501. Particle P executes uniform circular motion with radius \(R\) and frequency \(f\). Particle Q moves uniformly in a circle of radius \(4R\). If the magnitudes of their centripetal accelerations are equal, which relation must hold?
ⓐ. \(f_Q=\dfrac f4,\quad v_Q=v_P,\quad T_Q=4T_P\)
ⓑ. \(f_Q=f,\quad v_Q=4v_P,\quad T_Q=T_P\)
ⓒ. \(f_Q=2f,\quad v_Q=8v_P,\quad T_Q=\dfrac12T_P\)
ⓓ. \(f_Q=\dfrac f2,\quad v_Q=2v_P,\quad T_Q=2T_P\)
Correct Answer: \(f_Q=\dfrac f2,\quad v_Q=2v_P,\quad T_Q=2T_P\)
Explanation: The centripetal acceleration in terms of frequency is
\[
a_c=4\pi^2f^2R.
\]
For P,
\[
a_P=4\pi^2f^2R.
\]
For Q,
\[
a_Q=4\pi^2f_Q^2(4R).
\]
The accelerations are equal, so
\[
4\pi^2f^2R=16\pi^2f_Q^2R.
\]
Therefore
\[
f_Q^2=\frac{f^2}{4},
\]
and frequency being non-negative gives
\[
f_Q=\frac f2.
\]
Now use
\[
v=2\pi fR.
\]
Thus
\[
v_Q=2\pi\left(\frac f2\right)(4R)=2v_P.
\]
Since \(T=1/f\),
\[
T_Q=\frac{1}{f/2}=2T_P.
\]
Increasing radius does not by itself determine whether speed, frequency or acceleration increases; the imposed condition of equal centripetal acceleration must first be used.
502. Two particles P and Q move counterclockwise on concentric circular paths of radii \(R\) and \(3R\), respectively. They have the same constant linear speed \(v\) and start on the same radial line. When P completes exactly one revolution for the first time, which set of results is correct?
ⓐ. Q has also completed one revolution; their separation is \(2R\), and their relative speed is zero.
ⓑ. Q has completed three revolutions; their separation is \(2R\), and \(a_Q=3a_P\).
ⓒ. Q has completed \(\dfrac13\) revolution; their separation is \(\sqrt{13}R\), their relative speed is \(\sqrt3\,v\), and \(a_Q=\dfrac13a_P\).
ⓓ. Q has completed \(\dfrac13\) revolution; their separation is \(4R\), and their relative speed is \(2v\).
Correct Answer: Q has completed \(\dfrac13\) revolution; their separation is \(\sqrt{13}R\), their relative speed is \(\sqrt3\,v\), and \(a_Q=\dfrac13a_P\).
Explanation: Equal linear speed does not imply equal angular speed when the radii differ. Since
\[
\omega=\frac{v}{R},
\]
P has angular speed \(v/R\), while Q has angular speed \(v/(3R)\). Therefore Q's frequency is one-third that of P. During the time in which P completes one revolution, Q completes
\[
\frac13
\]
of a revolution, corresponding to an angular displacement of \(120^\circ\). P has returned to its initial radial direction, so the angle between their position vectors is \(120^\circ\). Their separation \(d\) satisfies
\[
d^2=R^2+(3R)^2-2(R)(3R)\cos120^\circ.
\]
Since \(\cos120^\circ=-\frac12\),
\[
d^2=R^2+9R^2+3R^2=13R^2,
\]
giving
\[
d=\sqrt{13}R.
\]
Their tangent velocities have equal magnitude \(v\), and the angle between those velocity vectors is also \(120^\circ\). Hence
\[
v_{\rm rel}=\sqrt{v^2+v^2-2v^2\cos120^\circ}=\sqrt3\,v.
\]
Finally,
\[
a_P=\frac{v^2}{R},\qquad a_Q=\frac{v^2}{3R}=\frac13a_P.
\]
503. Read the experimental situation below.
In a ruler-drop reaction-time test, the lower end of a ruler is initially level with a student's fingers. In the first trial the ruler is released from rest and the student catches it after it has fallen \(20\,\text{cm}\). Take \(g=10\,\text{m s}^{-2}\). In a second trial, the student's reaction time is assumed unchanged, but at the instant timing begins the ruler already has a downward speed of \(0.50\,\text{m s}^{-1}\).
How far does the ruler fall during the second trial before being caught?
ⓐ. \(40\,\text{cm}\)
ⓑ. \(25\,\text{cm}\)
ⓒ. \(20\,\text{cm}\)
ⓓ. \(30\,\text{cm}\)
Correct Answer: \(30\,\text{cm}\)
Explanation: In the first trial the ruler starts from rest, so the fall distance determines the student's reaction time. Using downward as positive,
\[
s=\frac12gt^2.
\]
With
\[
s=0.20\,\text{m},\qquad g=10\,\text{m s}^{-2},
\]
we obtain
\[
0.20=5t^2.
