401. The position vector of a particle is
\[
\vec r(t)=3\cos(2t)\hat i+3\sin(2t)\hat j
\]
in metres, where \(t\) is in seconds. Which statement is correct at \(t=\pi/4\,\text{s}\)?
ⓐ. \(\vec r=-3\hat j,\ \vec v=6\hat i,\ \vec a=12\hat j\)
ⓑ. \(\vec r=3\hat i,\ \vec v=6\hat j,\ \vec a=-12\hat i\), period \(=2\pi\,\text{s}\)
ⓒ. \(\vec r=3\hat j,\ \vec v=-3\hat i,\ \vec a=-6\hat j\)
ⓓ. \(\vec r=3\hat j,\ \vec v=-6\hat i\,\text{m s}^{-1},\ \vec a=-12\hat j\,\text{m s}^{-2},\ T=\pi\,\text{s}\)
Correct Answer: \(\vec r=3\hat j,\ \vec v=-6\hat i\,\text{m s}^{-1},\ \vec a=-12\hat j\,\text{m s}^{-2},\ T=\pi\,\text{s}\)
Explanation: The position has the standard circular form
\[
\vec r=R\cos\omega t\,\hat i+R\sin\omega t\,\hat j.
\]
Hence
\[
R=3\,\text{m},\qquad \omega=2\,\text{rad s}^{-1}.
\]
At \(t=\pi/4\,\text{s}\),
\[
\omega t=\frac{\pi}{2},
\]
so
\[
\vec r=3\hat j.
\]
Differentiating,
\[
\vec v=-6\sin(2t)\hat i+6\cos(2t)\hat j.
\]
Therefore
\[
\vec v=-6\hat i
\]
at the stated instant. The acceleration is
\[
\vec a=-12\cos(2t)\hat i-12\sin(2t)\hat j,
\]
giving
\[
\vec a=-12\hat j.
\]
The speed is \(R\omega=6\,\text{m s}^{-1}\), and the period is
\[
T=\frac{2\pi}{2}=\pi\,\text{s}.
\]
402. A particle moves uniformly on a circle of radius \(5\,\text{m}\). Starting from a point A, the magnitude of its displacement becomes \(10\,\text{m}\) for the first time after \(2\,\text{s}\). Which set of results is correct?
ⓐ. Period \(=2\,\text{s}\), speed \(=5\pi\,\text{m s}^{-1}\), centripetal acceleration \(=5\pi^2\,\text{m s}^{-2}\)
ⓑ. Period \(=4\,\text{s}\), speed \(=\dfrac{5\pi}{2}\,\text{m s}^{-1}\), centripetal acceleration \(=\dfrac{5\pi^2}{4}\,\text{m s}^{-2}\)
ⓒ. Period \(=8\,\text{s}\), speed \(=\dfrac{5\pi}{4}\,\text{m s}^{-1}\), centripetal acceleration \(=\dfrac{5\pi^2}{16}\,\text{m s}^{-2}\)
ⓓ. Period \(=4\,\text{s}\), speed \(=5\,\text{m s}^{-1}\), centripetal acceleration \(=5\,\text{m s}^{-2}\)
Correct Answer: Period \(=4\,\text{s}\), speed \(=\dfrac{5\pi}{2}\,\text{m s}^{-1}\), centripetal acceleration \(=\dfrac{5\pi^2}{4}\,\text{m s}^{-2}\)
Explanation: The largest possible displacement between two points on a circle of radius \(5\,\text{m}\) is its diameter:
\[
2R=10\,\text{m}.
\]
Therefore displacement magnitude \(10\,\text{m}\) means the particle is at the point diametrically opposite A. Reaching that point for the first time corresponds to half a revolution. Since this takes \(2\,\text{s}\),
\[
\frac T2=2\,\text{s}.
\]
Hence
\[
T=4\,\text{s}.
\]
The distance travelled in half a revolution is
\[
\pi R=5\pi\,\text{m}.
\]
Thus
\[
v=\frac{5\pi}{2}\,\text{m s}^{-1}.
\]
The centripetal acceleration is
\[
a_c=\frac{v^2}{R}
=\frac{(25\pi^2/4)}{5}
=\frac{5\pi^2}{4}\,\text{m s}^{-2}.
\]
The displacement condition therefore determines the angular position before the speed and acceleration are inferred.
403. Consider the following Assertion and Reason.
Assertion: The average acceleration of a particle over one complete revolution in uniform circular motion is zero.
Reason: After one complete revolution, the particle's final velocity vector is identical to its initial velocity vector.
ⓐ. Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
ⓑ. Both Assertion and Reason are true, but Reason is not the correct explanation of Assertion.
ⓒ. Assertion is true, but Reason is false.
ⓓ. Assertion is false, but Reason is true.
Correct Answer: Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
Explanation: Average acceleration over a time interval is defined as
\[
\vec a_{\rm avg}=\frac{\Delta\vec v}{\Delta t}.
