JEE Physics Motion In A Plane MCQs Set 25

Answer: (b) \( (1+\sqrt5)\ \text{s} \)

Explanation:
1. Initial vertical velocity \( 20\sin30^\circ=10 \) m/s upward: \( -20=10t-5t^2 \).
2. So \( t^2-2t-4=0 \) and \( t=1+\sqrt5\ \text{s} \) (positive root).

Teacher's Note:
1. Use the full vertical displacement equation with minus the wall height.
2. Quadratic roots give one negative value, which is discarded.

Answer: (d) 2 s

Explanation:
1. The velocity is \( (5\sqrt3,\ 15) \) m/s (components along the horizontal and vertical).
2. Striking OB perpendicularly means the velocity is along the normal to OB, at \( 30^\circ \) below the horizontal, so \( v_y=-v_x\tan30^\circ=-5 \) m/s.
3. From \( 15-10t=-5 \): \( t=2 \) s.

Teacher's Note:
1. Horizontal velocity stays constant, so the final direction of the velocity gives the final vertical component immediately.

Answer: (c) \( 10\ \text{m/s} \)

Explanation:
1. At impact \( v_x=5\sqrt3 \) and \( v_y=15-10(2)=-5 \) m/s.
2. Speed \( =\sqrt{75+25}=10\ \text{m/s} \).

Teacher's Note:
1. Use the time of flight found from the perpendicular-strike condition, then recombine the components.

Answer: (b) 5 m

Explanation:
1. In 2 s the particle moves \( 5\sqrt3\times2=10\sqrt3 \) m horizontally and \( 15(2)-5(4)=10 \) m vertically.
2. Write P as \( (-p\cos30^\circ,\ p\sin30^\circ) \) and Q as \( (q\cos60^\circ,\ q\sin60^\circ) \). Then \( \frac q2+\frac{\sqrt3p}{2}=10\sqrt3 \) and \( \frac{\sqrt3q}{2}-\frac p2=10 \).
3. Solving gives \( p=10 \) m, so the height is \( p\sin30^\circ=5 \) m.

Teacher's Note:
1. Use the displacement between P and Q in both coordinates and the known directions of the two planes to locate P and Q.

Answer: (a) 20 m

Explanation:
1. The displacement from P to Q is \( (10\sqrt3,\ 10) \) m.
2. Distance \( =\sqrt{300+100}=20 \) m.

Teacher's Note:
1. The displacement is the straight-line vector between the launch and landing points, computed from the two components.

Answer: (d) 400 m

Explanation:
1. Both shots take the same time \( t \). Let the speeds be \( v_T,v_F \): \( v_T\cos30^\circ\,t=v_F\cos60^\circ\,t=200\sqrt3 \).
2. Vertical positions: from the foot \( v_F\sin60^\circ\,t-5t^2=100 \); with \( v_Ft=400\sqrt3 \) this gives \( 600-5t^2=100 \), so \( t=10 \) s.
3. Then \( v_T=40 \) m/s, \( v_F=40\sqrt3 \) m/s, and from the top: \( H+40\sin30^\circ(10)-500=100 \), giving \( H=400 \) m.

Teacher's Note:
1. Equal times and the same horizontal distance link the two shots.
2. Solve the foot shot first because it has no unknown height.

Answer: (c) 10 s

Explanation:
1. From the foot: \( v_F\cos60^\circ\,t=200\sqrt3 \) gives \( v_Ft=400\sqrt3 \).
2. Then \( v_F\sin60^\circ\,t=600 \), so \( 600-5t^2=100 \) and \( t=10 \) s.

Teacher's Note:
1. The vertical equation involves only t once the product of speed and time is known from the horizontal equation.

Answer: (b) 30 m

Explanation:
1. Before: \( H=\frac{u^2\sin^245^\circ}{2g}=120 \) gives \( u^2=4800 \).
2. After the bounce the kinetic energy is halved, so \( u'^2=2400 \).
3. New height \( \frac{2400\sin^230^\circ}{20}=30\ \text{m} \).

Teacher's Note:
1. Kinetic energy is proportional to speed squared, so halving the energy halves u squared.
2. The new angle then sets the vertical share.

Answer: (d) \( y^2=x^2+\text{constant} \)

Explanation:
1. \( \frac{dy}{dx}=\frac{v_y}{v_x}=\frac xy \).
2. So \( y\,dy=x\,dx \) and integrating, \( y^2=x^2+\text{constant} \).

Teacher's Note:
1. The slope of the path is the ratio of velocity components.
2. Separate the variables and integrate.

Answer: (c) \( \pi\dfrac{v^4}{g^2} \)

Explanation:
1. Water reaches farthest at \( 45^\circ \), with maximum range \( R_{max}=\frac{v^2}{g} \).
2. Wet area \( =\pi R_{max}^2=\frac{\pi v^4}{g^2} \).

Teacher's Note:
1. The wet region is a circle whose radius is the maximum range.
2. This is the same idea as bullets fired in all directions.

