Q: The blocks 1 and 2 in the arrangement have mass m each. The strings AB and BC are T_{1} and T_{2} respectively. The system is in equilibrium with a constant horizontal force mg . Then prove \tan \theta_{1} = \dfrac {1}{2} , \tan \theta_{2} = 1 , T_{1} = \sqrt {5} mg , T_{2} = \sqrt {2} mg .

\sum F_x=0\\mg=T_2\sin \theta_2=0\\mg=T_2\sin \theta_2(\theta=45°)\\T_2\sqrt2 mg Ring of man =m Radius =R x-y plane I_0 Uniform external magnetic field of strength \bar{B}=B_0(2i-2j+5k) t=0 B_0=constant \int{Edl}=-\cfrac{dQ}{dt}=\pi r^2\cfrac{dB}{dt} Let \lambda=9/2\pi R be the charge per unit length of the ringThe torque is given by N=\int{\lambda Edl} Angular momentum =I\omega=\int N dt On solving the aboveset of sequential equation R\lambda \pi^2 B=i\omega\\ R\pi R^2 Bq/2\pi R=mR^2\omega\\ \omega=\cfrac{9B}{2m}

Q: When a gun of mass M fires a bullet of mass m , the total energy released in the explosion is found to be E . The kinetic energy of the bullet is:

Using momentum conservation on bullet + gun as a system \Rightarrow initial momentum = Final momentum 0=-MV+mv \Rightarrow mv=MV ..........(1)Now total energy =E \Rightarrow \dfrac{1}{2}MV^{2}+\dfrac{1}{2}mv^{2}=E From (1) V=\dfrac{mv}{M} \Rightarrow\dfrac{1}{2}M\dfrac{m^{2}}{M^2}v^{2}+\dfrac{1}{2}mv^{2}=E \dfrac{1}{2}mv^{2}\left ( \dfrac{m}{M}+1 \right )=E \dfrac{1}{2}mv^{2}=\dfrac{EM}{M+m} \downarrow K.E bullet

Question 1

A rocket of mass  4000\  kg  is set for vertical firing. How much gas must be ejected per second so that the rocket may have initial upwards acceleration of magnitude  19.6\ m{ /s }^{ 2 } ? [Exhaust speed of fuel  =980m/s ]

Accepted Answer

For upward motion of rocket, thrust force must be greater in magnitude than its weight and net force will be equal to their difference, i.e.,  Upward \ \ thrust - Weight = F_{net}             -v_e \dfrac{dm}{dt} - mg = ma  -\dfrac{dm}{dt}=\dfrac{ma + mg}{v_e}  Here, negative sign is implying that mass of gas is decreasing with the passage of time and  v_e=980\ m/s  is the exhaust speed of fuel.Hence  -\dfrac{dm}{dt}=\dfrac{m(a+g)}{v_e} -\dfrac{dm}{dt}=\dfrac{4000(19.6 + 9.8)}{980} 	herefore -\dfrac{dm}{dt} = 120 kg/s

Question 2

With two forces acting at a point, the maximum effect is obtained when their resultant is 4N. If they  act at right angles, then their resultant is 3N. Then the forces are

Accepted Answer

Let P and Q are forces, we know that  R=\sqrt { { P }^{ 2 }+{ Q }^{ 2 }+2PQ\cos { 	heta  }  }  When  	heta ={ 0 }^{ 0 },R=4N=\sqrt { { P }^{ 2 }+{ Q }^{ 2 }+2PQ }  P+Q=4    ...(1)When  	heta ={ 90 }^{ 0 },R=3N { P }^{ 2 }+{ Q }^{ 2 }=9    ...(2)From equation (1)  { \left( P+Q ight)  }^{ 2 }=16\Rightarrow { P }^{ 2 }+{ Q }^{ 2 } + 2PQ=16\ \Rightarrow 9+2PQ=16\Rightarrow 2PQ=7 Now  { \left( P-Q ight)  }^{ 2 }={ P }^{ 2 }+{ Q }^{ 2 }-2PQ \Rightarrow { \left( P-Q ight)  }^{ 2 }=9-7\Rightarrow P-Q=\sqrt { 2 }     ...(3)On solving equation (1) and (3) \displaystyle P=\left( 2+\frac { 1 }{ 2 } \sqrt { 2 }  ight) N  and  \displaystyle Q=\left( 2-\frac { 1 }{ 2 } \sqrt { 2 }  ight) N