\]
Hence
\[
t^2=0.04,\qquad t=0.20\,\text{s}.
\]
The second trial has the same reaction time but a non-zero initial downward velocity. Therefore the rest-release expression cannot be reused unchanged. Instead,
\[
s=ut+\frac12gt^2.
\]
Substituting
\[
u=0.50\,\text{m s}^{-1},\qquad t=0.20\,\text{s},
\]
gives
\[
s=(0.50)(0.20)+5(0.20)^2.
\]
Thus
\[
s=0.10+0.20=0.30\,\text{m}=30\,\text{cm}.
\]
The extra \(10\,\text{cm}\) arises from the initial downward motion during the unchanged reaction interval.
504. Rain has ground-frame velocity
\[
\vec v_R=-12\hat j\ \text{m s}^{-1}.
\]
A cyclist starts from rest at \(t=0\) and accelerates uniformly eastward with
\[
\vec a_C=3\hat i\ \text{m s}^{-2}.
\]
At what instant does the rain appear to the cyclist to move at \(45^\circ\) west of the vertically downward direction, and what are the cyclist's displacement and the apparent speed of rain then?
ⓐ. \(t=2\,\text{s},\ 6\,\text{m},\ 6\sqrt5\,\text{m s}^{-1}\)
ⓑ. \(t=3\,\text{s},\ 13.5\,\text{m},\ 15\,\text{m s}^{-1}\)
ⓒ. \(t=4\,\text{s},\ 48\,\text{m},\ 24\,\text{m s}^{-1}\)
ⓓ. \(t=4\,\text{s},\ 24\,\text{m},\ 12\sqrt2\,\text{m s}^{-1}\)
Correct Answer: \(t=4\,\text{s},\ 24\,\text{m},\ 12\sqrt2\,\text{m s}^{-1}\)
Explanation: The cyclist's ground velocity at time \(t\) is
\[
\vec v_C=3t\,\hat i.
\]
The velocity of rain relative to the cyclist is
\[
\vec v_{R/C}=\vec v_R-\vec v_C.
\]
Therefore
\[
\vec v_{R/C}=-3t\,\hat i-12\hat j.
\]
The negative \(x\)-component means the rain appears to have a westward component, while the negative \(y\)-component is downward. For an apparent direction \(45^\circ\) west of vertical, the magnitudes of these two perpendicular components must be equal:
\[
3t=12.
\]
Thus
\[
t=4\,\text{s}.
\]
The cyclist started from rest with constant acceleration, so its eastward displacement is
\[
x=\frac12at^2=\frac12(3)(4^2)=24\,\text{m}.
\]
At that instant the relative-velocity components are \(-12\hat i\) and \(-12\hat j\), giving
\[
|\vec v_{R/C}|=\sqrt{12^2+12^2}=12\sqrt2\,\text{m s}^{-1}.
\]
The apparent rain direction is therefore controlled by the instantaneous observer velocity, while the cyclist's position requires the separate constant-acceleration relation.
505. An ideal projectile is launched from the origin. Its ground-frame velocity components are described by
The horizontal component \(v_x\) remains constant at \(12\,\text{m s}^{-1}\). The vertical component \(v_y\) is a straight line on a \(v_y-t\) graph, beginning at \(+18\,\text{m s}^{-1}\) at \(t=0\) and having constant slope \(-10\,\text{m s}^{-2}\).
An observer starts from the launch point at \(t=0\) and moves uniformly in the \(+x\)-direction at \(4\,\text{m s}^{-1}\). Relative to this observer, which set of results is correct up to the instant the projectile returns to its launch level?
ⓐ. Highest point occurs at \(t=3.6\,\text{s}\), its relative horizontal coordinate there is \(28.8\,\text{m}\), and the relative range is \(57.6\,\text{m}\).
ⓑ. Highest point occurs at \(t=1.8\,\text{s}\), but its height relative to the observer is \(7.2\,\text{m}\) because the observer is moving.
ⓒ. Highest point occurs at \(t=1.8\,\text{s}\), its relative coordinates are \((14.4\,\text{m},16.2\,\text{m})\), and the relative horizontal displacement on returning to launch level is \(28.8\,\text{m}\).
ⓓ. The observer's motion changes the projectile's vertical acceleration to less than \(10\,\text{m s}^{-2}\).
Correct Answer: Highest point occurs at \(t=1.8\,\text{s}\), its relative coordinates are \((14.4\,\text{m},16.2\,\text{m})\), and the relative horizontal displacement on returning to launch level is \(28.8\,\text{m}\).
Explanation: The slope of the \(v_y-t\) graph gives the vertical acceleration:
\[
a_y=-10\,\text{m s}^{-2}.
\]
The highest point occurs when
\[
v_y=18-10t=0,
\]
so
\[
t_H=1.8\,\text{s}.
\]
The observer moves horizontally at \(4\,\text{m s}^{-1}\), while the projectile's horizontal velocity is \(12\,\text{m s}^{-1}\). Hence the projectile's horizontal velocity relative to the observer is
\[
v_x'=12-4=8\,\text{m s}^{-1}.
