\(\vec{F} \ and \ \vec{v}\)are perpendicular
So, Power \(=\vec{F} \cdot \vec{v}=F v \cos 90^{\circ}\)
= Zero  (∵ cos 90° = 0)

Jet planes use atmospheric oxygen for burning fuel but rockets carry their own fuel and oxygen and do not depend on atmospheric oxygen.

\(|A+B|=|A-B|\)
\( \sqrt { A^{ 2 }+B^{ 2 }+2AB \cos\theta } =\sqrt { A^{ 2 }+B^{ 2 }-2AB \cos\theta } \)
\(\Rightarrow 4AB \cos\theta =0\Rightarrow \cos\theta =0\)
Hence cos θ = 90⁰ or θ = π/2

The graph shows that 'with the passage of time, total path length first increases and then decreases.
The path length always increases or remains constant with passage of time and it does not decrease with time as shown in figure. Thus, this graph is not possible.

Large number of readings make the result of experiment more accurate.

Weight of diamond = 5.42 g = 0.00542 Kg
Total weight = 1.2 + 0.00542
= 1.20542 kg = 1.2 kg

Horizontal acceleration of the train is not carried by the stone. Horizontal velocity of the stone will remain constant, during the fall of the stone.

Here M = 20 kg, m = 5 kg and \(\mu_k\) = 0.1
Here net pulling force F = mg - f_k = mg -\(\mu_k\)N

= mg -\(\mu_k\).Mg = 5 x 10 - 0.1 x 20 x 10
= 50 - 20 = 30 N
\(\therefore\)Acceleration of the system a = \({F\over (m+M)}\)
\(={30N\over (5+20)kg}\)
= 1.2 m s^-2
\(\therefore\)Tension in string T = mg - ma
= 5 x 10 - 5 x 1.2
= 50 - 6 = 44 N.

(i) [R.H.S] = \(\sqrt { \frac { [P] }{ D] } } \)
\(=\sqrt { \frac { M{ L }^{ -1 }{ T }^{ -2 } }{ M{ L }^{ -3 } } } =L{ T }^{ -1 }\)
[L.H.s] = [v]=LT^-1
[R.H.S] = [L.H.S]
Hence, the relation is correct.
(ii) \([R.H.S]=\frac { 1 }{ [l] } \sqrt { \frac { [F] }{ M] } } \)
\(=\frac { 1 }{ L } \sqrt { \frac { ML{ T }^{ -2 } }{ M{ L }^{ -1 } } } =\frac { 1 }{ L } L{ T }^{ -1 }={ T }^{ -1 }\)
[L.H.S] = \(\frac{1}{time}=\frac{1}{T}\) = T^-1

Graphs depicting variation of (i) gravitational potential energy (P.E.), (ii) kinetic energy (K.E.),and (iii) the total sum of potential and kinetic energies for a freely falling body are as shown in adjoining Fig. From the graphs, it is clear that:
(a) Gravitational potential energy decreases as the body falls downwards and is zero at the Earth.

(b) Kinetic energy increases as the body falls downwards and is maximum when the body just strikes the ground.
(c) The sum of kinetic and potential energies remains constant at all points during its free fall.

(a) We are given that u = 0 and velocity after 8 s is 42 m/s, so we can use v = u + at to find acceleration
a = \(\frac { v-u }{ t } =\frac { 42.0-u }{ 8.0 } \)= 5.25ms^-2
(b) distance travelled in 8.0 s, we can use , s = ut +\(\frac { 1 }{ 2 } \)at²
= 0 + \(\frac { 1 }{ 2 } \)\(\times \)5.25\(\times \)8² = 168m
(c) distance travelled in 8th second, we have, S_n =u+(2n-1)\(\frac { a }{ 2 } \)
= (2 x 8 - 1) \(\times \frac { 5.25 }{ 2 } \) = 39.375 m.

Here,mass of bullet m = 0.012 kg, initial speed of bullet \(\mu =70{ ms }^{ -1 }\) mass 0.4 kg 
As on collision, the bullet comes to rest w.r.t. block, it means that after collision bullet and block are moving with a common speed v.
From conservatio law of momentum mu = (M + m) v
\(\Rightarrow v=\frac { mu }{ (M+m) } =\frac { 0.012\times 70 }{ (0.4+0.012) } =2.04\quad m{ s }^{ -1 }\)
If the block now rises to a maximum height of h, then using conservtion law of mechanical energy, we have \(\frac { 1 }{ 2 } (M+m){ v }^{ 2 }-0=(M+m)\quad gh\)
\(\\ \Rightarrow h=\frac { { v }^{ 2 } }{ 2g } =\frac { { (2.04) }^{ 2 } }{ 2\times 9.8 } =0.212\quad m=21.2cm\)

The maximum speed v_max=μgr =0. 4 × 9. 8 × 30 =10. 84 m/s.

