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: 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: A running man has half the kinetic energy of that of a boy of half of his mass. The man speeds up by \$1m/s\$, so that he has same kinetic energy as that of the boy. The original speed of the man is?

\$\textbf{Step 1: Initial situation}\$Let mass of man in \$M\ kg\$ and speed is \$V\$ , mass of boy is \$M/2\$ and speed of boy is \$V_{1}\$\$K.E_{man} = \dfrac {1}{2} (K.E_{boy})\$\$\dfrac {1}{2} MV^{2} = \dfrac {1}{2} \left (\dfrac {1}{2} \dfrac {M}{2} V_{1}^{2}\right )\$ \$.... (1)\$\$\textbf{Step 2: Final situation}\$Man speeds up by \$1m/s\$ so new speed of man \$=V+1\$\$K.E_{man}=K.E_{boy}\$\$\dfrac {1}{2} M(V + 1)^{2} = \dfrac {1}{2} \left (\dfrac {M}{2}\right ) V_{1}^{2}\$ \$.... (2)\$\$\textbf{Step 3: Divide equation (1) by equation (2)}\$\$\dfrac {V^{2}}{(V + 1)^{2}} = \dfrac {1}{2}\$\$\Rightarrow V = \dfrac {1}{\sqrt {2} - 1}\$Hence option (C) is correct.

Q: A charged particle of mass m and charge q is released from rest in an electric field of constant magnitude E. The kinetic energy of the particle after time t is

A force is generated because of the elective field\$ F = qE \Rightarrow a = \dfrac{qE}{m}.\$As the particle starts from rest and acceleration is constant applying equation of motion.\$ V = u +at \$\$ V = 0+\dfrac{qEt}{m} \$\$ V = \dfrac{qET}{m}\$Then, \$KE \dfrac{1}{2} mv^2 = \dfrac{1}{2}m \left ( \dfrac{qEt}{m} \right )^2 = \dfrac{1}{2}m \dfrac{q^2 E^2 t^2}{m^2} = \dfrac{q^2 E^2 t^2}{2m}\$

Q: A body of mass 5 kg is moving with a momentum of 10 kg m/s. A force of 0.2 N acts on it in the direction of motion of the body for 10 sec. Find the increase in its kinetic energy.

\$t=10\$sec\$10=5\times v [P=mv]\$\$v=2m/s\$As \$F=0.2N\$\$a=\dfrac{0.2}{5}m/s^{2}=\dfrac{1}{25}m/s^{2}[a=\dfrac{F}{m}]\$at \$t=0, KE=\dfrac{1}{2}\times s\times (2)^{2} \quad [KE=\dfrac{1}{2}mv^{2}]\$\$KE_{i}=10J\$Now, at \$t=10\$sec, \$v_{f}=u+at\$\$v_{f}=g+\dfrac{1}{25}\times 10=2.4m/s^{2}\$\$\therefore KE_{f}=\dfrac{1}{2}\times 5\times (2.4)^{2}\$\$=14.4J\$\$\therefore \triangle KE=14.4-10\$\$=4.4J\$

Question 1

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 2

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 3

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 4

A ball whose kinetic energy is E, is thrown at an angle of $45^o$ with the horizontal, its kinetic energy at the highest point of its flight will be?

Accepted Answer

The kinetic energy of the ball at the highest point is as follows:the expression for kinetic energy is $KE=\frac{1}{2} mv^2$.So, the kinetic energy at the highest point is,$KE_h=\dfrac{1}{2} m(v\cos(45^\circ))^2$$KE_h=\dfrac{1}{2} m(v^2)(\cos(45^\circ))^2$$KE_h=E(\cos(45^\circ))^2$$KE_h=\dfrac{E}{2}$so, at the highest point the kinetic energy is $\dfrac{E}{2}$.

Question 5

Calculate the work required to be done to stop a car of $1500\ kg$ moving at the velocity of $60\ km/h$?

Accepted Answer

When a object is moving with a constant velocity, it possess K.E$K.E = \dfrac{1}{2} \times m \times v^{2}$So, in order to bring the object to rest i.e its same magnitude of energy is required so, as the final final energy comes to zero.$v =60 km/hr = 16.66 m/s$So, magnitude of work that needs to be done $ = \dfrac{1}{2} \times m \times v^{2}$Hence, work required to stop the car $= \dfrac{1}{2} \times 1500 \times 16.66^{2}$ $ =208166.7 J = 208.17 kJ$

Question 6

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 7

Derive the formula for kinetic energy.