\]
Its relative horizontal coordinate at the highest point is therefore
\[
x_H'=(8)(1.8)=14.4\,\text{m}.
\]
Horizontal frame motion does not alter the vertical coordinate, so the maximum height is
\[
H=\frac{18^2}{2(10)}=16.2\,\text{m}.
\]
For return to the launch level,
\[
T=\frac{2(18)}{10}=3.6\,\text{s}.
\]
Thus the relative horizontal displacement over the complete flight is
\[
R'=(8)(3.6)=28.8\,\text{m}.
\]
The moving observer changes the measured horizontal component of velocity and horizontal displacement, but not the common vertical acceleration or vertical height in this parallel translating frame.
506. A particle executes counterclockwise uniform circular motion of radius \(2\,\text{m}\). Its centripetal-acceleration magnitude is \(8\,\text{m s}^{-2}\). At \(t=0\) the particle is at the rightmost point of the circle. An observer moves with constant velocity \(+\hat j\,\text{m s}^{-1}\) relative to the ground. Which statement is correct?
ⓐ. \(f=\dfrac{2}{\pi}\,\text{Hz},\ v_{\rm rel}(0)=5\,\text{m s}^{-1},\ v_{\rm rel}(T/4)=\sqrt{17}\,\text{m s}^{-1}\)
ⓑ. \(f=\dfrac1\pi\,\text{Hz},\ v_{\rm rel}(0)=3\,\text{m s}^{-1},\ v_{\rm rel}(T/4)=3\,\text{m s}^{-1}\)
ⓒ. \(f=\pi\,\text{Hz},\ v_{\rm rel}(0)=5\,\text{m s}^{-1},\ v_{\rm rel}(T/4)=\sqrt5\,\text{m s}^{-1}\)
ⓓ. \(f=\dfrac1\pi\,\text{Hz},\ v_{\rm rel}(0)=3\,\text{m s}^{-1},\ v_{\rm rel}(T/4)=\sqrt{17}\,\text{m s}^{-1}\)
Correct Answer: \(f=\dfrac1\pi\,\text{Hz},\ v_{\rm rel}(0)=3\,\text{m s}^{-1},\ v_{\rm rel}(T/4)=\sqrt{17}\,\text{m s}^{-1}\)
Explanation: The centripetal relation first gives the particle's linear speed:
\[
a_c=\frac{v^2}{R}.
\]
Therefore
\[
8=\frac{v^2}{2},
\]
so
\[
v=4\,\text{m s}^{-1}.
\]
The angular speed is
\[
\omega=\frac{v}{R}=\frac42=2\,\text{rad s}^{-1}.
\]
Hence
\[
f=\frac{\omega}{2\pi}=\frac1\pi\,\text{Hz},
\qquad
T=\frac{2\pi}{\omega}=\pi\,\text{s}.
\]
At the rightmost point, counterclockwise tangential velocity is
\[
\vec v_P=4\hat j.
\]
Since the observer has velocity \(\hat j\),
\[
\vec v_{P/O}=4\hat j-\hat j=3\hat j,
\]
so the initial relative speed is \(3\,\text{m s}^{-1}\). After one-quarter period, the particle reaches the top point, where its ground-frame velocity is leftward:
\[
\vec v_P=-4\hat i.
\]
The observer still has velocity \(+\hat j\), so
\[
\vec v_{P/O}=-4\hat i-\hat j.
\]
Its magnitude is
\[
|\vec v_{P/O}|=\sqrt{4^2+1^2}=\sqrt{17}\,\text{m s}^{-1}.
\]
Thus a constant particle speed in the ground frame does not imply a constant speed relative to a translating observer when the particle's velocity direction continuously changes.
507. A particle has the following ground-frame velocity-time description.
From \(t=0\) to \(4\,\text{s}\), its velocity increases linearly from \(-2\,\text{m s}^{-1}\) to \(+6\,\text{m s}^{-1}\). From \(t=4\) to \(8\,\text{s}\), its velocity decreases linearly from \(+6\,\text{m s}^{-1}\) to \(-2\,\text{m s}^{-1}\). An observer moves throughout with constant velocity \(+2\,\text{m s}^{-1}\) along the same line.
Which description is correct in the observer's frame during \(0\) to \(8\,\text{s}\)?
ⓐ. The particle never reverses in the observer's frame because the observer's velocity is constant.
ⓑ. The particle reverses only at \(t=4\,\text{s}\), has relative displacement \(16\,\text{m}\), and relative distance \(16\,\text{m}\).
ⓒ. Relative reversals occur at \(t=2\) and \(6\,\text{s}\); relative displacement is zero, relative distance is \(16\,\text{m}\), and average relative speed is \(2\,\text{m s}^{-1}\).
ⓓ. The relative displacement is zero and therefore the relative distance is also zero.