Velocity of the aeroplane in the horizontal direction is 
\({ u }_{ oy }=600km/h=600\times \frac { 5 }{ 18 } =\frac { 500 }{ 3 } m/s\)
Velocity remains constant throughout the flight of the body.
u_oy = 0 and y = h = 1960m
Let t=time taken by the body to reach the ground
\(y={ u }_{ oy }t+\frac { 1 }{ 2 } g{ t }^{ 2 }\)
\(\\ Here,\quad y=h=1960m,{ u }_{ oy }=0\)
\(\\ \therefore \quad 1960=\frac { 1 }{ 2 } \times 9.8\times { t }^{ 2 }\)
\( \Rightarrow t=\sqrt { \frac { 1960 }{ 4.9 } } =\sqrt { 400 } =20s\)
Distance travelled by the body in the horizontal direction,
\(Ab=x={ v }_{ ax }t=\frac { 500 }{ 3 } \times 20\)
\(\\ =\frac { 10000 }{ 3 } =3333=3.33km\) 

Unit vector perpendicular to both
\(\overrightarrow{\mathrm{A}}=2 \hat{i}+\hat{j}+\hat{k} \text { and } \hat{i}-\hat{j}+2 \hat{k}\)
is given by \(\hat{n}=\frac{\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}}{|\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}|}\)
\(=\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}=\left|\begin{array}{rrr} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 1 \\ 1 & -1 & 2 \end{array}\right|\)
\(=\hat{i}[2-(-1)]-\hat{j}(4-1)+\hat{k}(-2-1)\)
\(=3 \hat{i}-3 \hat{j}-3 \hat{k}\)
Unit vector is \(\hat{n}=\frac{3 \hat{i}-3 \hat{j}-3 \hat{k}}{\sqrt{9+9+9}}\)
  \(\frac { 3\hat { i } -\hat { 3j } +3\hat { k } }{ \sqrt { 27 } } \) 

Given \(V={ 2t }^{ 2 }-3t-10m/s\)
\({ a }_{ in }=\frac { dv }{ dt } =4t-3m/{ s }^{ 2 }\)
If t = 5 \({ a }_{ in }=5\times 4-3=17m/{ s }^{ 2 }\) 

(a) One dimensional elastic collision is one in which both momentum and K.E. are conserved and the body moves in the same line of motion even after the collision.
If m₁, m₂ are the masses, u₁, u₂ are the initial velocities and v₁, v₂ are the final velocities, then
m₁u₁ + m₂u₂ = m₁v₁ + m₂ v₂               ...(i)
\(\frac{1}{2} m_{1} u_{1}^{2}+\frac{1}{2} m_{2} u_{2}^{2}=\frac{1}{2} m_{1} v_{1}^{2}+\frac{1}{2} m_{2} v_{2}^{2}\)         ...(ii)
\(\text { i.e., } \quad m_{1}\left(v_{1}^{2}-u_{1}^{2}\right)=m_{2}\left(v_{2}^{2}-u_{2}^{2}\right)\)  from (ii)
\(m_{1}\left(v_{1}-u_{1}\right)=m_{2}\left(v_{2}-u_{2}\right)\)  from (i)
Dividing both sides
v₁ + u₁ = v₂ + u₂
⇒ v₁ = v₂ + u ₂ - u₁
Substituting in (i) we have
m₁u₁ + m₂u₂ = ·m₁(v₂ + u₂ - u₁) + m₂ v₂
2m₁u₁ + u₂(m₂ - m₁) = v₂ (m₁ + m₂ )
\(\therefore \ v_{2}=\frac{u_{2}\left(m_{2}-m_{1}\right)+2 m_{1} u_{1}}{m_{1}+m_{2}}\)
Similarly, \(v_{1}=\frac{u_{1}\left(m_{1}-m_{2}\right)+2 m_{2} u_{2}}{m_{1}+m_{2}}\) 
(b) Given, m₁ = 9000 kg, u₁ = 36 km/h = =10m/s
m₂ = 9000 kg, u₂ = 0, v₁ = v2 = v
By conservation of momentum
m₁u₁ + m₂u₂ = (m₁ + m₂)v
\(9000\times 10\times 9000\times 0=\left( 9000+9000 \right) v\)   
or     \(v=\frac { 90000 }{ 18000 } =5m/s\)     
Total KE before collision = \(\frac { 1 }{ 2 } { m }_{ 1 }{ u }_{ 1 }^{ 2 }+\frac { 1 }{ 2 } { m }_{ 2 }{ u }_{ 2 }^{ 2 }\) 
 \(=\frac { 1 }{ 2 } \times 9000\times 10\times 10+0\)  
\(\\ =450000J\)
Total KE after collision=\(\frac { 1 }{ 2 } \left( { m }_{ 1 }+{ m }_{ 2 } \right) { v }^{ 2 }\)  
\(=\frac { 1 }{ 2 } \times 2\times 9000\times { \left( 5 \right) }^{ 2 }\) 
ஃ Initial K.E. ≠ Final K.E.
Hence, collision is inelastic.