Question 3

Two objects A and B, each of mass m,are connected by a light inextensible string. They are restricted to move on a frictionless ring of radius R in a vertical plane. The objects are released from rest at the position shown in figure. Then, the tension in the cord just after release is:

Accepted Answer

directions of 'a' and 'b' are different but the magnitude of 'a' and 'b' are the same.Since the length of the rope AB is same \begin{array}{l} a\cos  45=b\cos  45 \ a=b \ Also\, \, T\sin  45=mb..................\left( 1 ight) \, \, \, for\, \, A \ mg-T\sin  45=ma...............\left( 2 ight) \, \, for\, \, \, B \ 2ma=mg=>a=\dfrac { g }{ 2 }  \ substituting\, \, in\, \, \left( 1 ight)  \ \dfrac { T }{ { \sqrt { 2 }  } } =\dfrac { { mg } }{ 2 }  \ T=\dfrac { { mg } }{ { \sqrt { 2 }  } }  \end{array} Hence,Option  D  is correct.

Question 4

Three identical particles are joined together by a thread as shown in the figure. All the three particles are moving on a smooth horizontal plane about point  Q . If the speed of the outermost particle is  V_o , then the ratio of tensions,  T_1 : T_2 : T_3 , in the three sections of the string is (Assume that the string remains straight).

Accepted Answer

Let angular velocity of system is  w .For ball  c , v_0=3wL w=\dfrac{v_0}{3L} For ball  A , force due to rotation,  F_A=mw^2L=\dfrac{mv_0}{9L} Similarly, F_B=\dfrac{2mv_0}{9L} F_C=\dfrac{mv_0}{3L} So, tension in the string joining  B  and  C  is  \dfrac{mv_0}{3L} T_1=\dfrac{mv_0}{3L} T_2=\dfrac{2mv_0}{9L}+\dfrac{mv_0}{3L}=\dfrac{5mv_0}{9L} So,  T_1:T_2:T_3  will be \dfrac1{3}:\dfrac5{9}:\dfrac2{3} \implies 3:5:6

Question 5

The blocks  1  and  2  in the arrangement have mass  m  each. The strings  AB  and  BC  are  T_{1}  and  T_{2}  respectively. The system is in equilibrium with a constant horizontal force  mg . Then prove  	an 	heta_{1} = \dfrac {1}{2} ,  	an 	heta_{2} = 1 ,  T_{1} = \sqrt {5} mg ,  T_{2} = \sqrt {2} mg .

Accepted Answer

\sum F_x=0\mg=T_2\sin 	heta_2=0\mg=T_2\sin 	heta_2(	heta=45&#176;)\T_2\sqrt2 mg Ring of man  =m Radius =R x-y  plane  I_0  Uniform external magnetic field of strength  \bar{B}=B_0(2i-2j+5k) t=0 B_0=constant \int{Edl}=-\cfrac{dQ}{dt}=\pi r^2\cfrac{dB}{dt} Let  \lambda=9/2\pi R  be the charge per unit length of the ringThe torque is given by N=\int{\lambda Edl} Angular momentum =I\omega=\int N dt On solving the aboveset of sequential equation R\lambda \pi^2 B=i\omega\ R\pi R^2 Bq/2\pi R=mR^2\omega\ \omega=\cfrac{9B}{2m}

Question 6

Two masses of 10 kg and 20 kg are connected by a massless spring. A force of 200 N acts on the 20 kg mass away from the spring. At a certain instant, the acceleration of 10 kg mass is 12 m/s ^{2} . The acceleration of the 20 kg mass at that instant is

Accepted Answer

Let,  a_1, a_2  are the acceleration of masses  m_1, m_2  at instant  t  and let  F  be the force acts on the mass  m_2 .So the equation of motion of the system,  F=m_1a_1+m_2a_2 Given,  m_1=10\ kg, m_2=20\ kg, a_1=12\ m/s^2 , F=200\ N 	herefore   200=10	imes 12+20a_2  20a_2=200-120=80 a_2=4\ m/s^2