\]
In one complete revolution the particle returns not only to its original position but also to the same direction of motion. Since its speed is unchanged in uniform circular motion, the complete velocity vector is again the same. Therefore
\[
\Delta\vec v=\vec0.
\]
It follows that
\[
\vec a_{\rm avg}=\vec0.
\]
This does not mean that the instantaneous acceleration is zero during the revolution. At every instant the particle possesses centripetal acceleration
\[
a_c=\frac{v^2}{R}
\]
toward the centre. Thus the Assertion is true, and the Reason correctly explains why the finite-interval average acceleration vanishes.
404. Consider the following statements for a particle in uniform circular motion of radius \(R\) and speed \(v\).
Statement I: Its speed is constant but its velocity vector changes continuously.
Statement II: Its instantaneous acceleration is perpendicular to its instantaneous velocity.
Statement III: Over one complete revolution, its displacement is zero but the distance travelled is \(2\pi R\).
Statement IV: Over one complete revolution, its average velocity is zero but its average speed is \(v\).
Which set of statements is correct?
ⓐ. Statement I and Statement II only
ⓑ. Statement I, Statement II and Statement III only
ⓒ. Statement II, Statement III and Statement IV only
ⓓ. Statement I, Statement II, Statement III and Statement IV
Correct Answer: Statement I, Statement II, Statement III and Statement IV
Explanation: Uniform circular motion means that the magnitude of velocity remains fixed. Its direction changes continuously, so Statement I is correct. The instantaneous acceleration points toward the centre, whereas the velocity is tangent to the circle. Radius and tangent are perpendicular, making Statement II correct. After one complete revolution the final position equals the initial position, so displacement is zero. However, the actual path length is the circumference
\[
2\pi R,
\]
so Statement III is correct. Average velocity is displacement divided by time and is therefore zero over the full revolution. Average speed is total distance divided by the period:
\[
\frac{2\pi R}{T}=v.
\]
Hence Statement IV is also correct. Zero average velocity therefore coexists with continuous non-zero motion.
405. A particle moves clockwise with uniform speed \(10\,\text{m s}^{-1}\) on a circle of radius \(5\,\text{m}\) centred at the origin. At an instant its position is
\[
\vec r=3\hat i+4\hat j\ \text{m}.
\]
Which statement is correct?
ⓐ. \(\vec v=8\hat i-6\hat j,\ \vec a=12\hat i+16\hat j,\ \omega=0.5\,\text{rad s}^{-1}\)
ⓑ. \(\vec v=-8\hat i+6\hat j,\ \vec a=12\hat i+16\hat j,\ \omega=5\,\text{rad s}^{-1}\)
ⓒ. \(\vec v=6\hat i+8\hat j,\ \vec a=-20\hat j,\ \omega=2\,\text{rad s}^{-1}\)
ⓓ. \(\vec v=8\hat i-6\hat j\,\text{m s}^{-1},\ \vec a=-12\hat i-16\hat j\,\text{m s}^{-2},\ \omega=2\,\text{rad s}^{-1}\)
Correct Answer: \(\vec v=8\hat i-6\hat j\,\text{m s}^{-1},\ \vec a=-12\hat i-16\hat j\,\text{m s}^{-2},\ \omega=2\,\text{rad s}^{-1}\)
Explanation: The radial unit vector at the point is
\[
\hat r=\frac{3\hat i+4\hat j}{5}.
\]
For counterclockwise motion the tangent unit vector would be
\[
-\frac45\hat i+\frac35\hat j.
\]
The motion is clockwise, so the tangent direction is reversed:
\[
\hat t=\frac45\hat i-\frac35\hat j.
\]
Therefore the velocity is
\[
\vec v=10\hat t
=8\hat i-6\hat j.
\]
The centripetal acceleration magnitude is
\[
a_c=\frac{v^2}{R}=\frac{100}{5}=20\,\text{m s}^{-2}.
\]
It points toward the origin, opposite to \(\hat r\), so
\[
\vec a=-20\hat r=-12\hat i-16\hat j.
\]
Finally,
\[
\omega=\frac vR=\frac{10}{5}=2\,\text{rad s}^{-1}.
\]
406. Two particles start simultaneously from the same point on a circular track of radius \(R\). Each moves with constant speed \(v\), but they travel in opposite directions. When do they meet for the first time after starting, and what is their relative speed at that instant?
ⓐ. After \(\dfrac{\pi R}{2v}\), at a point \(90^\circ\) from the start, with relative speed \(v\)
ⓑ. After \(\dfrac{\pi R}{v}\), at the point diametrically opposite the start, with relative speed \(2v\)
ⓒ. After \(\dfrac{2\pi R}{v}\), back at the start, with relative speed zero
ⓓ. After \(\dfrac{\pi R}{v}\), at the starting point, with relative speed \(v\)
Correct Answer: After \(\dfrac{\pi R}{v}\), at the point diametrically opposite the start, with relative speed \(2v\)
Explanation: Each particle has angular speed
\[
\omega=\frac vR.
\]
Because they move in opposite directions, their relative angular speed is
\[
2\omega=\frac{2v}{R}.