(a) Consider the mass m attached to a string of length I. Let the lower most and top most points be marked A and B respectively. Consider a point P where the length 1 has turned by θ from the vertical line through A.

Centripetal force is provided by the tension and mg cos θ acting in opposite directions.
From the diagram, OA = OP = I,
OT = lcos θ, AT = l (1 - cos θ)
(i) If V_A and v_p are velocities at A and P respectively, using
v ² = u ² + 2as, we have
\(v_{P}^{2}=v_{A}^{2}-2 g(A T)\)
\(v_{P}^{2}=v_{A}^{2}-2 g l(1-\cos \theta)\)
Velocity at any point P,
\(v_{P}=\sqrt{v_{A}^{2}-2 g l(1-\cos \theta)}\)
(ii) At P \(T_{P}-\mathrm{mg} \cos \theta=\frac{m v_{P}^{2}}{l}\)
\(T_{P}=m g \cos \theta+\frac{m v_{P}^{2}}{l}\)
ஃ T_p = mgcos θ
\(+\frac{m}{l} \sqrt{v_{A}^{2}-2 g l(1-\cos \theta)}\)
By knowing V_A and θ, one can find the velocity and tension at any point.
(iii) To perform verticle circle, the tension should be non-zero till the mass reaches the top most point B
ஃ T_B ≥ 0 at θ = 180^o
ஃ T_B = 0 will be minimum.
Substituting,
TP = 0 at θ = π, we have
\(0=\mathrm{Mg} \cos \pi+\frac{M}{l}\left[v_{A}^{2}-2 g l(1-\cos \pi)\right]\)
\(0=-M g+\frac{M}{l}\left[v_{A}^{2}-2 g l(2)\right]\)
\(v_{A}^{2}-4 g l=g l, v_{A}^{2}=5 g l, v_{A}=\sqrt{5 g l}\)
So, the minimum velocity required at lowermost point to perform verticle circle is \(\sqrt{5 g l}\)
(b) Given: Mass ofa bullet (m) = 0.01 kg
Speed of bullet (v) = 500 m/s
Mass of block (M) = 2 kg
Length of string (I) = 5 m
Vertical height (h) = 0.1 m

Let V be the velocity acquired by block
\(\frac{1}{2} M V^{2}=M g h\)
\(\Rightarrow \ V=\sqrt{2 g h}\)
\(=\sqrt{2 \times 9.8 \times 0.1}=1.4 \mathrm{~m} / \mathrm{s}\)
If V' is the speed of bullet on emerging out of the block then by the law of conservation of energy.
mv + M x 0 = MV + m V'
\(\Rightarrow \ V^{\prime}=\frac{m v-M V}{m}\)
\(=\frac{0.01 \times 500-2 \times 1.4}{0.01}\)
\(V^{\prime}=\frac{5-2.8}{0.01}=\frac{2.2}{0.01}=220 \mathrm{~m} / \mathrm{s}\)