Answer: (b) 20 m

Explanation:
1. Vertical throw: \( \frac{u^2}{2g}=10 \), so \( u^2/g=20 \) m.
2. Maximum range \( =\frac{u^2}{g}=20 \) m.

Teacher's Note:
1. Maximum range is exactly twice the maximum height of a vertical throw with the same speed.

Answer: (a) \( y=2x-5x^2 \)

Explanation:
1. \( x=t \) and \( y=2t-5t^2 \).
2. Eliminating \( t \): \( y=2x-5x^2 \).

Teacher's Note:
1. With the components given, simply eliminate time.
2. Check using tan theta = 2 and the coefficient g over 2 u x squared.

Answer: (d) 20 cm

Explanation:
1. Range up the incline \( R=\frac{2u^2\sin\phi\cos(\phi+\beta)}{g\cos^2\beta} \) with \( \phi=15^\circ,\beta=30^\circ \).
2. So \( R=\frac{2(4)\sin15^\circ\cos45^\circ}{10\times0.75}\approx0.195\ \text{m} \).
3. This is close to 20 cm.

Teacher's Note:
1. Use the inclined-plane range formula with angles measured from the plane.
2. The answer is only approximate, so choose the nearest option.

Answer: (c) (2) and (3) only

Explanation:
1. Horizontal component \( u\cos\alpha \); vertical component \( u\sin\alpha-gt \). Magnitudes equal: \( u\sin\alpha-gt=\pm u\cos\alpha \).
2. So \( t=\frac ug(\sin\alpha-\cos\alpha) \) or \( t=\frac ug(\sin\alpha+\cos\alpha) \) (both positive for \( \alpha>45^\circ \)).
3. Hence statements (2) and (3) are correct.

Teacher's Note:
1. Remember that equal magnitudes allow both signs.
2. Statement (1) would be negative for angles above 45 degrees.

Answer: (b) (1) and (2) only

Explanation:
1. Trajectory: \( y=x\tan\theta-\frac{gx^2}{2u^2}(1+\tan^2\theta) \) with \( \frac{gx^2}{2u^2}=\frac{10\times576}{2\times576}=5 \).
2. At \( x=24,\ y=14 \): \( 14=24T-5(1+T^2) \), i.e. \( 5T^2-24T+19=0 \).
3. Roots \( T=1 \) and \( T=\frac{19}{5} \): statements (1) and (2).

Teacher's Note:
1. Use the trajectory form containing 1 plus tan squared theta, which turns the problem into a quadratic in tan theta.

Answer: (a) (2) and (4) only

Explanation:
1. Same maximum height means the same vertical launch component, so the same time of flight: \( T_1=T_2 \).
2. Greater range with equal time means a larger horizontal component for stone 2.
3. The speed is \( \sqrt{u_x^2+u_y^2} \) with the same \( u_y \), so \( u_1\lt u_2 \): statements (2) and (4).

Teacher's Note:
1. Height and time of flight depend only on the vertical component, and the range depends on the horizontal one as well.

Answer: (d) (1) and (3) only

Explanation:
1. Energy conservation: \( v=\sqrt{u^2+2gh} \) for every direction, so the final speeds are equal (statement 1 is true and statement 4 is false).
2. The particle thrown downward has the largest downward velocity throughout, so it lands first; the others take different times (statement 3 true, statement 2 false).

Teacher's Note:
1. Final speed depends only on the initial speed and the height dropped, while the time of flight depends on the direction.

Answer: (c) \( \frac{\sqrt2u}{g} \)

Explanation:
1. \( v_y=u-\frac{g}{\sqrt2}t=0 \).
2. So \( t=\frac{\sqrt2u}{g} \).

Teacher's Note:
1. Parallel to x means the y-component (normal to the plane) is zero, which is the top of the path as seen from the incline.

Answer: (b) \( \frac{u}{\sqrt2g} \)

Explanation:
1. \( v_x=\frac{g}{\sqrt2}t \) and \( v_y=u-\frac{g}{\sqrt2}t \).
2. Equal components: \( u=\frac{2g}{\sqrt2}t=\sqrt2gt \), so \( t=\frac{u}{\sqrt2g} \).

Teacher's Note:
1. A 45 degree direction means equal magnitudes of the two components.
2. Here both are positive.

Answer: (a) \( \frac{2u}{g} \)

Explanation:
1. Time of flight \( T=\frac{2u}{g/\sqrt2}=\frac{2\sqrt2u}{g} \) and range \( x_R=\frac12\cdot\frac{g}{\sqrt2}T^2=\frac{2\sqrt2u^2}{g} \).
2. Half the range: \( \frac{g}{2\sqrt2}t^2=\frac{\sqrt2u^2}{g} \), so \( t=\frac{2u}{g} \).

Teacher's Note:
1. Displacement along x grows with the square of time, so half the range is reached at 1 over root 2 of the flight time; 2 sqrt2 u over g divided by sqrt2 gives 2u over g, matching the direct calculation.