Correct Answer: Relative reversals occur at \(t=2\) and \(6\,\text{s}\); relative displacement is zero, relative distance is \(16\,\text{m}\), and average relative speed is \(2\,\text{m s}^{-1}\).
Explanation: The velocity measured by the moving observer is
\[
v'=v-2.
\]
Hence from \(0\) to \(4\,\text{s}\), \(v'\) rises linearly from \(-4\) to \(+4\,\text{m s}^{-1}\). It crosses zero halfway through that interval:
\[
t=2\,\text{s}.
\]
From \(4\) to \(8\,\text{s}\), \(v'\) falls linearly from \(+4\) to \(-4\,\text{m s}^{-1}\), crossing zero at
\[
t=6\,\text{s}.
\]
Thus the particle reverses relative direction twice. The signed area under the relative \(v'-t\) graph from \(0\) to \(4\,\text{s}\) is zero because equal negative and positive triangular areas cancel. The same is true from \(4\) to \(8\,\text{s}\), so the total relative displacement is zero. The distance in the first \(4\,\text{s}\) is
\[
2\left[\frac12(2)(4)\right]=8\,\text{m},
\]
and the second half contributes another \(8\,\text{m}\). Therefore the relative distance is \(16\,\text{m}\), giving average relative speed
\[
\frac{16}{8}=2\,\text{m s}^{-1}.
\]
508. A particle moves in a plane with one constant acceleration during a \(4\,\text{s}\) interval. Its displacement over the interval is
\[
\Delta\vec r=24\hat i+8\hat j\ \text{m},
\]
and its change in velocity is
\[
\Delta\vec v=-4\hat i+12\hat j\ \text{m s}^{-1}.
\]
The speeds at the beginning and end of the interval are equal. Which set of results is correct?
ⓐ. \(\vec v_0=4\hat i+8\hat j,\ \vec v_4=8\hat i-4\hat j,\ \vec a=\hat i-3\hat j,\ v_{\min}=2\sqrt{10}\,\text{m s}^{-1}\) at \(t=2\,\text{s}\)
ⓑ. \(\vec v_0=8\hat i-4\hat j,\ \vec v_4=4\hat i+8\hat j,\ \vec a=-\hat i+3\hat j,\ v_{\min}=2\sqrt{10}\,\text{m s}^{-1}\) at \(t=2\,\text{s}\)
ⓒ. \(\vec v_0=6\hat i+2\hat j,\ \vec v_4=2\hat i+14\hat j,\ \vec a=-\hat i+3\hat j,\ v_{\min}=2\sqrt{10}\,\text{m s}^{-1}\) at \(t=4\,\text{s}\)
ⓓ. The given displacement and change in velocity are inconsistent with equal endpoint speeds, so no such constant-acceleration motion exists.
Correct Answer: \(\vec v_0=8\hat i-4\hat j,\ \vec v_4=4\hat i+8\hat j,\ \vec a=-\hat i+3\hat j,\ v_{\min}=2\sqrt{10}\,\text{m s}^{-1}\) at \(t=2\,\text{s}\)
Explanation: For constant acceleration, the average velocity over the interval is the arithmetic mean of the endpoint velocities:
\[
\vec v_{\rm avg}=\frac{\vec v_0+\vec v_4}{2}.
\]
From the displacement,
\[
\vec v_{\rm avg}=\frac{24\hat i+8\hat j}{4}=6\hat i+2\hat j.
\]
Hence
\[
\vec v_0+\vec v_4=12\hat i+4\hat j.
\]
We also know
\[
\vec v_4-\vec v_0=-4\hat i+12\hat j.
\]
Solving the sum and difference equations gives
\[
\vec v_0=8\hat i-4\hat j
\]
and
\[
\vec v_4=4\hat i+8\hat j.
\]
Both have magnitude \(4\sqrt5\,\text{m s}^{-1}\), confirming the equal-speed condition. The acceleration is
\[
\vec a=\frac{\Delta\vec v}{4}=-\hat i+3\hat j.
\]
At the midpoint time,
\[
\vec v(2)=\frac{\vec v_0+\vec v_4}{2}=6\hat i+2\hat j.
\]
Since
\[
(6\hat i+2\hat j)\cdot(-\hat i+3\hat j)=0,
\]
the speed is minimum there, with value
\[
\sqrt{6^2+2^2}=2\sqrt{10}\,\text{m s}^{-1}.
\]
509. A target moves horizontally along the line \(y=15\,\text{m}\). At \(t=0\) it is at \(x=15\,\text{m}\) and moves in the \(+x\)-direction at \(5\,\text{m s}^{-1}\). Simultaneously, a projectile is launched from the origin. Take \(g=10\,\text{m s}^{-2}\). The projectile intercepts the target while its own velocity is directed \(45^\circ\) below the horizontal. Which set of launch parameters is correct?
