NCERT Solutions: Class 11 Physics Chapter 3 - Motion in a Plane

Scalars have magnitude only. Vectors have magnitude and direction and obey vector addition. Angular velocity is treated as an axial vector.

Work is a scalar. Electric current is also treated as a scalar quantity, although current density is a vector. The remaining listed quantities are vectors or axial vectors.

Impulse equals change in momentum, so it has direction. All other listed quantities are scalars.

Addition requires same physical nature and same mathematical type. A scalar cannot be added to a vector. Multiplication of scalars or a vector by a scalar is meaningful, with appropriate units. A component of a vector is a scalar, so it cannot be added to the vector itself.

A vector's magnitude is scalar. A vector component can be written as a scalar component along an axis or as a vector component, so saying each component is always scalar is not generally correct. Path length equals displacement magnitude only for motion without change of direction. Since path length is at least displacement magnitude, average speed is at least magnitude of average velocity. If three vectors add to zero, they can be arranged head-to-tail to form a triangle and therefore lie in one plane.

For any two vectors, the third side of a triangle cannot exceed the sum of the other two sides and cannot be less than their difference. Thus $|\mathbf a+\mathbf b|\le |\mathbf a|+|\mathbf b|$ and $|\mathbf a+\mathbf b|\ge ||\mathbf a|-|\mathbf b||$. Replacing $\mathbf b$ by $-\mathbf b$ gives the two inequalities for $|\mathbf a-\mathbf b|$. Equality in the upper limit occurs when the two added vectors are parallel in the same direction; equality in the lower limit occurs when they are antiparallel.

From $\mathbf a+\mathbf b+\mathbf c+\mathbf d=0$, $\mathbf a+\mathbf c=-(\mathbf b+\mathbf d)$, so their magnitudes are equal. Also $\mathbf a=-(\mathbf b+\mathbf c+\mathbf d)$, so $|\mathbf a|\le |\mathbf b|+|\mathbf c|+|\mathbf d|$. Finally $\mathbf b+\mathbf c=-(\mathbf a+\mathbf d)$, so it lies in the plane of $\mathbf a$ and $\mathbf d$ or on their line if they are collinear. The four vectors need not individually be zero.

Displacement depends only on initial and final positions. Since P and Q are diametrically opposite on a circle of radius $200\,\text{m}$, $PQ=2R=400\,\text{m}$ for all three. Only path B is the straight diameter, so only for B is path length equal to displacement magnitude.

The cyclist starts and ends at O, so net displacement and average velocity are zero. Total path length $=OP+\text{arc }PQ+QO=1+\frac{\pi}{2}(1)+1=2+\frac{\pi}{2}=3.571\,\text{km}$. Time $=10\,\text{min}=\frac{1}{6}\,\text{h}$. Average speed $=3.571/(1/6)=21.4\,\text{km h}^{-1}$.

Successive $500\,\text{m}$ displacements differ in direction by $60^\circ$, forming sides of a regular hexagon. After $3$ turns, the vector sum has magnitude $1000\,\text{m}$. After $6$ turns, the hexagon closes and displacement is zero. After $8$ turns, the net is the same as the first two side vectors, with magnitude $\sqrt{500^2+500^2+2(500)(500)\cos60^\circ}=500\sqrt3\,\text{m}$.

Time $=28\,\text{min}=28/60\,\text{h}$. Average speed $=23/(28/60)=49.3\,\text{km h}^{-1}$. Magnitude of average velocity $=10/(28/60)=21.4\,\text{km h}^{-1}$. They differ because total path length differs from displacement magnitude.

For maximum range without hitting the ceiling, the highest point just reaches $25\,\text{m}$. Thus $u_y^2/(2g)=25$, so $u_y=\sqrt{2gH}=\sqrt{490}=22.1\,\text{m s}^{-1}$. With total speed $40\,\text{m s}^{-1}$, $u_x=\sqrt{40^2-u_y^2}=\sqrt{1110}=33.3\,\text{m s}^{-1}$. Time of flight $=2u_y/g=4.52\,\text{s}$. Range $=u_xT\approx33.3\times4.52=1.51\times10^2\,\text{m}$.