(a) (i) When an object is projected with velocity u making an angle θ with horizontal direction.
\(R_{1}=\frac{u^{2} \sin 2 \theta}{g}\)              .............(i)
When an object is projected with u making an angle (90^o - θ)
\(R_{2}=\frac{u^{2} \sin 2\left(90^{\circ}-\theta\right)}{g}\)
\(=\frac{u^{2}}{g} \sin \left(180^{\circ}-2 \theta\right)\)
\(=\frac{u^{2}}{g} \sin 2 \theta\)                .............(ii)
From (i) and (ii)
R₁ = R₂
ஃ The horizontal range is same for two complementary angles.
(ii) For angle of projection θ
Height attained \(=\frac{u^{2} \sin ^{2} \theta}{2 g}=H\)
for angle of projection (90° - θ)
Height attained = \(\frac{u^{2} \sin ^{2}\left(90^{\circ}-\theta\right)}{2 g}\)
\(=\frac{u^{2} \cos ^{2} \theta}{2 g}=H^{\prime}\)
\(\frac{H}{H^{\prime}}=\frac{u^{2} \sin ^{2} \theta}{2 g} \times \frac{2 g}{u^{2} \cos ^{2} \theta}\)
= tan² θ : 1
(b) Horizontal range \(=\frac{u^{2} \sin 2 \theta}{g}\)
Maximum height = \(\frac{u^{2} \sin ^{2} \theta}{2 g}\)
Horizontal range = Maximum height
\(\frac{u^{2} \sin 2 \theta}{g}=\frac{u^{2} \sin ^{2} \theta}{2 g}\)
2 x 2 sinθ cosθ = sin² θ
4 = tanθ
or θ = tan^-1 4

When a body is projected in the air in any direction, then the body is called a projectile.
(a) Suppose a body is projected with velocity u at an angle θ with the horizontal, P(x, y) is any point on its trajectory at time t.
Horizontal component of velocity is unaffected by gravity, but the vertical component (u sin θ) changes due to gravity.
ஃ x = (u cos θ)t.
\(y=(u \sin \theta) t-\frac{1}{2} g t^{2}\)
\(=u \sin \theta \times \frac{x}{u \cos \theta}-\frac{1}{2} g\left(\frac{x}{u \cos \theta}\right)^{2}\)
\(y=x \tan \theta-\frac{g x^{2}}{2 u^{2} \cos ^{2} \theta}\)                 ............(i)
(b) The greatest vertical distance attained by the projectile above the horizontal plane from the point of projection is called maximum height.
Maximum height, LN = H
(i) At maximum height
v = 0
\(\therefore \quad v^{2}-u_{y}^{2}=-2 g H, \text { where }\)
u_y = u sin θ 
or (u sinθ )² = 2 gH
\(\text { or } \quad H=\frac{u^{2} \sin ^{2} \theta}{2 g}\)
(ii) At maximum height
v = 0
ஃ 0 = u sin θ - gt
\(\text { or } \quad t=\frac{u \sin \theta}{g}\)
But time of flight
\(\mathrm{T}=2 t=\frac{2 u \sin \theta}{g}\)
(iii) When the body returns to the same horizontal level y = 0
\(\therefore \quad 0=x \tan \theta-\frac{g x^{2}}{2 u^{2} \cos ^{2} \theta} \quad[\text { From }(i)]\)
\(\text { or } x \tan \theta=\frac{g x^{2}}{2 u^{2} \cos ^{2} \theta}\)
\(\text { or } \quad x=\frac{2 u^{2} \sin \theta \cos \theta}{g}=\frac{u^{2} \sin 2 \theta}{g}\)
But coordinates of M are (R, 0). Putting x = R,
we have
\(\mathrm{R}=\frac{u^{2} \sin 2 \theta}{g}\)
(c) θ = 45°
(d) When an object is projected with velocity u making an angle θ with horizontal direction
\(R_{1}=\frac{u^{2} \sin 2 \theta}{g}\)             ...........(i)
When an object is projected with u making an angle (90^o - θ)
\(\mathrm{R}_{2}=\frac{u^{2} \sin 2\left(90^{\circ}-\theta\right)}{g}\)
\(=\frac{u^{2}}{g} \sin \left(180^{\circ}-2 \theta\right)\)
\(=\frac{u^{2}}{g} \sin 2 \theta\)                       ............(ii)
from (i) and (ii) R₁ = R₂
ஃ The horizontal range IS same for two complementary angles.