Question 7

Two blocks of masses  m  and  M  are placed on a horizontal frictionless table connected by light spring as shown in the figure. Mass  M  is pulled to the right with a force  F . If the acceleration of mass  m  is  a , the acceleration of mass  M  will be

Accepted Answer

For the block with mass  m , the force experienced is the spring force. 	herefore  Spring force  =  Mass  	imes  acceleration =  ma For block with mass  M , forces are spring force and the external  F 	herefore  acceleration  = \dfrac {total \ force}{mass} = \dfrac{F-ma}{M}

Question 8

When a gun of mass  M  fires a bullet of mass  m , the total energy released in the explosion is found to be  E . The kinetic energy of the bullet is:

Accepted Answer

&#10;&#10;&#10;&#9;&#10;&#9;&#10;&#9;&#10;&#9;&#10;&#10;&#10;Using momentum&#10;conservation on bullet + gun as a system \Rightarrow  &#10;initial  momentum =  Final &#10;momentum 0=-MV+mv \Rightarrow mv=MV ..........(1)Now&#10;total energy  =E \Rightarrow&#10;\dfrac{1}{2}MV^{2}+\dfrac{1}{2}mv^{2}=E From (1) V=\dfrac{mv}{M} \Rightarrow\dfrac{1}{2}M\dfrac{m^{2}}{M^2}v^{2}+\dfrac{1}{2}mv^{2}=E \dfrac{1}{2}mv^{2}\left&#10;( \dfrac{m}{M}+1 \right&#10;)=E \dfrac{1}{2}mv^{2}=\dfrac{EM}{M+m} \downarrow&#10; K.E bullet

Question 9

Two balls of equal masses are thrown upwards along the same vertical line at an interval of  2  seconds with the same initial velocity of  39.2ms^{-1} . The total time of flight of each ball, if they collide at a certain height, and the collision is perfectly inelastic, will be

Accepted Answer

When the particles collide the height of both particles will be same,For first particle time will be  t  and for second particle time will be  t-2 h=ut-\dfrac { 1 }{ 2 } g{ t }^{ 2 }=u(t-2)-\dfrac { 1 }{ 2 } g{ (t-2) }^{ 2 } ut-\dfrac { 1 }{ 2 } g{ t }^{ 2 }=ut-2u-\dfrac { 1 }{ 2 } g{ (t^2-4t+4) } 0=-2u+2gt-2g t=\dfrac{u}{g}+1 t=5 s By solving the quadratic equation in  t  we get the time of first collision to be  5 s. h= 39.2 	imes 5 -\dfrac{1}{2}g 	imes5^2=20g- 12.5g =7.5 g  \ meter  Now by momentum conservation, velocity of particles after collision m(u-gt)+m(u-g(t-2)) =mv' mu-5mg+mu-3mg = mv' v' =2u-8g =0 m/s a=-g Now for collision with ground let time taken be  { t }_{ 2 } h=\dfrac{1}{2}gt_2^2 \Rightarrow t= \sqrt{\dfrac{2h}{g}} =\sqrt{\dfrac{2	imes 7.5g}{g}} =\sqrt{15} Thus, times of total time of flights will be  5+\sqrt { 15 }  s and   3+\sqrt { 15 }  s

Question 10

A body is acted on by a force given by  F = (10 + 2t )  N . The impulse received by the body during the first four seconds is :

Accepted Answer

Given force,  F=(10+2t)N Impulse,  J=\int_{0}^{4}F \cdot dt =\int_{0}^{4}(10+2t)\cdot dt=\left[10t+t^2\right]^{4}_{0}=\left[10(4)+4^2\right]=56Ns .