\]
For their next coincidence, their relative angular displacement must become \(2\pi\). Thus
\[
t=\frac{2\pi}{2v/R}
=\frac{\pi R}{v}.
\]
During this time each particle travels
\[
s=vt=\pi R,
\]
which is half the circumference. Hence both reach the point diametrically opposite the start. At that common point their tangent directions are opposite because the particles are travelling in opposite senses. Therefore their velocity vectors are equal in magnitude and opposite in direction. The relative speed is consequently
\[
|\vec v_1-\vec v_2|=2v.
\]
The particles meet even though their relative velocity at the meeting instant is non-zero.
407. Particle P moves uniformly on a circle of radius \(R\) with speed \(v\). Particle Q moves uniformly on another circle of radius \(4R\) with speed \(2v\). Which comparison is correct?
ⓐ. \(a_Q=4a_P,\ \omega_Q=\omega_P/2,\ T_Q=4T_P\)
ⓑ. \(a_Q=2a_P,\ \omega_Q=2\omega_P,\ T_Q=T_P/2\)
ⓒ. \(a_Q=a_P/2,\ \omega_Q=\omega_P,\ T_Q=T_P\)
ⓓ. \(a_Q=a_P,\ \omega_Q=\omega_P/2,\ T_Q=2T_P\)
Correct Answer: \(a_Q=a_P,\ \omega_Q=\omega_P/2,\ T_Q=2T_P\)
Explanation: For P,
\[
a_P=\frac{v^2}{R}.
\]
For Q,
\[
a_Q=\frac{(2v)^2}{4R}
=\frac{4v^2}{4R}
=\frac{v^2}{R}.
\]
Thus the centripetal accelerations have equal magnitudes. The angular speed is
\[
\omega=\frac vr.
\]
Therefore
\[
\omega_Q=\frac{2v}{4R}
=\frac12\frac vR
=\frac{\omega_P}{2}.
\]
The period is inversely proportional to angular speed:
\[
T=\frac{2\pi}{\omega}.
\]
Hence
\[
T_Q=2T_P.
\]
Doubling the speed does not necessarily increase centripetal acceleration when the radius is changed simultaneously. The combination \(v^2/R\), not speed alone, controls the acceleration magnitude.
408. A particle executes uniform circular motion about the origin. At a certain instant,
\[
\vec v=v\hat i
\]
and
\[
\vec a=-\frac{v^2}{R}\hat j.
\]
Which statement correctly identifies the particle's position and its state after one-quarter period?
ⓐ. Initially \(\vec r=R\hat j\) and the motion is clockwise; after one-quarter period \(\vec r=R\hat i,\ \vec v=-v\hat j,\ \vec a=-\dfrac{v^2}{R}\hat i\)
ⓑ. Initially \(\vec r=-R\hat j\) and the motion is counterclockwise; after one-quarter period \(\vec r=R\hat i,\ \vec v=v\hat j\)
ⓒ. Initially \(\vec r=R\hat i\); after one-quarter period the velocity becomes zero
ⓓ. Initially \(\vec r=R\hat j\) and the motion is counterclockwise; after one-quarter period \(\vec r=-R\hat i,\ \vec v=-v\hat j\)
Correct Answer: Initially \(\vec r=R\hat j\) and the motion is clockwise; after one-quarter period \(\vec r=R\hat i,\ \vec v=-v\hat j,\ \vec a=-\dfrac{v^2}{R}\hat i\)
Explanation: In uniform circular motion, the acceleration always points from the particle toward the centre. Since
\[
\vec a=-\frac{v^2}{R}\hat j,
\]
the centre lies directly below the particle. With the centre at the origin, the position must therefore be
\[
\vec r=R\hat j.
\]
At the top of the circle, a velocity in the \(+\hat i\) direction corresponds to clockwise motion. After one-quarter period, clockwise motion carries the particle from the topmost point to the rightmost point. Hence
\[
\vec r=R\hat i.
\]
At the rightmost point, the clockwise tangent points downward, giving
\[
\vec v=-v\hat j.
\]
The centripetal acceleration points toward the origin, which is leftward:
\[
\vec a=-\frac{v^2}{R}\hat i.
\]
Thus the given instantaneous vectors are sufficient to infer both the location and the sense of circular motion.
409. A projectile is launched from level ground with speed \(20\,\text{m s}^{-1}\) at \(60^\circ\) above the horizontal. Take \(g=10\,\text{m s}^{-2}\). At the instant when its velocity becomes perpendicular to its launch velocity, which set of results is correct?
ⓐ. \(t=\dfrac{4}{\sqrt3}\,\text{s},\ \vec r=\dfrac{40}{\sqrt3}\hat i+\dfrac{40}{3}\hat j\,\text{m},\ v=\dfrac{20}{\sqrt3}\,\text{m s}^{-1}\)
ⓑ. \(t=\dfrac{2}{\sqrt3}\,\text{s},\ \vec r=\dfrac{20}{\sqrt3}\hat i+\dfrac{20}{3}\hat j\,\text{m},\ v=\dfrac{10}{\sqrt3}\,\text{m s}^{-1}\)
ⓒ. \(t=2\sqrt3\,\text{s},\ \vec r=20\sqrt3\,\hat i\,\text{m},\ v=10\,\text{m s}^{-1}\)
ⓓ. The velocity can never become perpendicular to the launch velocity before landing.