Answer: (d) \( \frac{2\sqrt2u}{g} \)

Explanation:
1. Normal motion: \( 0=ut-\frac12\frac{g}{\sqrt2}t^2 \).
2. So \( T=\frac{2\sqrt2u}{g} \).

Teacher's Note:
1. The stone lands when its normal coordinate returns to zero; only the normal component of gravity matters here.

Answer: (c) \( \sqrt{\frac{g}{2b}} \)

Explanation:
1. Compare with \( y=x\tan\theta-\frac{gx^2}{2u_x^2} \): \( b=\frac{g}{2u_x^2} \).
2. So \( u_x=\sqrt{\frac{g}{2b}} \).

Teacher's Note:
1. The coefficient of x squared in the trajectory involves g over twice the square of the horizontal speed.

Answer: (b) \( \sqrt{\frac{2a^2}{bg}} \)

Explanation:
1. Range \( R=\frac ab \) (where \( y=0 \)) and \( u_x=\sqrt{g/2b} \).
2. So \( T=\frac{R}{u_x}=\frac ab\sqrt{\frac{2b}{g}}=\sqrt{\frac{2a^2}{bg}} \).

Teacher's Note:
1. Time of flight is range divided by the constant horizontal speed.

Answer: (a) \( \frac{a^2}{4b} \)

Explanation:
1. The top is at \( x=\frac{a}{2b} \) (set \( dy/dx=0 \)).
2. Then \( y_{max}=a\frac{a}{2b}-b\frac{a^2}{4b^2}=\frac{a^2}{4b} \).

Teacher's Note:
1. Set the slope of the path to zero to find where the top occurs, then substitute back.

Answer: (c) 4 s

Explanation:
1. Equal heights at 1 s and 3 s are symmetric about the highest point, which is therefore reached at \( \frac{1+3}{2}=2 \) s.
2. Time of flight (landing at launch level) is twice that: \( 4 \) s.

Teacher's Note:
1. The path is symmetric about the top, so two equal-height instants average to the time of the highest point.

Answer: (b) \( \sqrt{\dfrac{2gh}{2+\cot^2\theta}} \)

Explanation:
1. Perpendicular strike: velocity \( (v_0,\ -gt) \) is normal to the plane (direction \( (\cos\theta,\sin\theta) \) along the plane), so \( v_0\cos\theta=gt\sin\theta \) and \( gt=v_0\cot\theta \).
2. Height of impact on the plane equals \( v_0t\tan\theta=\frac{v_0^2}{g} \) and the stone has dropped \( \frac12gt^2=\frac{v_0^2\cot^2\theta}{2g} \).
3. So \( h=\frac{v_0^2}{g}+\frac{v_0^2\cot^2\theta}{2g} \), giving \( v_0^2=\frac{2gh}{2+\cot^2\theta} \).

Teacher's Note:
1. Two conditions are needed: the direction condition (perpendicular strike) and the position condition (landing on the plane).
2. Together they give the speed.

Answer: (a) 1100 m

Explanation:
1. Trajectory through \( (x,105) \): \( 105=xT-\frac{10x^2}{2(110)^2}(1+T^2) \), a quadratic in \( T=\tan\theta \).
2. Real roots exist only if the discriminant is non-negative, which gives \( x^2\left(1-\frac{420}{2420}\right)\ge\frac{4x^4}{2420^2} \).
3. The limit is \( x=1100 \) m.

Teacher's Note:
1. The maximum reachable distance corresponds to the discriminant being zero, i.e. only one launch angle works.

Answer: (d) \( s(\tan\beta-\tan\alpha) \)

Explanation:
1. Both shots take the same time to reach horizontal distance \( s \), so they have the same horizontal speed component, hence the same gravity drop \( \frac{gs^2}{2u_x^2} \) below their lines of sight.
2. Without gravity the shots would meet at the point where the two straight lines cross: bottom line height \( s\tan\beta \), top line height \( H+s\tan\alpha \). Equating, \( H=s(\tan\beta-\tan\alpha) \).

Teacher's Note:
1. Because the shots have the same gravitational drop, treat the straight-line directions only.
2. Lines of aim from both points must meet at the target position plus the common drop.

Answer: (b) \( (\sqrt5-1)R \)

Explanation:
1. Flight time \( T \) is unchanged by the wind, and the extra shift is \( \Delta x=\sqrt2uT=2R \) (since \( R=\frac{u}{\sqrt2}T \) for the \( 45^\circ \) launch).
2. So the landing circle simply shifts by \( 2R \) along the wind direction, with centre at distance \( 2R \) from A, perpendicular to AB.
3. Distance from B to the new centre \( =\sqrt{R^2+(2R)^2}=\sqrt5R \); the nearest landing point is \( \sqrt5R-R=(\sqrt5-1)R \).

Teacher's Note:
1. The wind adds a uniform drift to every landing point, so the whole circle of landing points moves.
2. The shortest distance from an outside point to a circle is the distance to the centre minus the radius.