Question 11

Two blocks A and B of masses  2.9 kg  and  1.9 kg  respectively are suspended with the help of two inextensible wires each of the length  1m  from a rigid support as shown in the figure. The mass of the upper wire is negligible and that of the lower wire  0.2 kg/m . This system of block, wires and rigid support is moving upwards with an acceleration of  0.2 m/s^{2} . The tension at the midpoint C and D of the lower and the upper wires will be respectively:

Accepted Answer

For Tension at C:  (M_B+m_{BC})a=T_C-M_Ag  Given:   a=0.2 m/s^2 ,  m_{BC}=\dfrac{1}{2}	imes0.2=0.1 kg , for wire BC. T_C=(1.9+0.1)(0.2+9.8)=20 N For tension at D:  (M_A+M_B+m_{AB})a=T_D-(M_A+M_B+m_w)g  Given:  m_{AB}=0.2 kg   for wire AB. T_D=(2.9+1.9+0.2)(0.2+9.8)=50 N 	herefore T_C=20N	ext{ and } T_D=50N

Question 12

In the arrangement shown in figure above the mass of ball 1 is  \eta=1.8  times as great as that of rod 2. The length of the latter is  l=100\:cm . The masses of the pulleys and the threads, as well as the friction, are negligible. The ball is set on the same level as the lower end of the rod and then released. If t be the time taken by the ball to reach the opposite end of bar, find 5t

Accepted Answer

Suppose, the ball goes up with acceleration  w_1  and the rod comes down with the acceleration  w_2 .As the length of the thread is constant,  2w_1=w_2  (1)From Newton's second law in projection form along vertically upward for the ball and vertically downward for the rod respectively gives, T-mg=mw_1  (2)and  Mg-T^\prime=Mw_2  (3)but  T=2T  (because pulley is massless) (4)From equations (1), (2), (3) and (4) \displaystyle w_1=\frac{(2M-m)g}{m+4M}=\frac{(2-\eta)g}{\eta+4}  (in upward direction)and  \displaystyle w_2=\frac{2(2-\eta)g}{\eta+4}  (downwards)From kinematical equation in projection form, we get \displaystyle l=\frac{1}{2}(w_1+w_2)t^2 as,  w_1  and  w_2  are in the opposite direction.Putting the values of  w_1  and  w_2 , the sought time becomes \displaystyle t=\sqrt{\frac{2l(\eta+4)}{3(2-\eta)g}}=1.4\:s \implies 5t=7

Question 13

In the figure shown  C  is a fixed wedge. A block  B  is kept on the inclined surface of the wedge  C . Another block  A  is inserted in a slot in the block  B  as shown in figure. A light inextensible string passes over a light pulley which is fixed to the block  B  through a light rod. One end of the string is fixed and other end of the string is fixed to  A. S  is a fixed support on the wedge. All the surfaces are smooth. Masses of  A  and  B  are same. Then the magnitude of acceleration of  A  is  \dfrac {\sqrt {x}}{3} m/s^{2} . Then  x  is  (\sin 37^{\circ} = 3/5) .

Accepted Answer

Lets find relation between accelerations of two blocks using string constrain. x+y=L Differentiating twice w.r.t time. a_A+a_B=0 let  a  be the magnitude of acceleration for block  B . 2mg\sin	heta-T=2maightarrow(1) For block  A T-mg\cos	heta=maightarrow(2) From here we can get, T=\dfrac{14mg}{15} acceleration of  A=\dfrac4{3}ms^{-2} acceleration of  B=\dfrac4{3}ms^{-2} So, resultant acceleration of  A  is  a_A=\sqrt{\left(\dfrac{4}{3}ight)^2+\left(\dfrac{4}{3}ight)^2} a_A=\dfrac{\sqrt{32}}{3}ms^{-2}

Question 14

Two uniform rods tied together with the help of a string are balanced as shown in the figure. The minimum coefficient of friction for which the system will remain in equilibrium in the position is  \dfrac{2}{x} . The value of   'x'   is

Accepted Answer

First balance the force in horizontal and vertical direction:T = &#956;N&#956;N .... (i)mg = N ......(ii)From (i) and (ii) we get: T=\mu mg                                  For the rods to be in rest the torque must be zero.Taking torque about point 'P'. 	au =0\ 	au =mg\frac { l }{ 2 } cos{ 37 }^{ 0 }-Tlsin{ 37 }^{ 0 }=0\ \Rightarrow mg\frac { l }{ 2 } cos{ 37 }^{ 0 }=Tlsin{ 37 }^{ 0 } By putting value of T: T=\mu mg     We get: mg\dfrac { l }{ 2 } cos{ 37 }^{ 0 }=\mu mglsin{ 37 }^{ 0 }\ \dfrac { 1 }{ 2 } cos{ 37 }^{ 0 }=\mu sin{ 37 }^{ 0 }\ \Rightarrow \mu =\dfrac { 2 }{ 3 }  \mu =\dfrac { 2 }{ 3 } =\dfrac { 2 }{ x }  Therefore, x=3