Correct Answer: \(t=\dfrac{4}{\sqrt3}\,\text{s},\ \vec r=\dfrac{40}{\sqrt3}\hat i+\dfrac{40}{3}\hat j\,\text{m},\ v=\dfrac{20}{\sqrt3}\,\text{m s}^{-1}\)
Explanation: The launch components are
\[
u_x=20\cos60^\circ=10\,\text{m s}^{-1},
\]
\[
u_y=20\sin60^\circ=10\sqrt3\,\text{m s}^{-1}.
\]
At time \(t\),
\[
\vec v=10\hat i+(10\sqrt3-10t)\hat j.
\]
The velocity is perpendicular to the initial velocity when
\[
\vec u\cdot\vec v=0.
\]
Therefore
\[
(10)(10)+(10\sqrt3)(10\sqrt3-10t)=0.
\]
This gives
\[
400-100\sqrt3\,t=0,
\]
so
\[
t=\frac4{\sqrt3}\,\text{s}.
\]
The coordinates are
\[
x=10t=\frac{40}{\sqrt3}\,\text{m},
\]
and
\[
y=10\sqrt3\,t-5t^2
=40-\frac{80}{3}
=\frac{40}{3}\,\text{m}.
\]
At this instant,
\[
v_y=10\sqrt3-\frac{40}{\sqrt3}=-\frac{10}{\sqrt3}\,\text{m s}^{-1}.
\]
Hence
\[
v=\sqrt{10^2+\left(\frac{10}{\sqrt3}\right)^2}
=\frac{20}{\sqrt3}\,\text{m s}^{-1}.
\]
410. A projectile is launched from level ground with horizontal and vertical velocity components \(15\,\text{m s}^{-1}\) and \(20\,\text{m s}^{-1}\), respectively. Take \(g=10\,\text{m s}^{-2}\). During ascent and descent its velocity makes angles \(+45^\circ\) and \(-45^\circ\), respectively, with the horizontal. Which statement is correct?
ⓐ. Both events occur at height \(5\,\text{m}\) and are separated horizontally by \(30\,\text{m}\).
ⓑ. Both events occur at maximum height \(20\,\text{m}\).
ⓒ. Height \(=\dfrac{35}{4}\,\text{m}\) at both events; time separation \(=3\,\text{s}\), horizontal separation \(=45\,\text{m}\).
ⓓ. The two events occur at different heights because the vertical velocity has opposite signs.
Correct Answer: Height \(=\dfrac{35}{4}\,\text{m}\) at both events; time separation \(=3\,\text{s}\), horizontal separation \(=45\,\text{m}\).
Explanation: The horizontal velocity remains
\[
v_x=15\,\text{m s}^{-1}.
\]
For the velocity to make \(45^\circ\) with the horizontal,
\[
|v_y|=v_x=15\,\text{m s}^{-1}.
\]
During ascent,
\[
v_y=+15\,\text{m s}^{-1},
\]
whereas during descent,
\[
v_y=-15\,\text{m s}^{-1}.
\]
Using
\[
v_y^2=u_y^2-2gy,
\]
we get
\[
225=400-20y.
\]
Hence
\[
y=\frac{175}{20}=\frac{35}{4}\,\text{m}.
\]
The time required for \(v_y\) to change from \(+15\) to \(-15\,\text{m s}^{-1}\) is
\[
\Delta t=\frac{30}{10}=3\,\text{s}.
\]
Therefore the horizontal separation is
\[
\Delta x=v_x\Delta t=(15)(3)=45\,\text{m}.
\]
The opposite signs of the vertical velocities distinguish ascent from descent but do not change the height corresponding to the same \(v_y^2\).
411. The velocity components of a projectile are represented by the following description.
The \(v_x-t\) graph is a horizontal line at \(15\,\text{m s}^{-1}\) from \(t=0\) to \(t=4\,\text{s}\). The \(v_y-t\) graph is a straight line from \(+20\,\text{m s}^{-1}\) at \(t=0\), through zero at \(t=2\,\text{s}\), to \(-20\,\text{m s}^{-1}\) at \(t=4\,\text{s}\). The projectile is at its launch level at \(t=0\) and \(t=4\,\text{s}\).
Which set of results follows from the graphs?
