answer

We know that force is equal to rate of change of momentum. $\implies F = \dfrac{dp}{dt}$$\implies  F \propto dp$From the given equation we can easily observed thatlarger the change in momentum in a given time, larger will be the force.So , option (D) is incorrect statement.

Question 10

State and derive work energy theorem.

Accepted Answer

Work energy theorem states that the change in kinetic energy of an object is equal to the net-work done on it by the net force.Let us suppose that a body is initially at rest and a force. $\vec F $is applied on the body to displace it through$ d \vec  s$along the direction of the force. Then, a small amount of work done is given bydw = $\vec F . d \vec s $ = FdsAlso, according to Newton's second law of motion, we haveF = ma where a is acceleration produced (in the direction of force) on applying the force. Therefore, dw = Mada = $M \dfrac{dv}{dt} ds$Now, work done by the force in order to increase its velocity from u (initial velocity) to v (final velocity) is given byW = $\int_u^v Mvdv $ = $M \int_u^v vdv$     = $M|\dfrac{v^2}{2}|_u^v$W = $\dfrac{1}{2} Mv^2 - \frac{1}{2}Mu^2$Hence, work done on a body by a force is equal to the change in its kinetic energy.