Question 15

A small disc  A  is placed on an inclined plane forming an angle  \alpha  with the horizontal (shown in figure above) and is imparted an initial velocity  v_0 . Find how the velocity of the disc depends on the angle  \varphi  if the friction coefficient  k=	an{\alpha}  and at the initial moment  \displaystyle\varphi_0=\frac{\pi}{2} .

Accepted Answer

The acceleration of the disc along the plane is determined by the projection of the force of gravity on this plane  F_x=mg\sin{\alpha}  and the friction force  fr=kmg\cos{\alpha} . In our case  k=	an{\alpha}  and therefore  fr=F_x=mg\sin{\alpha} Let us find the projection of the acceleration on the direction of the tangent to the trajectory and on the x-axis  mw_t=F_x\cos{\varphi}-fr=mg\sin{\alpha}(\cos{\varphi}-1) mw_x=F_x-fr\cos{\varphi}=mg\sin{\alpha}(1-\cos{\varphi}) It is seen from this that  w_t=-w_x , which means that the velocity  v  and its projection  v_x  differ only by a constant value  C  which does not change with time, i.e.  v=v_x+C ,Where  v_x=v\cos{\varphi} . The constant  C  is found from the initial condition  v=v_0 , whence  C=v_0  since  \displaystyle\varphi=frac{\pi}{2}  initially. Finally we obtain  \displaystyle v=\frac{v_0}{1+\cos{\varphi}} In the course of time  \varphiightarrow0  and  \displaystyle vightarrow\frac{v_0}{2} . (Motion then is unaccelerated.)

Question 16

Inside a lift moving vertically downwards with an acceleration  g , two blocks of mass  m  and  2m  are placed as shown in the figure. The top block is given a sharp hit at an angle  	heta   so as to move it. What is the value of  	heta   for which both blocks move together without sliding on each other?

Accepted Answer

Lets calculate the effective  g  first {g}_{eff} =   downward acceleration of the lift - gravitational acceleration \implies {g}_{eff} = g -g = 0  Horizontal and vertical components of the applied force (F) {F}_{H} = Fcos	heta {F}_{V} = F sin	heta Lets assume that both blocks are moving together with velocity V.Apply momentum conservation in the positive x direction on the system  {F}_{H} dt = mV + (2m)V {F}_{H} dt = 3mV . . . . .(1)Now apply momentum conservation in the positive x direction 'm' {F}_{H}dt - fdt =mV  . . . . . (2);  where  f  is friction force on 'm'By solving equation 1 and equation 2 fdt = 2mV Contact force between m and am  N = F sin	heta . . . . . . (3)To move both the blocks together \mu Ndt > 2mV \mu Fsin	heta dt > 2(\dfrac{Fcos	heta dt}{3} ) tan	heta > \frac{2}{3\mu} \implies 	heta > 53^0

Question 17

A vehicle can travel along a curved road with a speed of   \displaystyle 18\ km\ {h} ^{-1}  . If the total centripetal force is provided by the banking of the road and the radius of the curvature is  10  m, calculate the banking angle of the road.