ⓐ. \(g=10\,\text{m s}^{-2},\ H=20\,\text{m},\ R=60\,\text{m},\ u=25\,\text{m s}^{-1}\)
ⓑ. \(g=5\,\text{m s}^{-2},\ H=40\,\text{m},\ R=60\,\text{m},\ u=25\,\text{m s}^{-1}\)
ⓒ. \(g=10\,\text{m s}^{-2},\ H=40\,\text{m},\ R=30\,\text{m},\ u=25\,\text{m s}^{-1}\)
ⓓ. \(g=20\,\text{m s}^{-2},\ H=10\,\text{m},\ R=60\,\text{m},\ u=25\,\text{m s}^{-1}\)
Correct Answer: \(g=10\,\text{m s}^{-2},\ H=20\,\text{m},\ R=60\,\text{m},\ u=25\,\text{m s}^{-1}\)
Explanation: The slope of the \(v_y-t\) graph is the vertical acceleration:
\[
a_y=\frac{-20-(+20)}{4}=-10\,\text{m s}^{-2}.
\]
Therefore
\[
g=10\,\text{m s}^{-2}.
\]
The projectile reaches its highest point when
\[
v_y=0,
\]
which occurs at \(t=2\,\text{s}\). The vertical displacement up to that instant equals the area under the positive part of the \(v_y-t\) graph:
\[
H=\frac12(2)(20)=20\,\text{m}.
\]
The horizontal velocity is constant at \(15\,\text{m s}^{-1}\). Over the total \(4\,\text{s}\) flight,
\[
R=(15)(4)=60\,\text{m}.
\]
The initial velocity components are \(15\) and \(20\,\text{m s}^{-1}\), so
\[
u=\sqrt{15^2+20^2}=25\,\text{m s}^{-1}.
\]
The two velocity-component graphs therefore encode the acceleration, maximum height, range and launch speed.
412. A projectile is launched from the top of a cliff. It strikes the ground \(20\,\text{m}\) below the launch level after \(4\,\text{s}\). Its highest point is horizontally \(30\,\text{m}\) from the launch point. Take \(g=10\,\text{m s}^{-2}\). Which statement is correct?
ⓐ. \(\vec u=15\hat i+20\hat j\,\text{m s}^{-1},\ x_{\rm impact}=60\,\text{m},\ v_{\rm impact}=5\sqrt{20}\,\text{m s}^{-1}\)
ⓑ. \(\vec u=20\hat i+15\hat j\,\text{m s}^{-1},\ x_{\rm impact}=80\,\text{m},\ v_{\rm impact}=5\sqrt{41}\,\text{m s}^{-1}\)
ⓒ. \(\vec u=20\hat i+20\hat j\,\text{m s}^{-1},\ x_{\rm impact}=80\,\text{m},\ v_{\rm impact}=20\sqrt2\,\text{m s}^{-1}\)
ⓓ. \(\vec u=10\hat i+15\hat j\,\text{m s}^{-1},\ x_{\rm impact}=40\,\text{m},\ v_{\rm impact}=5\sqrt{17}\,\text{m s}^{-1}\)
Correct Answer: \(\vec u=20\hat i+15\hat j\,\text{m s}^{-1},\ x_{\rm impact}=80\,\text{m},\ v_{\rm impact}=5\sqrt{41}\,\text{m s}^{-1}\)
Explanation: Taking the launch point as \(y=0\), the impact point has
\[
y=-20\,\text{m}.
\]
The vertical equation at \(t=4\,\text{s}\) is
\[
-20=4u_y-\frac12(10)(4^2).
\]
Thus
\[
-20=4u_y-80,
\]
so
\[
u_y=15\,\text{m s}^{-1}.
\]
The highest point is reached after
\[
t_H=\frac{u_y}{g}=1.5\,\text{s}.
\]
Its horizontal coordinate is \(30\,\text{m}\), hence
\[
u_x=\frac{30}{1.5}=20\,\text{m s}^{-1}.
\]
The horizontal impact distance is therefore
\[
x=(20)(4)=80\,\text{m}.
\]
At impact,
\[
v_x=20\,\text{m s}^{-1},
\]
and
\[
v_y=15-40=-25\,\text{m s}^{-1}.
\]
Hence
\[
v=\sqrt{20^2+25^2}=5\sqrt{41}\,\text{m s}^{-1}.
\]
413. Projectile P is launched from level ground with speed \(u\) at \(30^\circ\) in a gravitational field of magnitude \(g\). Projectile Q is launched from level ground with the same speed \(u\) in a gravitational field of magnitude \(2g\). Both return to their launch levels, and their maximum heights are equal. Which comparison is correct?
ⓐ. \(\theta_Q=30^\circ,\ R_Q/R_P=1/2,\ T_Q/T_P=1/2\)
ⓑ. \(\theta_Q=45^\circ,\ R_Q/R_P=1/\sqrt3,\ T_Q/T_P=1/\sqrt2\)
ⓒ. \(\theta_Q=60^\circ,\ R_Q/R_P=1,\ T_Q/T_P=\sqrt3/2\)
ⓓ. \(\theta_Q=45^\circ,\ R_Q/R_P=\sqrt3,\ T_Q/T_P=\sqrt2\)
Correct Answer: \(\theta_Q=45^\circ,\ R_Q/R_P=1/\sqrt3,\ T_Q/T_P=1/\sqrt2\)
Explanation: For a same-level projectile, the maximum height is
\[
H=\frac{u^2\sin^2\theta}{2g_{\rm eff}}.