Accepted Answer

Let an object of mass m, move with uniform velocity u. Let us displace it by s, due to constant force F, acting on it. Work done on the object of mass 'm' is $W=F \times  s$ ...(i) Due to the force, velocity changes to v and the acceleration produced is 'a'. Relationship between v, u, a and s can be given by formula $v^2- u^2 = 2as $$\therefore s=\dfrac{v^2-u^2}{2a}$..........(ii)$F=ma$............(iii)Substituting (ii) and (iii) in (i) we get$W =F \times s$$=ma\times \dfrac{v^2-u^2}{2a}$$W=\dfrac{1}{2}m(v^2-u^2)$If u=0,(object starts at rest)$W=\dfrac{1}{2}mv^2$Work done=Change in kinetic energy$\therefore E_k=\dfrac{1}{2}mv^2$

Question 8

A running man has half the kinetic energy of that of a boy of half of his mass. The man speeds up by $1m/s$, so that he has same kinetic energy as that of the boy. The original speed of the man is?

Accepted Answer

$\textbf{Step 1: Initial situation}$Let mass of man in $M\ kg$ and speed is $V$ , mass of boy is $M/2$ and speed of boy is $V_{1}$$K.E_{man} = \dfrac {1}{2} (K.E_{boy})$$\dfrac {1}{2} MV^{2} = \dfrac {1}{2} \left (\dfrac {1}{2} \dfrac {M}{2} V_{1}^{2}\right )$                                                                                             $.... (1)$$\textbf{Step 2: Final situation}$Man speeds up by $1m/s$ so new speed of man $=V+1$$K.E_{man}=K.E_{boy}$$\dfrac {1}{2} M(V + 1)^{2} = \dfrac {1}{2} \left (\dfrac {M}{2}\right ) V_{1}^{2}$                                                                                     $.... (2)$$\textbf{Step 3: Divide equation (1) by equation (2)}$$\dfrac {V^{2}}{(V + 1)^{2}} = \dfrac {1}{2}$$\Rightarrow V = \dfrac {1}{\sqrt {2} - 1}$Hence option (C) is correct.

Question 9

A charged particle of mass m and charge q is released from rest in an electric field of constant magnitude E. The kinetic energy of the particle after time t is

Accepted Answer

A force is generated because of the elective field$ F = qE \Rightarrow a = \dfrac{qE}{m}.$As the particle starts from rest and acceleration is constant applying equation of motion.$ V = u +at $$ V = 0+\dfrac{qEt}{m} $$ V = \dfrac{qET}{m}$Then, $KE \dfrac{1}{2} mv^2 = \dfrac{1}{2}m \left ( \dfrac{qEt}{m} \right )^2 = \dfrac{1}{2}m \dfrac{q^2 E^2 t^2}{m^2} = \dfrac{q^2 E^2 t^2}{2m}$

Question 10

A body of mass 5 kg is moving with a momentum of 10 kg m/s. A force of 0.2 N acts on it in the direction of motion of the body for 10 sec. Find the increase in its kinetic energy.

Accepted Answer

$t=10$sec$10=5\times v [P=mv]$$v=2m/s$As $F=0.2N$$a=\dfrac{0.2}{5}m/s^{2}=\dfrac{1}{25}m/s^{2}[a=\dfrac{F}{m}]$at $t=0, KE=\dfrac{1}{2}\times s\times (2)^{2} \quad [KE=\dfrac{1}{2}mv^{2}]$$KE_{i}=10J$Now, at $t=10$sec, $v_{f}=u+at$$v_{f}=g+\dfrac{1}{25}\times 10=2.4m/s^{2}$$\therefore KE_{f}=\dfrac{1}{2}\times 5\times (2.4)^{2}$$=14.4J$$\therefore \triangle KE=14.4-10$$=4.4J$

Introduction to gravitational potential energy

Language

Rate

REVISE WITH CONCEPTS

Gravitational Potential Energy

Gravitational Potential Energy- Medium

QUICK SUMMARY WITH STORIES