Maximum range on level ground is $R_{\max}=u^2/g=100\,\text{m}$. Maximum vertical height with the same speed is $H=u^2/(2g)=R_{\max}/2=50\,\text{m}$.

Radius $r=0.80\,\text{m}$ and frequency $f=14/25=0.56\,\text{s}^{-1}$. Angular speed $\omega=2\pi f=3.52\,\text{rad s}^{-1}$. Centripetal acceleration $a=\omega^2r=(3.52)^2(0.80)=9.9\,\text{m s}^{-2}$ toward the centre.

$v=900\,\text{km h}^{-1}=250\,\text{m s}^{-1}$ and $r=1.00\,\text{km}=1000\,\text{m}$. Thus $a_c=v^2/r=250^2/1000=62.5\,\text{m s}^{-2}$. Dividing by $g\approx9.8\,\text{m s}^{-2}$ gives $a_c/g\approx6.4$.

a. In non-uniform circular motion there is a tangential acceleration in addition to radial acceleration, so net acceleration is not always purely radial. b. Instantaneous velocity is tangent to the path. c. In one complete cycle of uniform circular motion, initial and final velocities are the same, so average acceleration $\Delta\mathbf v/\Delta t=0$.

Differentiate position: $\mathbf v=d\mathbf r/dt=3.0\hat i-4.0t\hat j$ and $\mathbf a=d\mathbf v/dt=-4.0\hat j$. At $t=2.0\,\text{s}$, $\mathbf v=3\hat i-8\hat j$. Magnitude $=\sqrt{3^2+8^2}=\sqrt{73}=8.54\,\text{m s}^{-1}$. Direction satisfies $\tan\theta=8/3$, so $\theta\approx69^\circ$ below $+x$.

For $x$: $x=0+0t+\frac12(8)t^2=4t^2$. Setting $x=16$ gives $t=2.0\,\text{s}$. For $y$: $y=10t+\frac12(2)t^2=20+4=24\,\text{m}$. Velocity at that time is $\mathbf v=16\hat i+(10+4)\hat j=16\hat i+14\hat j$, so speed $=\sqrt{16^2+14^2}=21.3\,\text{m s}^{-1}$.

The vectors $\hat i+\hat j$ and $\hat i-\hat j$ have components $(1,1)$ and $(1,-1)$, so each has magnitude $\sqrt2$ and directions $45^\circ$ and $-45^\circ$. Unit vectors along them are $(\hat i+\hat j)/\sqrt2$ and $(\hat i-\hat j)/\sqrt2$. Scalar components of $\mathbf A$ along these directions are $\mathbf A\cdot(\hat i+\hat j)/\sqrt2=(2+3)/\sqrt2=5/\sqrt2$ and $\mathbf A\cdot(\hat i-\hat j)/\sqrt2=(2-3)/\sqrt2=-1/\sqrt2$.

Average velocity is defined as displacement divided by time interval, so b is true. Average acceleration is change in velocity divided by time interval, so e is true. Relations a, c and d require special conditions such as constant acceleration and are not valid for arbitrary motion.

A scalar need not be conserved; temperature is scalar but can change. Scalars can be negative, such as electric potential or Celsius temperature. Scalars may have dimensions, such as mass or energy. Scalar fields can vary with position, such as temperature in a room. A scalar is independent of coordinate-axis orientation, so e is true.

Assuming the two aircraft positions are symmetrically placed about the point closest to the observer, the horizontal distance covered in $10.0\,\text{s}$ is $2h\tan(15^\circ)=2(3400)(0.268)=1.82\times10^3\,\text{m}$. Speed $=1.82\times10^3/10.0=1.82\times10^2\,\text{m s}^{-1}$, or about $656\,\text{km h}^{-1}$.