\]
For P,
\[
H_P=\frac{u^2\sin^230^\circ}{2g}
=\frac{u^2}{8g}.
\]
For Q, the gravitational acceleration is \(2g\), so
\[
H_Q=\frac{u^2\sin^2\theta_Q}{4g}.
\]
The condition \(H_Q=H_P\) gives
\[
\frac{\sin^2\theta_Q}{4}=\frac18,
\]
hence \(\sin^2\theta_Q=1/2\). For an upward launch angle between \(0^\circ\) and \(90^\circ\),
\[
\theta_Q=45^\circ.
\]
Now
\[
R_P=\frac{u^2\sin60^\circ}{g}
=\frac{\sqrt3\,u^2}{2g},
\]
whereas
\[
R_Q=\frac{u^2\sin90^\circ}{2g}
=\frac{u^2}{2g}.
\]
Therefore \(R_Q/R_P=1/\sqrt3\). Similarly, \(T_P=u/g\) and \(T_Q=u/(\sqrt2\,g)\), so \(T_Q/T_P=1/\sqrt2\). Thus option B is correct.
414. In the ground frame, a projectile is launched from the origin with velocity
\[
20\hat i+20\hat j\ \text{m s}^{-1}.
\]
An observer moves east with constant velocity \(10\hat i\,\text{m s}^{-1}\). Take \(g=10\,\text{m s}^{-2}\). Which statement correctly describes the projectile in the observer's frame?
ⓐ. Relative trajectory \(y=2x'-\dfrac{x'^2}{20}\), relative range \(=40\,\text{m}\), highest point at \(x'=20\,\text{m}\).
ⓑ. Its trajectory is unchanged as \(y=x'-\dfrac{x'^2}{40}\), and its relative range is \(80\,\text{m}\).
ⓒ. It rises vertically because the observer removes the entire horizontal velocity.
ⓓ. Its time of flight changes from \(4\,\text{s}\) to \(2\,\text{s}\).
Correct Answer: Relative trajectory \(y=2x'-\dfrac{x'^2}{20}\), relative range \(=40\,\text{m}\), highest point at \(x'=20\,\text{m}\).
Explanation: Relative to the observer, the projectile's horizontal velocity is
\[
v'_x=20-10=10\,\text{m s}^{-1}.
\]
The observer has no vertical velocity, so the vertical motion is unchanged:
\[
y=20t-5t^2.
\]
The relative horizontal coordinate is
\[
x'=10t.
\]
Therefore
\[
t=\frac{x'}{10}.
\]
Substituting into the vertical equation gives
\[
y=2x'-\frac{x'^2}{20}.
\]
The same-level time of flight remains
\[
T=\frac{2(20)}{10}=4\,\text{s},
\]
because a uniform horizontal change of observer does not alter the vertical motion. Hence the relative horizontal displacement during the flight is
\[
x'=10(4)=40\,\text{m}.
\]
The highest point occurs at \(t=2\,\text{s}\), so its relative horizontal coordinate is
\[
x'=20\,\text{m}.
\]
The geometrical trajectory therefore depends on the chosen uniformly moving frame even though the vertical acceleration remains the same.
415. Projectile P is launched from level ground in gravitational field \(g\) with speed \(u\) at \(45^\circ\). Projectile Q is launched from level ground in gravitational field \(2g\). The two projectiles have the same horizontal range and the same time of flight. Which comparison is correct?
ⓐ. \(u_Q=u_P\), \(\theta_Q=45^\circ\), \(H_Q=H_P\)
ⓑ. \(u_Q=2u_P,\ \theta_Q=45^\circ,\ H_Q=4H_P\)
ⓒ. \(u_Q=\sqrt{\dfrac52}\,u_P,\ \tan\theta_Q=2,\ H_Q=2H_P\)
ⓓ. \(u_Q=\dfrac{u_P}{\sqrt2},\ \tan\theta_Q=\dfrac12,\ H_Q=\dfrac{H_P}{2}\)
Correct Answer: \(u_Q=\sqrt{\dfrac52}\,u_P,\ \tan\theta_Q=2,\ H_Q=2H_P\)
Explanation: Let P have equal horizontal and vertical launch components
\[
u_{xP}=u_{yP}=k.
\]
Since both projectiles have the same range and same time of flight,
\[
u_{xQ}=\frac RT=k.
\]
For P,
\[
T=\frac{2k}{g}.
\]
For Q under gravitational acceleration \(2g\),
\[
T=\frac{2u_{yQ}}{2g}=\frac{u_{yQ}}{g}.
\]
Equating the times gives
\[
u_{yQ}=2k.
\]
Thus
\[
\tan\theta_Q=\frac{2k}{k}=2.
\]
The speeds are
\[
u_P=\sqrt2\,k
\]
and
\[
u_Q=\sqrt5\,k,
\]
so
\[
\frac{u_Q}{u_P}=\sqrt{\frac52}.
