JEE Physics Motion In A Plane MCQs Set 24

Answer: (a) \( \tan\theta:1 \)

Explanation:
1. Time of flight is \( T=\frac{2u\sin\theta}{g} \), so \( T_1=\frac{2u\sin\theta}{g} \) and \( T_2=\frac{2u\cos\theta}{g} \).
2. Hence \( T_1:T_2=\sin\theta:\cos\theta=\tan\theta:1 \).

Teacher's Note:
1. Complementary angles give the same range but different times of flight and heights.
2. The time ratio is tan theta : 1 while the ratio of maximum heights is tan squared theta : 1.

Answer: (d) \( \tan^{-1}4 \)

Explanation:
1. Range \( R=\frac{2u^2\sin\theta\cos\theta}{g} \) and height \( H=\frac{u^2\sin^2\theta}{2g} \).
2. Setting \( R=H \): \( 4\cos\theta=\sin\theta \), so \( \tan\theta=4 \) and \( \theta\approx76^\circ \).

Teacher's Note:
1. Divide R by H to eliminate the speed: R/H = 4 cot theta, so R = H gives tan theta = 4.
2. This angle is much steeper than the 45 degrees of maximum range.

Answer: (c) \( \frac{2R}{g} \)

Explanation:
1. The two angles are \( \theta \) and \( 90^\circ-\theta \): \( T_1=\frac{2u\sin\theta}{g},\ T_2=\frac{2u\cos\theta}{g} \).
2. \( T_1T_2=\frac{4u^2\sin\theta\cos\theta}{g^2}=\frac{2}{g}\cdot\frac{u^2\sin2\theta}{g}=\frac{2R}{g} \).

Teacher's Note:
1. Keep the combination u squared times sin 2 theta over g together, because that is the range.
2. This product relation is a standard trick for two-angle problems.

Answer: (b) 50 m

Explanation:
1. Maximum range \( u^2/g=100\ \text{m} \).
2. Vertical throw: \( H=\frac{u^2}{2g}=\frac{100}{2}=50\ \text{m} \).

Teacher's Note:
1. Maximum range is u squared over g, and the greatest height reached by a vertical throw is half of it.
2. Remember the pair R max = u squared / g and H max = u squared / 2g.

Answer: (a) \( 5\sqrt2\ \text{m/s} \)

Explanation:
1. Range \( R=\frac{u^2\sin90^\circ}{g}=40\ \text{m} \) and time \( T=\frac{2u\sin45^\circ}{g}=2\sqrt2\ \text{s} \).
2. The player must cover \( 60-40=20 \) m in \( 2\sqrt2 \) s, so \( v=\frac{20}{2\sqrt2}=5\sqrt2\ \text{m/s} \).

Teacher's Note:
1. Find where and when the ball lands first, then treat the runner as a separate uniform-speed problem.
2. Both motions share only the time.

Answer: (d) 980 m

Explanation:
1. \( 490=\tfrac12(9.8)t^2\Rightarrow t=10\ \text{s} \).
2. Horizontal distance \( x=98\times10=980\ \text{m} \).

Teacher's Note:
1. Vertical fall time depends only on the height, not on the horizontal speed.
2. Then the horizontal distance is simply speed times that time.

Answer: (c) \( 98\sqrt2\ \text{m/s} \)

Explanation:
1. \( v_x=98 \) m/s and \( v_y=gt=9.8\times10=98 \) m/s.
2. Speed \( =\sqrt{98^2+98^2}=98\sqrt2\ \text{m/s} \), at \( 45^\circ \) below the horizontal.

Teacher's Note:
1. The horizontal component never changes, while the vertical component grows as g t.
2. Combine them with Pythagoras for the final speed.

Answer: (b) 61.25 m

Explanation:
1. Vertical component \( u_y=30\sin30^\circ=15 \) m/s, rise above the tower \( \frac{15^2}{2\times10}=11.25 \) m.
2. Height above ground \( 50+11.25=61.25\ \text{m} \).

Teacher's Note:
1. Maximum height is measured from the point of projection first, then the height of that point is added.
2. Do not forget the tower height.

Answer: (a) 5 s

Explanation:
1. Take upward positive: \( -50=15t-5t^2 \), i.e. \( t^2-3t-10=0 \).
2. Factorising, \( (t-5)(t+2)=0 \), so \( t=5 \) s.

Teacher's Note:
1. Use the full displacement equation with the displacement equal to minus the tower height, and reject the negative root.
2. The initial vertical velocity is 15 m/s upward.

Answer: (d) \( 2\ \text{m/s} \)

Explanation:
1. Compare with \( y=x\tan\theta-\frac{gx^2}{2u^2\cos^2\theta} \): \( \tan\theta=\sqrt3 \), so \( \theta=60^\circ \).
2. \( \frac{g}{2u^2\cos^2\theta}=\frac g2\Rightarrow u\cos\theta=1 \), hence \( u=\frac{1}{\cos60^\circ}=2\ \text{m/s} \).

Teacher's Note:
1. Match the coefficient of x to tan theta and the coefficient of x squared to g over 2 u squared cos squared theta.
2. Both unknowns then follow directly.