ⓐ. \(t=2\,\text{s},\ \vec u=15\hat i+20\hat j\,\text{m s}^{-1},\ |\vec u|=25\,\text{m s}^{-1}\)
ⓑ. \(t=4\,\text{s},\ \vec u=10\hat i+15\hat j\,\text{m s}^{-1},\ |\vec u|=5\sqrt{13}\,\text{m s}^{-1}\)
ⓒ. \(t=3\,\text{s},\ \vec u=15\hat i+15\hat j\,\text{m s}^{-1},\ |\vec u|=15\sqrt2\,\text{m s}^{-1}\)
ⓓ. \(t=3\,\text{s},\ \vec u=10\hat i+20\hat j\,\text{m s}^{-1},\ |\vec u|=10\sqrt5\,\text{m s}^{-1}\)
Correct Answer: \(t=3\,\text{s},\ \vec u=10\hat i+20\hat j\,\text{m s}^{-1},\ |\vec u|=10\sqrt5\,\text{m s}^{-1}\)
Explanation: Let the projectile launch components be \(u_x\) and \(u_y\), and let the interception time be \(t\). Horizontally, interception requires
\[
u_xt=15+5t.
\]
At interception the projectile moves \(45^\circ\) below the horizontal, so its vertical velocity is the negative of its horizontal velocity:
\[
u_y-10t=-u_x.
\]
Therefore
\[
u_y=10t-u_x.
\]
The vertical coordinate at interception is \(15\,\text{m}\):
\[
15=u_yt-5t^2.
\]
Substituting for \(u_y\),
\[
15=(10t-u_x)t-5t^2=5t^2-u_xt.
\]
Using the horizontal relation \(u_xt=15+5t\),
\[
15=5t^2-15-5t.
\]
Thus
\[
t^2-t-6=0,
\]
giving the physical solution
\[
t=3\,\text{s}.
\]
Then
\[
u_x=\frac{15+5(3)}{3}=10\,\text{m s}^{-1}
\]
and
\[
u_y=30-10=20\,\text{m s}^{-1}.
\]
Hence the launch speed is
\[
u=\sqrt{10^2+20^2}=10\sqrt5\,\text{m s}^{-1}.
\]
510. Projectile P is launched with speed \(u\) at angle \(\theta\) and returns to its launch level. Its time of flight, range and maximum height are \(T,R,H\), respectively. Projectile Q is launched with the same speed and angle from the top of a cliff of height \(3H\) above its landing level. Both experience the same uniform downward acceleration \(g\). Which statement is correct for Q?
ⓐ. Its time of flight is \(\dfrac{3T}{2}\), but its horizontal range remains \(R\).
ⓑ. Its time of flight is \(2T\), its range is \(2R\), and \(\phi=\theta\).
ⓒ. Its time of flight is \(\dfrac T2\), and its range is unchanged.
ⓓ. \(T_Q=\dfrac{3T}{2},\ R_Q=\dfrac{3R}{2},\ \tan\phi=2\tan\theta\)
Correct Answer: \(T_Q=\dfrac{3T}{2},\ R_Q=\dfrac{3R}{2},\ \tan\phi=2\tan\theta\)
Explanation: Write the launch components as
\[
u_x=u\cos\theta,\qquad u_y=u\sin\theta.
\]
For P,
\[
T=\frac{2u_y}{g},\qquad H=\frac{u_y^2}{2g}.
\]
For Q, take the launch point as vertical origin. At impact,
\[
y=-3H.
\]
Thus
\[
-3\frac{u_y^2}{2g}=u_yt-\frac12gt^2.
\]
Multiplying through and defining \(z=gt/u_y\) gives
\[
z^2-2z-3=0.
\]
The physical root is
\[
z=3,
\]
so
\[
t_Q=\frac{3u_y}{g}=\frac{3T}{2}.
\]
The horizontal speed is unchanged, hence
\[
R_Q=u_xt_Q=\frac32R.
\]
At impact,
\[
v_y=u_y-gt_Q=u_y-3u_y=-2u_y.
\]
Therefore
\[
\tan\phi=\frac{2u_y}{u_x}=2\tan\theta.
\]
The changed landing level alters both the flight duration and the impact direction.
511. Projectile P is launched from the origin at \(t=0\) with velocity
\[
\vec u_P=10\hat i+20\hat j\ \text{m s}^{-1}.
\]
At \(t=1\,\text{s}\), projectile Q is launched from the point
\[
25\hat i\ \text{m}
\]
with velocity
\[
\vec u_Q=-5\hat i+25\hat j\ \text{m s}^{-1}.
\]
Both subsequently have acceleration \(-10\hat j\,\text{m s}^{-2}\). Which statement is correct?
ⓐ. Collision at \(t=2\,\text{s}\), \(\vec r=20\hat i+20\hat j\,\text{m}\); relative speed after Q launches \(=15\sqrt2\,\text{m s}^{-1}\).
