JEE Physics Motion In A Plane MCQs Set 16

Answer: (a) tangential to its path

Explanation:
1. Velocity is the rate of change of the position vector, and the displacement during a very short interval lies along the tangent to the path.
2. So the instantaneous velocity is always tangential to the path.

Teacher's Note:
1. Acceleration, not velocity, can have a component normal to the path.
2. In uniform circular motion velocity is tangential and acceleration is radial.

Answer: (d) All of the above

Explanation:
1. \( AB\cos\theta=m \) and \( AB\sin\theta=n \).
2. \( \tan\theta=\frac nm \), and \( AB=\sqrt{m^2+n^2} \).
3. \( \sin\theta=\frac{n}{\sqrt{m^2+n^2}} \), \( \cos\theta=\frac{m}{\sqrt{m^2+n^2}} \). All three expressions give the same angle.

Teacher's Note:
1. Dividing the cross-product magnitude by the dot product gives tanθ.
2. Squaring and adding gives (AB)² = m² + n².

Answer: (b) a dθ

Explanation:
1. The change in the vector is the chord of the arc of radius a: for a small angle the chord is almost the arc length.
2. \( |\Delta\vec a|=2a\sin\frac{d\theta}{2}\approx a\,d\theta \)

Teacher's Note:
1. For small angles sin x ≈ x, so the chord equals the arc.
2. This result is used to get the centripetal acceleration v dθ/dt = vω.

Answer: (d) \( \frac{5\pi}6 \)

Explanation:
1. \( \vec B=\vec C-\vec A \) and A is perpendicular to C, so \( B^2=C^2+A^2 \).
2. With C = B/2: \( A^2=B^2-\frac{B^2}4=\frac{3B^2}{4} \), so \( A=\frac{\sqrt3}2B \).
3. \( \vec A\cdot\vec B=\vec A\cdot(\vec C-\vec A)=-A^2=-\frac34B^2 \).
4. \( \cos\theta=\frac{-\frac34B^2}{AB}=\frac{-\frac34}{\frac{\sqrt3}{2}}=-\frac{\sqrt3}2\Rightarrow\theta=\frac{5\pi}6 \)

Teacher's Note:
1. Use A·C = 0 to simplify A·B.
2. cosθ = −√3/2 corresponds to 150°, that is 5π/6.

Answer: (a) the point O

Explanation:
1. Time of flight (vertical motion only): \( T=\frac{2u\sin\theta}{g} \).
2. Horizontal motion: \( x=u\cos\theta\,T-\frac12g\cot\theta\,T^2=\frac{2u^2\sin\theta\cos\theta}{g}-\frac{2u^2\sin\theta\cos\theta}{g}=0 \).
3. So the ball returns to the starting point O.

Teacher's Note:
1. Horizontal and vertical motions are independent, so T depends only on the vertical motion.
2. Substitute T in the horizontal displacement formula and the two terms cancel.

Answer: (b) 16 m

Explanation:
1. The range is the non-zero value of x for which y = 0 (the projectile returns to the level of projection).
2. \( 12x-\frac34x^2=0\Rightarrow x\left(12-\frac34x\right)=0\Rightarrow x=16\ \text{m} \)

Teacher's Note:
1. Setting y = 0 gives the two points on the ground, x = 0 (the start) and x = R.
2. The value of g is not needed because the trajectory equation is already given.

Answer: (d) equal path lengths are traversed in equal intervals

Explanation:
1. Constant speed means the path length covered in a time interval is v0 × (interval), so equal path lengths are covered in equal intervals.
2. The direction of velocity may change (for example, circular motion), so the average velocity can be zero over a full circle, the average acceleration need not vanish, and displacements in equal intervals need not be equal.

Teacher's Note:
1. Speed is the rate of change of path length; velocity is the rate of change of displacement.
2. Uniform circular motion is a good counter-example to the other options.