Accepted Answer

Equation of motion: N\cos 	heta=mg \implies N=\cfrac{mg}{\cos 	heta} and,  N\sin 	heta\ge \cfrac{mV^2}{r} \implies mg\cfrac{\sin 	heta}{\cos 	heta}\ge \cfrac{mV^2}{r} \implies g	an 	heta\ge \cfrac{V^2}{r} \implies 	an 	heta\ge \cfrac{V^2}{rg} \implies 	an 	heta_{max}=\cfrac{V^2}{rg} Velocity,  V=18   {km/h} \implies V=\cfrac{18	imes 10^3}{3600} \implies V=5   {m/s} r=10   m , \implies 	an 	heta=\cfrac{25}{100} \implies 	an	heta=0.25 \implies 	heta=	an^{-1}{0.25} \implies 	heta=14^o

Question 18

A circular road of radius  1000\ m  has banking angle  45^{o} . The maximum safe speed of a car having mass  2000\ kg  will be, if coefficient of friction between tyre and road is  0.5 :

Accepted Answer

Given that:  r= 1000m, 	heta=45^{0}, \mu=0.5, m=2000kg We know that maximum safe velocity for a banked is given by : v=\sqrt{ \dfrac{rg ( 	an 	heta+\mu)}{1- \mu tan 	heta}} hence on putting the given values, we get v=\sqrt{\dfrac{1000 	imes9.8 	imes(	an45 +0.5)}{1- 0.5	an45}} or  v=172m/s hence option A is correct

Question 19

A bullet of mass 5 g travelling at a speed of 120 m s ^{-1}  penetrates deeply into a fixed target and is brought to rest in 0.01 s. Calculate the average force exerted on the bullet.

Accepted Answer

Mass of the bullet  =  m = 5g = 5 	imes 10^{-3} kg Initial velocity of the bullet  =  u =  120  m  s^{-1} Final velocity of the bullet  =  v = 0  m  s^{-1} Time taken by the bullet to come to rest = t = 0.01  s Force exerted on the bullet  =  F  = ?We know, thatF  =  ma \Rightarrow = \displaystyle m \left ( \frac{v-u}{t} ight )          ........... (1) \displaystyle \left [ \because a = \frac{v-u}{t} ight ] Substituting the above values in (1),We get, F = \displaystyle 5 	imes 10^{-3} \left ( \frac{0-120}{0.01} ight ) =(5 	imes 10^{-3}) 	imes (-12000) = -5 	imes 12 = -60 N As 'F' is negative, it indicates that the retarding force is 60 N.Therefore, the force exerted on the bullet by the target to stop it is 60 N.

Question 20

The accelerations of the blocks A and B respectively as seen in the situation shown in the figure are (pulleys & strings are massless and pulleys are frictionless, acceleration due to gravity is  g )

Accepted Answer

We can see that two segments of string holding the mass 5kg is balanced by a string carrying mass  4 kg . If acceleration is a on  5   kg  side, it is  2 a  on  4   kg  side (Pulley constraint problem)Let us consider the movement of 4kg mass to be downward. Therefore. 4g - T = 4 (2a)          ........(i) On the other side, we have 2T - 5g = 5(a)             ........(ii) Solving i and ii, we haveAcceleration of block  A= \dfrac{2g}{7}  downwards  and  acceleration of block  B= \dfrac{g}{7}  upwards  Answer is A

Question 21

In the arrangement shown in the figure, if the surface is smooth, the acceleration of the hanging block  m_{2}  will be ( m_{1}  is on the table)

Accepted Answer

Let,  a_1, a_2   are acceleration of the masses  m_1,m_2 . and in this system,  a_1=a_2+a_2=2a_2  For block m on the table, the tension, i.e,   T=m_1a_1  ....(1) Equation of motion of the hanging block is  m_2a_2=m_2g-2T   .....(2) m_2a_2=m_2g-2m_1a_1=m_2g-4m_1a_2 a_2(m_2+4m_1)=m_2g a_2=\dfrac{m_2g}{4m_1+m_2}

Question 22

The upper surface of block C is horizontal and its right part is inclined to the horizontal at angle  37^0 . The mass of blocks A and B are  {m}_{1}=1.4 kg  and  {m}_{2}=5.5 kg , respectively. Neglect friction and mass of the pulley. Calculate acceleration a with which block C should be moved to the right so that A and B can remain stationary relative to it.