ⓑ. Collision at \(t=1.5\,\text{s}\), \(\vec r=15\hat i+15\hat j\,\text{m}\); relative speed after Q launches \(=15\sqrt2\,\text{m s}^{-1}\)
ⓒ. No collision occurs because P was launched earlier; after Q launches their relative speed remains \(15\sqrt2\,\text{m s}^{-1}\)
ⓓ. Relative acceleration after Q launches is \(-20\hat j\,\text{m s}^{-2}\); they therefore cannot have constant relative speed or collide at \(t=2\,\text{s}\)
Correct Answer: Collision at \(t=2\,\text{s}\), \(\vec r=20\hat i+20\hat j\,\text{m}\); relative speed after Q launches \(=15\sqrt2\,\text{m s}^{-1}\).
Explanation: At the instant Q is launched, P has already been moving for \(1\,\text{s}\). Its position is
\[
\vec r_P(1)=10\hat i+(20-5)\hat j
=10\hat i+15\hat j.
\]
Its velocity then is
\[
\vec v_P(1)=10\hat i+10\hat j.
\]
Thus Q's initial position relative to P is
\[
\vec r_{Q/P}=15\hat i-15\hat j.
\]
Immediately after Q is launched, the relative velocity is
\[
\vec v_{Q/P}=(-5\hat i+25\hat j)-(10\hat i+10\hat j)
=-15\hat i+15\hat j.
\]
Both projectiles have exactly the same acceleration, so
\[
\vec a_{Q/P}=0.
\]
Hence their relative motion is uniform. After \(1\,\text{s}\),
\[
\vec r_{Q/P}=(15\hat i-15\hat j)+(-15\hat i+15\hat j)(1)=\vec0.
\]
Therefore the collision occurs at global time
\[
t=2\,\text{s}.
\]
P's position then is
\[
\vec r_P(2)=20\hat i+(40-20)\hat j
=20\hat i+20\hat j.
\]
The constant relative-speed magnitude is
\[
\sqrt{15^2+15^2}=15\sqrt2\,\text{m s}^{-1}.
\]
512. Two non-zero vectors satisfy
\[
|\vec A|=6,\qquad \vec R=\vec A+\vec B,\qquad |\vec R|=8,
\]
and \(\vec R\perp\vec A\). Which set of results is correct?
ⓐ. \(B=2,\ \cos\theta=1,\ |\vec A\times\vec B|=0\)
ⓑ. \(B=10,\ \cos\theta=-\dfrac35,\ |\vec A\times\vec B|=48\)
ⓒ. \(B=10,\ \cos\theta=\dfrac35,\ |\vec A\times\vec B|=48\)
ⓓ. \(B=14,\ \cos\theta=-1,\ |\vec A\times\vec B|=0\)
Correct Answer: \(B=10,\ \cos\theta=-\dfrac35,\ |\vec A\times\vec B|=48\)
Explanation: Since
\[
\vec R=\vec A+\vec B,
\]
we may write
\[
\vec B=\vec R-\vec A.
\]
The vectors \(\vec R\) and \(\vec A\) are perpendicular, so
\[
B^2=R^2+A^2.
\]
Therefore
\[
B=\sqrt{8^2+6^2}=10.
\]
Now take the scalar product of \(\vec A\) with \(\vec B\):
\[
\vec A\cdot\vec B
=\vec A\cdot(\vec R-\vec A).
\]
Since \(\vec A\cdot\vec R=0\),
\[
\vec A\cdot\vec B=-A^2=-36.
\]
Thus
\[
\cos\theta=\frac{-36}{(6)(10)}=-\frac35.
\]
For the cross product,
\[
\vec A\times\vec B
=\vec A\times(\vec R-\vec A)
=\vec A\times\vec R.
\]
Because \(\vec A\perp\vec R\),
\[
|\vec A\times\vec B|=(6)(8)=48.
\]
The required angle between \(\vec A\) and \(\vec B\) is therefore obtuse.
513. Let
\[
\vec A=7\hat i+\hat j,\qquad
\vec B=3\hat i+4\hat j,
\]
and define
\[
\vec C=\vec A+\lambda\vec B.
\]
The value of \(\lambda\) is chosen so that \(\vec C\perp\vec B\). Which statement is correct?
ⓐ. \(\lambda=+1,\ \vec C=10\hat i+5\hat j,\ |\vec C|=5\)
ⓑ. \(\lambda=-1,\ \vec C=4\hat i-3\hat j,\ |\vec C|=5,\ \vec C\times\vec B=25\hat k\)
ⓒ. \(\lambda=-\dfrac12,\ |\vec C|=\dfrac52,\ \vec C\times\vec B=\vec0\)
ⓓ. \(\lambda=-1,\ \vec C=4\hat i-3\hat j,\ \vec C\parallel\vec B\)
Correct Answer: \(\lambda=-1,\ \vec C=4\hat i-3\hat j,\ |\vec C|=5,\ \vec C\times\vec B=25\hat k\)
Explanation: The perpendicularity condition is
\[
\vec C\cdot\vec B=0.
\]
Therefore
\[
(\vec A+\lambda\vec B)\cdot\vec B=0.