Answer: (c) 53

Explanation:
1. Use \( y=x\tan\theta\left(1-\frac xR\right) \): \( \frac R4=\frac{3R}{4}\tan\theta\cdot\frac14 \).
2. So \( \tan\theta=\frac43 \) and \( \theta=53^\circ \).

Teacher's Note:
1. The range form of the trajectory equation removes the speed entirely.
2. Recognise tan theta = 4/3 as the 3-4-5 triangle angle of about 53 degrees.

Answer: (b) \( y=-\dfrac{b}{a^2}x^2 \)

Explanation:
1. \( x=at \) and \( y=-bt^2 \).
2. Eliminating \( t=x/a \): \( y=-\frac{b}{a^2}x^2 \), a downward parabola.

Teacher's Note:
1. To find a path, eliminate time between the two coordinate equations.
2. The path is independent of how fast the point moves along it.

Answer: (a) \( 0^\circ<\theta<180^\circ \)

Explanation:
1. Gravity points straight down. While rising the velocity has an upward part, so the angle with gravity is more than \( 90^\circ \).
2. At the top it is exactly \( 90^\circ \), and while falling it is less than \( 90^\circ \); it never reaches \( 0^\circ \) or \( 180^\circ \) because the horizontal component never vanishes.

Teacher's Note:
1. Track the vertical component of velocity through the flight: up, zero, down.
2. The horizontal component keeps the velocity from ever being exactly parallel to gravity.

Answer: (d) \( v\sin\theta \), horizontal, in the plane of projection

Explanation:
1. At the two instants the heights are equal, so the vertical displacement is zero and the average velocity is horizontal.
2. The horizontal velocity is constant and equals \( v\sin\theta \) (angle measured from the vertical), so the average velocity is \( v\sin\theta \) along the horizontal in the plane of projection.

Teacher's Note:
1. Average velocity is displacement divided by time, so equal heights remove the vertical part.
2. Read the angle carefully: it is measured from the vertical here, so the horizontal component is v sin theta.

Answer: (c) \( 2ac^2\,\hat j \)

Explanation:
1. \( y=ax^2\Rightarrow v_y=2ax\,v_x=2acx \).
2. With \( x=ct \): \( a_y=\frac{d v_y}{dt}=2ac\cdot c=2ac^2 \) and \( a_x=0 \).

Teacher's Note:
1. Differentiate the path equation twice with respect to time, using the constant horizontal velocity.
2. The acceleration comes out constant and vertical, as it must be for a projectile.

Answer: (b) \( \dfrac{\sqrt3u^2}{2g} \)

Explanation:
1. Speed at top is \( u\cos\theta=\frac u2 \), so \( \theta=60^\circ \).
2. Range \( R=\frac{u^2\sin120^\circ}{g}=\frac{\sqrt3u^2}{2g} \).

Teacher's Note:
1. At the top only the horizontal component survives, which gives cos theta immediately.
2. Then use the range formula with sin 2 theta.

Answer: (a) \( 18\ \text{m/s} \)

Explanation:
1. \( \tan45^\circ=\frac{v_y}{v_x}=1 \).
2. The horizontal component stays \( 18 \) m/s, so \( v_y=18\ \text{m/s} \).

Teacher's Note:
1. The horizontal velocity is unchanged for the whole flight, so the angle fixes the vertical component at once.
2. Equal components mean a 45 degree strike.

Answer: (d) \( 15\sqrt3\ \text{m/s} \)

Explanation:
1. Height of the tower \( h=\tfrac12(10)(3)^2=45\ \text{m} \).
2. Since AB makes \( 30^\circ \) with the horizontal, \( x=\frac{45}{\tan30^\circ}=45\sqrt3\ \text{m} \).
3. Speed \( u=\frac{45\sqrt3}{3}=15\sqrt3\ \text{m/s} \).

Teacher's Note:
1. The line AB is the displacement vector, so its slope is vertical drop over horizontal range.
2. Use the fall time to get the drop first.

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

Explanation:
1. Bomb falls for \( t=\sqrt{2\times80/10}=4 \) s and lands \( 20\times4=80 \) m ahead of the release point. The hunter stops after \( 10\times2=20 \) m.
2. The bullet must cover \( 60 \) m horizontally and return to the ground in the remaining \( 2 \) s: \( u_x=30 \) m/s and \( \frac{2u_y}{g}=2\Rightarrow u_y=10 \) m/s.
3. Speed \( =\sqrt{30^2+10^2}=10\sqrt{10}\ \text{m/s} \).

Teacher's Note:
1. Split the situation into the bomb's flight and the bullet's flight, linked by time and place of meeting.
2. The bullet starts and ends at ground level, so its time of flight fixes its vertical speed.

Answer: (b) \( \dfrac{v^2}{g(1+\sin\theta)} \)

Explanation:
1. Up the incline, \( R_{max}=\frac{u^2}{g(1+\sin\theta)} \), reached when the firing direction bisects the angle between the incline and the vertical.
2. So the maximum range is \( \frac{v^2}{g(1+\sin\theta)} \), smaller than on level ground.