Accepted Answer

Since block  A  and  B  remain stationary relative to block  C , it means motions of blocks C or acceleration of all three blocks is same as a horizontally (rightwards) . Now considering FBD w.r.t C. For block B, N_{2}=m_{2} g    	ext {   or, } N_{2}=55 	ext { newton } \ T=m_{2} a       	ext {     -(1) } For block  A , N_{1} \cos 37^{\circ}+T \sin 37^{\circ}=m_{1} g 	ext { -(2) } N_{1} \sin 37^{\circ}-T \cos 37^{\circ}=m_{1} a 	ext { -(3) } solving above equations, T=10 \mathrm{~N} \ N_{1}=20 \mathrm{~N} Putting the value of 'T' and 'N,' 10=5.5 	imes a \ 	herefore a=\frac{20}{11} m s^{-2}

Question 23

A particle moves in the  xy  plane under action of a force  F  such that the value of its linear momentum  (P)  at any time  t  is  P_{x}=2\cos t, P_{y}=2\sin t . The angle  \Theta   between  P  and  F  at that time  t  will be :

Accepted Answer

|\vec{P}| = \sqrt{P_x ^2 + P_y ^2}= 2 Since  P_x  and  P_y  change with time and  |\vec{P}|   is constant, it is a case of uniform circular motion. v=\cfrac {P}{m} v_y = \dfrac{2}{m} \sin t v_x =  \dfrac{2}{m} \cos t a=\cfrac {dv}{dt}  a_x = - \dfrac{2}{m}  \sin t a_y =  \dfrac{2}{m} \cos t  \vec{v} .\vec{a}=- \cos t \sin t + \sin t \cos t =0 Hence angle between  \vec{a} \:\ and\ \:  \vec{p}  is   = 90^0 .

Question 24

A bullet of mass 100 g is fired from a gun of mass 20 kg with a velocity of  100 ms^{-1} . Calculate the velocity of recoil of the gun:

Accepted Answer

&#10;&#10;&#10;&#9;&#10;&#9;&#10;&#9;&#10;&#9;&#10;&#10;&#10;Mass of bullet,&#10; m=100 g=\dfrac {100}{1000}kg=\dfrac {1}{10}kg Velocity of&#10;bullet,  v=100 ms^{-1} Mass of gun,  M=20 kg Let recoil&#10;velocity of gun  =V Before firing, the system (gun+bullet) is&#10;at rest, therefore, initial momentum of the system  =0 Final&#10;momentum of the system  =  momentum of bullet + momentum of&#10;gun =mv+MV=\dfrac {1}{10}	imes 100+20 V=10+20V Applying&#10;law of conservation of momentum, Final momentum  =  Initial&#10;momentumi.e.,  10+20 V=0 20 V=-10 or  V=-\dfrac&#10;{10}{20}=-0.5 ms^{-1} .Negative sign shows that the direction of&#10;recoil velocity of the gun is opposite to the direction of the&#10;velocity of the bullet.&#10;&#10;

Question 25

A  5000\ kg  rocket is set for vertical firing. The exhaust speed is  800\ ms^{-1}.  To give an upward acceleration of  20\ ms^{-2},  the amount of gas ejected per second to supply the needed thrust is  \left ( g = 10\ ms^{-2} \right )

Accepted Answer

Given that,&#10;&#10;Mass of rocket  m=5000\,Kg  &#10;&#10;Acceleration  a=20\,m/{{s}^{2}} &#10;&#10;Speed  v=800\,m/s  &#10;&#10; g=10\,m/{{s}^{2}}  &#10;&#10;Now, thrust force on the rocket&#10;&#10; {{F}_{t}}={{v}_{r}}\left( \dfrac{-dm}{dt} \right)  &#10;&#10;Net force on the rocket&#10;&#10;    {{F}_{net}}={{F}_{t}}-W&#10;  &#10;&#10;   ma={{v}_{r}}\left(&#10;\dfrac{-dm}{dt} \right)-mg   &#10;&#10;   \left( \dfrac{-dm}{dt}&#10;\right)=\dfrac{m\left( g+a \right)}{{{v}_{r}}}   &#10;&#10;Rate of gas ejected per second&#10;&#10;    \left( \dfrac{-dm}{dt}&#10;\right)=\dfrac{5000\left( 10+20 \right)}{800}   &#10;&#10;   \dfrac{-dm}{dt}=187.5\,kg/s&#10;  &#10;&#10;Hence, the amount of gas ejected per second&#10;is  187.5\ kg/s

Laws Of Motion