\]
Now
\[
\vec A\cdot\vec B=(7)(3)+(1)(4)=25,
\]
while
\[
B^2=3^2+4^2=25.
\]
Hence
\[
25+25\lambda=0,
\]
giving
\[
\lambda=-1.
\]
Thus
\[
\vec C=\vec A-\vec B=4\hat i-3\hat j.
\]
Its magnitude is
\[
|\vec C|=\sqrt{4^2+(-3)^2}=5.
\]
Finally,
\[
\vec C\times\vec B
=(4\hat i-3\hat j)\times(3\hat i+4\hat j).
\]
The \(k\)-component is
\[
4(4)-(-3)(3)=25,
\]
so
\[
\vec C\times\vec B=25\hat k.
\]
Adding a suitable multiple of \(\vec B\) has removed the component of \(\vec A\) parallel to \(\vec B\).
514. A river current flows east at \(3\,\text{m s}^{-1}\). A floating log is initially \(20\,\text{m}\) east of a boat and simply drifts with the current. The boat can move at \(5\,\text{m s}^{-1}\) relative to the water. It first heads due east relative to the water until it catches the log, then immediately heads due west relative to the water until it returns to its original fixed point on the bank. Which set of results is correct?
ⓐ. The boat catches the log but cannot return to \(x=0\) because the eastward current prevents any westward ground motion
ⓑ. Catch after \(2.5\,\text{s}\) at \(x=20\,\text{m}\); return leg \(=2.5\,\text{s}\); total time \(=5\,\text{s}\)
ⓒ. Catch after \(4\,\text{s}\) at \(x=20\,\text{m}\); return leg \(=4\,\text{s}\); total time \(=8\,\text{s}\)
ⓓ. Catch after \(4\,\text{s}\) at \(x=32\,\text{m}\); return leg \(=16\,\text{s}\); total time \(=20\,\text{s}\)
Correct Answer: Catch after \(4\,\text{s}\) at \(x=32\,\text{m}\); return leg \(=16\,\text{s}\); total time \(=20\,\text{s}\)
Explanation: The log moves east with the water at
\[
3\,\text{m s}^{-1}.
\]
While chasing it, the boat's ground speed is
\[
5+3=8\,\text{m s}^{-1}
\]
eastward. The boat gains on the log at the water-relative boat speed:
\[
8-3=5\,\text{m s}^{-1}.
\]
With an initial separation of \(20\,\text{m}\),
\[
t_{\rm catch}=\frac{20}{5}=4\,\text{s}.
\]
The catch point is
\[
x=(8)(4)=32\,\text{m}
\]
east of the fixed starting point. On the return leg the boat points west relative to the water. Its ground speed westward is
\[
5-3=2\,\text{m s}^{-1}.
\]
Therefore returning \(32\,\text{m}\) requires
\[
t_{\rm return}=\frac{32}{2}=16\,\text{s}.
\]
The total elapsed time is
\[
4+16=20\,\text{s}.
\]
Although the current does not affect the boat-log closing speed during the chase, it strongly affects the time needed to return to a fixed bank point.
515. In a ground frame, a particle moves along the \(x\)-axis according to
\[
x_P=2+6t-t^2,
\]
while the origin of an observer's frame moves according to
\[
x_O=-4+2t.
\]
Both axes point in the same direction. Which statement is correct?
ⓐ. Maximum separation \(10\,\text{m}\) at \(t=2\,\text{s}\); meeting at \(2+\sqrt{10}\,\text{s}\) with relative speed \(2\sqrt{10}\,\text{m s}^{-1}\).
ⓑ. Relative speed is zero at \(t=2\,\text{s}\), separation is \(10\,\text{m}\), and the particle then remains permanently at rest relative to the observer
ⓒ. Maximum relative separation occurs at \(t=4\,\text{s}\), equals \(6\,\text{m}\), and the particle never meets the moving origin
ⓓ. They meet at \(t=2\,\text{s}\) because relative speed is zero there; the relative separation at that instant is also zero
Correct Answer: Maximum separation \(10\,\text{m}\) at \(t=2\,\text{s}\); meeting at \(2+\sqrt{10}\,\text{s}\) with relative speed \(2\sqrt{10}\,\text{m s}^{-1}\).
Explanation: The particle's coordinate relative to the moving origin is
\[
X=x_P-x_O.
\]
Therefore
\[
X=(2+6t-t^2)-(-4+2t)
=6+4t-t^2.
\]
Its relative velocity is
\[
V=\frac{dX}{dt}=4-2t.
\]
Thus the separation is stationary at
\[
t=2\,\text{s}.
\]
Because \(X(t)\) is a downward-opening parabola, this stationary value is a maximum:
\[
X_{\max}=6+8-4=10\,\text{m}.
\]
A zero relative velocity at that instant means only momentary relative rest; it does not mean the relative position remains fixed. They meet when
\[
X=0.
\]
Hence
\[
t^2-4t-6=0,
\]
giving the positive solution
\[
t=2+\sqrt{10}\,\text{s}.