Answer: (a) \( \frac{ut}{\sqrt3} \)

Explanation:
1. Horizontal distance covered: \( AC=u\cos60^\circ\cdot t=\frac{ut}{2} \).
2. The line AB makes 30° with the horizontal, so \( AB=\frac{AC}{\cos30^\circ}=\frac{ut/2}{\sqrt3/2}=\frac{ut}{\sqrt3} \)

Teacher's Note:
1. The horizontal velocity of a projectile is constant, u cosθ.
2. Resolve AB into its horizontal projection AC using the 30° angle.

Answer: (c) \( \frac{t^2+T^2}{t^2-T^2} \)

Explanation:
1. Let the man's speed be v, river speed u and width d.
2. Crossing straight: \( d=\sqrt{v^2-u^2}\,t \). Downstream: \( d=(v+u)T \).
3. Equate and square: \( (v^2-u^2)t^2=(v+u)^2T^2\Rightarrow(v-u)t^2=(v+u)T^2 \).
4. \( v(t^2-T^2)=u(t^2+T^2)\Rightarrow\frac vu=\frac{t^2+T^2}{t^2-T^2} \)

Teacher's Note:
1. For the straight crossing, the man heads upstream so that the downstream component of his velocity cancels the river's.
2. Cancel the common factor (v + u) before solving for v/u.

Answer: (d) \( \vec V_{A/B}=\vec V_{B/A} \)

Explanation:
1. Relative velocity of A with respect to B is \( \vec V_A-\vec V_B \), and the relative velocity of B with respect to A is \( \vec V_B-\vec V_A=-\vec V_{A/B} \).
2. So the two are equal in magnitude but opposite in direction, and \( \vec V_{A/B}=\vec V_{B/A} \) is not correct.

Teacher's Note:
1. Options (a) and (b) are valid identities, and (c) is true for magnitudes only.
2. Relative velocity vectors of two bodies always point opposite ways.

Answer: (b) \( 2\hat i-3\hat j \)

Explanation:
1. The horizontal component of velocity stays constant: 2 m/s along +x.
2. At the same level the vertical component has the same magnitude but points downward: −3 m/s.
3. Hence \( \vec v_B=2\hat i-3\hat j \).

Teacher's Note:
1. By symmetry of the parabola, the speed at the same height is the same.
2. Only the vertical component reverses.

Answer: (a) \( \frac{a}{\sqrt{v^2-v_1^2}} \)

Explanation:
1. For A to meet B, A's velocity component perpendicular to AB must equal v1, so \( v\sin\phi=v_1 \).
2. The component along AB covers the distance a in time t: \( v\cos\phi\cdot t=a \).
3. \( v\cos\phi=\sqrt{v^2-v_1^2}\Rightarrow t=\frac{a}{\sqrt{v^2-v_1^2}} \)

Teacher's Note:
1. To catch a moving target the pursuer must match its sideways motion.
2. Resolve v along and perpendicular to AB.

Answer: (d) both (a) and (b)

Explanation:
1. \( \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \), so the path is an ellipse.
2. At \( t=\frac\pi{2p} \): \( \vec v=(-ap\sin pt,\ bp\cos pt)=(-ap,\,0) \) and \( \vec a=-p^2(x,y)=(0,\,-bp^2) \), which are perpendicular.
3. In general \( \vec a\cdot\vec v=p^3(a^2-b^2)\sin pt\cos pt\ne0 \), so the acceleration is not always perpendicular to the velocity.

Teacher's Note:
1. Eliminate t using cos² + sin² = 1 to find the path.
2. Check perpendicularity with a dot product.

Answer: (c) 90°

Explanation:
1. Square both sides: \( A^2+B^2+2AB\cos\theta=A^2+B^2-2AB\cos\theta \).
2. \( 4AB\cos\theta=0\Rightarrow\theta=90^\circ \)

Teacher's Note:
1. The diagonals of a parallelogram are equal only when it is a rectangle.
2. The dot product of the two vectors is zero.

Answer: (b) \( \theta=\frac\pi2 \)

Explanation:
1. \( C^2=A^2+B^2+2AB\cos\theta \). Comparing with \( C^2=A^2+B^2 \) gives \( 2AB\cos\theta=0 \).
2. \( \theta=\frac\pi2 \)

Teacher's Note:
1. This is the Pythagoras relation for a right-angled triangle of vectors.
2. A and B are mutually perpendicular.