\]
Also,
\[
H_P=\frac{k^2}{2g},
\]
while
\[
H_Q=\frac{(2k)^2}{2(2g)}=\frac{k^2}{g}=2H_P.
\]
Equal range and time therefore do not require equal launch speed or angle when the gravitational field changes.
416. Two particles A and B move counterclockwise on the same circle of radius \(R\). A has angular speed \(3\omega\) and starts from the rightmost point, while B has angular speed \(\omega\) and starts from the leftmost point. When do they meet for the first time, and what is their relative speed at that instant?
ⓐ. \(t=\dfrac{\pi}{4\omega}\), they meet at the topmost point, and relative speed \(=\omega R\)
ⓑ. \(t=\dfrac{\pi}{\omega}\), they meet at the rightmost point, and relative speed \(=4\omega R\)
ⓒ. \(t=\dfrac{\pi}{2\omega}\), they meet at the bottommost point, and relative speed \(=2\omega R\)
ⓓ. They never meet because they start diametrically opposite.
Correct Answer: \(t=\dfrac{\pi}{2\omega}\), they meet at the bottommost point, and relative speed \(=2\omega R\)
Explanation: Initially B is ahead of A by an angular separation
\[
\pi.
\]
Both move in the same sense, so the relative angular speed is
\[
3\omega-\omega=2\omega.
\]
The first meeting occurs when A gains the initial angular separation:
\[
2\omega t=\pi.
\]
Hence
\[
t=\frac{\pi}{2\omega}.
\]
During this time A advances through
\[
3\omega t=\frac{3\pi}{2},
\]
which carries it from the rightmost point to the bottommost point. B advances from its initial angle \(\pi\) by
\[
\omega t=\frac{\pi}{2},
\]
also reaching angle \(3\pi/2\). At the common point both tangent velocities have the same direction because both move counterclockwise. Their speeds are \(3\omega R\) and \(\omega R\). Therefore their relative speed is
\[
3\omega R-\omega R=2\omega R.
\]
417. Two particles P and Q move counterclockwise with the same angular speed \(\omega\) on concentric circles of radii \(R\) and \(2R\), respectively. At every instant they lie on the same radial line from the common centre. Which statement is correct?
ⓐ. Their separation is constant at \(R\), so their relative velocity and relative acceleration are both zero.
ⓑ. Their separation changes between \(R\) and \(3R\) during each revolution.
ⓒ. Their relative speed is \(3\omega R\), and their relative acceleration is zero.
ⓓ. \(v_{\rm rel}=\omega R,\ a_{\rm rel}=\omega^2R\), and both relative vectors rotate with angular speed \(\omega\).
Correct Answer: \(v_{\rm rel}=\omega R,\ a_{\rm rel}=\omega^2R\), and both relative vectors rotate with angular speed \(\omega\).
Explanation: Because the two particles always lie on the same radial line, their position vectors may be written as
\[
\vec r_P=R\hat r
\]
and
\[
\vec r_Q=2R\hat r.
\]
Thus
\[
\vec r_{Q/P}=R\hat r,
\]
so the separation magnitude remains constant at \(R\). Their tangential speeds are
\[
v_P=\omega R
\]
and
\[
v_Q=2\omega R.
\]
Since the tangent directions are identical at each instant,
\[
|\vec v_{Q/P}|=\omega R.
\]
Their centripetal accelerations are
\[
\vec a_P=-\omega^2R\hat r
\]
and
\[
\vec a_Q=-2\omega^2R\hat r.
\]
Therefore
\[
\vec a_{Q/P}=-\omega^2R\hat r,
\]
with magnitude \(\omega^2R\). The relative position, velocity and acceleration directions rotate continuously, so constant separation does not imply zero relative velocity.
418. A particle moves uniformly on a circle. During an interval it turns through \(120^\circ\). What is the ratio of the magnitude of its average velocity to its constant speed?
ⓐ. \(\dfrac{2\pi}{3\sqrt3}\)
ⓑ. \(\dfrac{\sqrt3}{2}\)
ⓒ. \(\dfrac{3\sqrt3}{2\pi}\)
ⓓ. \(\dfrac3\pi\)
Correct Answer: \(\dfrac{3\sqrt3}{2\pi}\)
Explanation: Let the radius be \(R\) and the constant speed be \(v\). The angular displacement is
\[
\theta=\frac{2\pi}{3}.
\]
The distance travelled is the arc length
\[
s=R\theta=\frac{2\pi R}{3}.
\]
Therefore the elapsed time is
\[
\Delta t=\frac{s}{v}
=\frac{2\pi R}{3v}.
\]
The displacement magnitude is the chord corresponding to \(120^\circ\):
\[
|\Delta\vec r|=2R\sin60^\circ=\sqrt3R.
\]
Hence
\[
|\vec v_{\rm avg}|
=\frac{\sqrt3R}{2\pi R/(3v)}
=\frac{3\sqrt3}{2\pi}v.
\]
Therefore
\[
\frac{|\vec v_{\rm avg}|}{v}
=\frac{3\sqrt3}{2\pi}.