Teacher's Note:
1. Sin theta is positive for an upward slope, which reduces the range; the downward-slope version has 1 minus sin theta.
2. For theta equal to zero both reduce to u squared over g.

Answer: (a) \( \dfrac{2\sqrt2\,v^2}{g} \)

Explanation:
1. Take coordinates \( x=vt,\ y=\tfrac12gt^2 \) (downward). Landing on the plane: \( \tfrac12gt^2=vt\tan45^\circ \), so \( t=\frac{2v}{g} \).
2. Then \( x=\frac{2v^2}{g} \) and \( y=\frac{2v^2}{g} \), so the distance along the plane is \( \sqrt2\cdot\frac{2v^2}{g}=\frac{2\sqrt2v^2}{g} \).

Teacher's Note:
1. Find the time when the vertical drop equals the horizontal distance times tan of the incline angle.
2. The distance along the slope is the straight-line distance, not just the horizontal part.

Answer: (d) 3 m

Explanation:
1. Perpendicular to the incline: \( u_\perp=10\sin37^\circ=6 \) m/s and \( a_\perp=g\cos53^\circ=6\ \text{m/s}^2 \) (towards the plane).
2. Maximum distance \( \frac{u_\perp^2}{2a_\perp}=\frac{36}{12}=3\ \text{m} \).

Teacher's Note:
1. On an incline, take axes along and perpendicular to the plane.
2. The maximum distance from the plane is where the perpendicular velocity becomes zero.

Answer: (c) \( H_A+H_C=H_B \)

Explanation:
1. Height above the incline \( H=\frac{u^2\sin^2\phi}{2g\cos\theta} \), with \( \phi \) the angle with the incline.
2. So \( H_A\propto\sin^2\alpha,\ H_B\propto1,\ H_C\propto\cos^2\alpha \).
3. Since \( \sin^2\alpha+\cos^2\alpha=1 \), \( H_A+H_C=H_B \).

Teacher's Note:
1. Only the perpendicular component matters for the height above the incline.
2. Complementary angles give sin squared and cos squared, which add up to one.

Answer: (b) \( [(8+2t)\hat i-15\hat j]\ \text{m/s} \)

Explanation:
1. Only the \( x \) velocity changes: \( v_x=8+2t \).
2. The \( y \) velocity stays \( -15 \) m/s because there is no acceleration in that direction.

Teacher's Note:
1. Treat each axis separately with its own initial velocity and acceleration.
2. A zero acceleration component means that velocity component is constant.

Answer: (a) \( 60^\circ \)

Explanation:
1. To remain below the plane until the hit, the shell and plane must have the same horizontal velocity: \( 400\cos\theta=200 \).
2. So \( \cos\theta=\tfrac12 \) and \( \theta=60^\circ \).

Teacher's Note:
1. To hit a body moving horizontally at constant speed, the projectile must match its horizontal velocity.
2. The height only decides whether the shell can reach it in time.

Answer: (d) the maximum horizontal range

Explanation:
1. \( R=\frac{2u^2\sin\theta\cos\theta}{g},\ h=\frac{u^2\sin^2\theta}{2g} \).
2. \( \frac{R^2}{8h}=\frac{u^2\cos^2\theta}{g} \) and \( 2h=\frac{u^2\sin^2\theta}{g} \).
3. Sum \( =\frac{u^2}{g} \), the maximum horizontal range.

Teacher's Note:
1. Substitute the standard formulas and use sin squared plus cos squared equals one.
2. The quantity u squared over g is the range at 45 degrees.

Answer: (c) \( \sqrt{3gh} \)

Explanation:
1. Horizontal component \( u\cos60^\circ=\frac u2 \). At the roof the vertical component has magnitude \( \frac{u}{2}\tan30^\circ=\frac{u}{2\sqrt3} \) (downward).
2. Vertical motion: \( v_y^2=u^2\sin^260^\circ-2gh \) gives \( \frac{u^2}{12}=\frac{3u^2}{4}-2gh \).
3. So \( 2gh=\frac{2u^2}{3} \) and \( u=\sqrt{3gh} \).

Teacher's Note:
1. The horizontal component of velocity is the link between launch and impact angles.
2. Then apply the vertical equation v squared = u squared minus 2 g h.

Answer: (b) \( \tan^{-1}\sqrt5 \)

Explanation:
1. First case: \( v_x=u\cos30^\circ \) and at the ground \( v_y=v_x\tan45^\circ=u\cos30^\circ \). Then \( v_y^2=u^2\sin^230^\circ+2gh \) gives \( \frac{3u^2}{4}=\frac{u^2}{4}+2gh \), so \( 2gh=\frac{u^2}{2} \).
2. Second case: \( v_x=\frac u2 \) and \( v_y^2=\frac{3u^2}{4}+\frac{u^2}{2}=\frac{5u^2}{4} \).
3. So \( \tan\phi=\frac{\sqrt5u/2}{u/2}=\sqrt5 \).

Teacher's Note:
1. The first case is used only to find the combination 2gh in terms of u squared.
2. Strike speed and angle then follow from conservation of horizontal velocity.