Answer: (d) 90°

Explanation:
1. \( \vec a\cdot\vec b=2(-3)+3(0)+1(6)=-6+6=0 \).
2. The dot product is zero, so the vectors are perpendicular.

Teacher's Note:
1. A zero dot product of two non-zero vectors means an angle of 90°.
2. Missing components are zero.

Answer: (c) 10 metres (towards west)

Explanation:
1. Take east as x and north as y. Displacements: \( (0,30) \), \( (20,0) \) and \( 30\sqrt2(-\cos45^\circ,-\sin45^\circ)=(-30,-30) \).
2. Sum: \( (20-30,\ 30-30)=(-10,\ 0) \).
3. The displacement is 10 m towards the west.

Teacher's Note:
1. South-west makes 45° with both south and west.
2. Add the x and y components separately.

Answer: (a) \( \frac{4v^2}{5g} \)

Explanation:
1. \( R=\frac{2v^2\sin\theta\cos\theta}{g} \) and \( H=\frac{v^2\sin^2\theta}{2g} \).
2. \( R=2H\Rightarrow2\sin\theta\cos\theta=\sin^2\theta\Rightarrow\tan\theta=2 \), so \( \sin\theta=\frac2{\sqrt5},\ \cos\theta=\frac1{\sqrt5} \).
3. \( R=\frac{2v^2}{g}\cdot\frac2{\sqrt5}\cdot\frac1{\sqrt5}=\frac{4v^2}{5g} \)

Teacher's Note:
1. Use the standard formulas for range and maximum height.
2. Check units: v²/g has the dimension of length.

Answer: (b) \( \frac{x}{2u} \)

Explanation:
1. Horizontal velocity components: left particle \( u\sqrt3\cos30^\circ=\frac{3u}{2} \), right particle \( u\cos60^\circ=\frac u2 \), directed towards each other.
2. Closing speed \( =\frac{3u}2+\frac u2=2u \).
3. \( t=\frac{x}{2u} \)

Teacher's Note:
1. Horizontal velocities stay constant, so the gap closes at the sum of the horizontal speeds.
2. Vertical motion does not affect the horizontal separation.

Answer: (b) 1500 m

Explanation:
1. Time to fall: \( 490=\frac12(9.8)t^2\Rightarrow t=10\ \text{s} \).
2. Horizontal distance: \( 150\times10=1500\ \text{m} \)

Teacher's Note:
1. The bag keeps the horizontal velocity of the aeroplane.
2. Vertical and horizontal motions are independent.

Answer: (a) At the starting point

Explanation:
1. Angular momentum about the point of projection is \( \vec L=m(\vec r\times\vec v) \).
2. At the starting point \( \vec r=0 \), so L = 0, which is the minimum possible value; it grows as the projectile moves away.

Teacher's Note:
1. The magnitude of L cannot be negative, so zero is the minimum.
2. L increases with time, being m × (horizontal speed) × (height) − m × (vertical speed) × (range) in magnitude.

Answer: (c) The velocity at the highest point is directed along the horizontal

Explanation:
1. At the highest point the vertical component of velocity is zero, so the velocity is horizontal.
2. With air resistance the trajectory is not symmetric, the time of rise is less than the time of fall, and mechanical energy keeps decreasing.

Teacher's Note:
1. Drag opposes velocity, so it acts both against rising and falling.
2. Without drag all of (a), (b) and (d) would be true.

Answer: (c) 10 m/s

Explanation:
1. \( v_y=\frac{dy}{dt}=8-10t\Rightarrow v_{y0}=8\ \text{m/s} \), \( v_x=6\ \text{m/s} \).
2. \( u=\sqrt{8^2+6^2}=10\ \text{m/s} \)

Teacher's Note:
1. Differentiate each coordinate and evaluate at t = 0.
2. 6-8-10 is a Pythagorean triple.