\]
The ratio is less than one because the chord joining the endpoints is shorter than the circular arc travelled.
419. A particle moves with constant speed \(6\,\text{m s}^{-1}\) on a circle. Starting from a point A, the magnitude of its displacement becomes equal to the radius of the circle for the first time after \(2\,\text{s}\). Which set of results is correct?
ⓐ. Angular displacement \(=\pi\), so the particle reaches the diametrically opposite point after \(2\,\text{s}\).
ⓑ. Angular displacement \(=\dfrac{\pi}{2}\), \(\omega=\dfrac{\pi}{4}\,\text{rad s}^{-1}\), \(R=\dfrac{24}{\pi}\,\text{m}\), and \(T=8\,\text{s}\).
ⓒ. Angular displacement \(=\dfrac{2\pi}{3}\), \(R=\dfrac{18}{\pi}\,\text{m}\), and \(a_c=2\pi\,\text{m s}^{-2}\).
ⓓ. Angular displacement \(=\dfrac{\pi}{3}\), \(\omega=\dfrac{\pi}{6}\,\text{rad s}^{-1}\), \(R=\dfrac{36}{\pi}\,\text{m}\), \(T=12\,\text{s}\), and \(a_c=\pi\,\text{m s}^{-2}\).
Correct Answer: Angular displacement \(=\dfrac{\pi}{3}\), \(\omega=\dfrac{\pi}{6}\,\text{rad s}^{-1}\), \(R=\dfrac{36}{\pi}\,\text{m}\), \(T=12\,\text{s}\), and \(a_c=\pi\,\text{m s}^{-2}\).
Explanation: If the particle turns through an angle \(\theta\), its displacement magnitude is the chord
\[
d=2R\sin\frac{\theta}{2}.
\]
The first time \(d=R\),
\[
2R\sin\frac{\theta}{2}=R,
\]
so
\[
\sin\frac{\theta}{2}=\frac12.
\]
For the first positive occurrence, \(\theta/2=\pi/6\), hence
\[
\theta=\frac{\pi}{3}.
\]
Since this angular displacement occurs in \(2\,\text{s}\),
\[
\omega=\frac{\theta}{\Delta t}=\frac{\pi}{6}\,\text{rad s}^{-1}.
\]
Using \(v=\omega R\),
\[
R=\frac{6}{\pi/6}=\frac{36}{\pi}\,\text{m}.
\]
The period is
\[
T=\frac{2\pi}{\omega}=12\,\text{s}.
\]
Finally,
\[
a_c=v\omega=6\left(\frac{\pi}{6}\right)=\pi\,\text{m s}^{-2}.
\]
The key geometric step is that a displacement equal to the radius corresponds to a \(60^\circ\) chord, not to a quarter or half revolution.
420. A particle undergoing uniform circular motion is observed at equal one-second intervals. During each one-second interval, the magnitude of its angular displacement is less than \(\pi\,\text{rad}\).
| Time | Position relative to centre |
|---|
| \(0\,\text{s}\) | \(R\hat i\) |
| \(1\,\text{s}\) | \(R\hat j\) |
| \(2\,\text{s}\) | \(-R\hat i\) |
Which statement is correct?
ⓐ. The motion is clockwise with \(\omega=\pi\,\text{rad s}^{-1}\), \(v=\pi R\), and period \(2\,\text{s}\).
ⓑ. The motion is counterclockwise with \(\omega=\dfrac{\pi}{2}\,\text{rad s}^{-1}\), \(v=\dfrac{\pi R}{2}\), and \(a_c=\dfrac{\pi^2R}{4}\).
ⓒ. The angular speed is \(\dfrac{\pi}{4}\,\text{rad s}^{-1}\), and the period is \(8\,\text{s}\).
ⓓ. The data are incompatible with uniform circular motion because the position vector changes direction.
Correct Answer: The motion is counterclockwise with \(\omega=\dfrac{\pi}{2}\,\text{rad s}^{-1}\), \(v=\dfrac{\pi R}{2}\), and \(a_c=\dfrac{\pi^2R}{4}\).
Explanation: The stated angular-displacement bound rules out a larger turn between successive observations. From \(R\hat i\) to \(R\hat j\), the allowed turn is therefore \(90^\circ\) counterclockwise in one second. Therefore
\[
\Delta\theta=\frac{\pi}{2}
\]
in
\[
\Delta t=1\,\text{s}.
\]
Hence
\[
\omega=\frac{\pi}{2}\,\text{rad s}^{-1}.
\]
The next observation at \(-R\hat i\) confirms another equal \(90^\circ\) counterclockwise turn in the next second. The linear speed is
\[
v=\omega R=\frac{\pi R}{2}.
\]
The centripetal acceleration magnitude is
\[
a_c=\omega^2R
=\frac{\pi^2R}{4}.
\]
The period is
\[
T=\frac{2\pi}{\omega}=4\,\text{s}.
\]
Thus equally spaced position vectors reveal both the sense of rotation and the angular speed.