Figure shows the forces acting on masses m₁ , m₂ and m₃ . Tension forces acting on pulley also shown in figure.
Since masses m₁ and m₂ are moving downward with acceleration a , by applying Newtons second law,
we get the following equations.
( m₁ + m₂ ) g - T₁ = m₁ a .........(1)
T₁ - T₂ = m₂ a  ......(2)
since masses m₂ and m₃ are connected by same string , Tension forces T₂ and T₃ are same
T₂ = T₃  ...........(3)
By applying Newton's second law on block of mass m₃ , we get
T₃ - m₃ g = m₃ a  .......(4)
By adding all the above equations (1), (2), (3) and (4), we get
( m₁ + m₂ - m₃ )g = ( m₁ + m₂ + m₃ ) a  ....... (5)
Hence acceleration a is obtained from above equation as , 
\(a=\left(\frac{m_1+m_2-m_3}{m_1+m_2+m_3}\right) g \) ....(6)
Tension force T₁ is obtained from (1) as ,  T₁ = (  m₁ + m₂ ) g- m₁ a  ......(7)
By substituting for acceleration a in eqn.(7) , we get T₁ after simplification as
\(T_1=\left(\frac{m_1 m_2+m_2^2+2 m_1 m_3+m_2 m_3}{m_1+m_2+m_3}\right) g\)
Tension force T₃ is obtained from (4) as ,  T₃ =  m₃(a+g) .........(8)
By substituting for acceleration a in eqn.(8) , we get T₃ after simplification as
\(T_3=2\left(\frac{\left(m_1+m_2\right) m_3}{m_1+m_2+m_3}\right) g \)
T = T₂ + T₃  and T₂ = T₃ .  Hence we get T as
\(T=4\left(\frac{\left(m_1+m_2\right) m_3}{m_1+m_2+m_3}\right) g\)
\(T_1=\frac{56}{9} g, T_2=\frac{56}{9} g, T_3=\frac{16}{9} g, a=\frac{5}{9} g\)

\(\frac { 1 }{ 2 } { mv }^{ 2 }=mgh,\ v=\sqrt { 2gh } \)
\(=\sqrt { 2\times 10\times 10 } { ms }^{ -1 }=14.14\ { ms }^{ -1 }\)
\({ v }_{ B }={ v }_{ C }=14.14\ { ms }^{ -1 },\ l=\frac { 1 }{ 2 } (g\ sin\ \theta ){ t }^{ 2 }\)
\(sin\ \theta =\frac { h }{ l } ,l=\frac { h }{ sin\ \theta } \)
\(\therefore \ \frac { h }{ sin\ \theta } =\frac { 1 }{ 2 } g\ sin\ \theta \ { t }^{ 2 }\ or\ t=\sqrt { \frac { 2h }{ g } } .\frac { 1 }{ sin\ \theta } \)
\(\therefore \ \frac { h }{ sin\theta } =\frac { 1 }{ 2 } g\ sin\ \theta \ { t }^{ 2 }\ or\ t=\sqrt { \frac { 2h }{ g } } .\frac { 1 }{ sin\ \theta } \)
\({ t }_{ B }=\sqrt { \frac { 2\times 10 }{ 10 } } .\frac { 1 }{ sin\quad 30° } =2\sqrt { 2 } s\)
\({ t }_{ C }=\sqrt { \frac { 2\times 10 }{ 10 } } .\frac { 1 }{ sin60° } =\frac { 2\sqrt { 2 } }{ \sqrt { 3 } } s.\)

(i) When a body is moving in straight line in a specific direction then both displacement and distance would be equal in magnitude.
(ii) Average speed is the average distance travelled per unit time
\(\text { or } v=\frac{\text { total distance travelled }}{\text { total time taken }}\)
Average velocity is the average displacement covered per unit time. It is a vector quantity.
\(\bar{v}=\frac{\text { net displacement }}{\text { time taken }}\)
(iii) 

(iv) Area under velocity-time graph and time axis is the measure of displacement in that particular interval of time.
(v) Slope of Position-time at any instant represent instantaneous velocity while that of velocity-time represent instantaneous acceleration.
(vi) A body thrown upward then at highest point body comes to rest momentarily but still acceleration is acrting downward.
(vii) During one complete revolution average displacement and velocity is zero while acceleration is acting radially toward the center of circular path.

B) If both assertion and reason are true but reason is not the correct explanation of the assertion.
When automobile moves in circular path then reaction on inner wheel and outer wheel will be different.\(R_{\text {inner }}=\frac{M}{2}\left[g-\frac{v^{2} h}{r a}\right] \text { and } R_{\text {outer }}=\frac{M}{2}\left[g+\frac{v^{2} h}{r a}\right]\)In critical condition \(v_{\mathrm{safe}}=\sqrt{\frac{g r a}{h}}\) .If v is equal or more that thus critical value then reaction on inner wheel becomes zero. So it leaves the ground first            

D) If the assertion and reason both are false.
Equation of motion can be applied if the acceleration is in opposite direction to that of velocity and uniform motion mean the acceleration is zero.

(a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.