\]
At that instant,
\[
V=4-2(2+\sqrt{10})=-2\sqrt{10}\,\text{m s}^{-1},
\]
so the relative speed is \(2\sqrt{10}\,\text{m s}^{-1}\).
516. Train A moves east at \(10\,\text{m s}^{-1}\). A ball is projected from A with velocity
\[
6\hat i+8\hat j\ \text{m s}^{-1}
\]
relative to the train. At the same instant, train B passes the launch point moving east at \(4\,\text{m s}^{-1}\). Take \(g=10\,\text{m s}^{-2}\). The ball later returns to its original vertical level. Which set of horizontal displacements during the flight is correct?
ⓐ. Ground frame: \(16\,\text{m}\); relative to A: \(0\); relative to B: \(9.6\,\text{m}\)
ⓑ. Ground frame: \(9.6\,\text{m}\); relative to A: \(25.6\,\text{m}\); relative to B: \(6.4\,\text{m}\)
ⓒ. Ground frame: \(25.6\,\text{m}\); relative to A: \(9.6\,\text{m}\); relative to B: \(19.2\,\text{m}\)
ⓓ. All three frames measure the same horizontal displacement because the vertical acceleration is identical.
Correct Answer: Ground frame: \(25.6\,\text{m}\); relative to A: \(9.6\,\text{m}\); relative to B: \(19.2\,\text{m}\)
Explanation: Relative to train A, the launch velocity components are \(6\,\text{m s}^{-1}\) horizontally and \(8\,\text{m s}^{-1}\) vertically. Adding A's ground velocity gives the ground-frame launch velocity:
\[
\vec u_G=16\hat i+8\hat j.
\]
The vertical motion is common to all uniformly translating frames. The time to return to the original vertical level is
\[
T=\frac{2u_y}{g}
=\frac{16}{10}=1.6\,\text{s}.
\]
In the ground frame, the horizontal displacement is
\[
x_G=(16)(1.6)=25.6\,\text{m}.
\]
Relative to train A, the ball's horizontal velocity remains \(6\,\text{m s}^{-1}\), so
\[
x_A'=(6)(1.6)=9.6\,\text{m}.
\]
Relative to train B, its horizontal velocity is
\[
16-4=12\,\text{m s}^{-1},
\]
giving
\[
x_B'=(12)(1.6)=19.2\,\text{m}.
\]
Thus the flight time is unchanged by these uniform horizontal frame changes, but the measured horizontal displacement is frame-dependent.
517. Particles P and Q start on the same radial line and move counterclockwise on concentric circles. P has radius \(R\) and frequency \(f\), while Q has radius \(2R\) and frequency \(2f\). At the first instant after \(t=0\) when their radius vectors are perpendicular, which set of results is correct?
ⓐ. \(t=\dfrac{1}{2f}\), separation \(=3R\), relative speed \(=6\pi fR\), and \(a_Q/a_P=4\)
ⓑ. \(t=\dfrac{1}{4f}\), separation \(=\sqrt5R\), relative speed \(=2\pi fR\sqrt{17}\), and \(a_Q/a_P=8\)
ⓒ. \(t=\dfrac{1}{4f}\), separation \(=R\), and the relative speed is zero
ⓓ. \(t=\dfrac{1}{8f}\), separation \(=\sqrt5R\), and \(a_Q/a_P=2\)
Correct Answer: \(t=\dfrac{1}{4f}\), separation \(=\sqrt5R\), relative speed \(=2\pi fR\sqrt{17}\), and \(a_Q/a_P=8\)
Explanation: The angular speeds are
\[
\omega_P=2\pi f
\]
and
\[
\omega_Q=4\pi f.
\]
Their relative angular speed is therefore
\[
\omega_Q-\omega_P=2\pi f.
\]
Starting from the same radial line, the first \(90^\circ\) angular separation occurs when
\[
(2\pi f)t=\frac{\pi}{2}.
\]
Thus
\[
t=\frac{1}{4f}.
\]
At that instant the position vectors of lengths \(R\) and \(2R\) are perpendicular, so their separation is
\[
d=\sqrt{R^2+(2R)^2}=\sqrt5R.
\]
Their linear speeds are
\[
v_P=2\pi fR,
\]
and
\[
v_Q=2\pi(2f)(2R)=8\pi fR=4v_P.
\]
The tangent velocity directions are also perpendicular, hence
\[
v_{\rm rel}=\sqrt{v_P^2+v_Q^2}
=v_P\sqrt{17}
=2\pi fR\sqrt{17}.
\]
Finally,
\[
\frac{a_Q}{a_P}
=\frac{\omega_Q^2(2R)}{\omega_P^2R}
=8.
\]
518. A particle moves counterclockwise in uniform circular motion of radius \(5\,\text{m}\) and speed \(10\,\text{m s}^{-1}\). At an instant its position relative to the centre is
\[
\vec r=3\hat i+4\hat j\ \text{m}.
\]
It then leaves the circular path and subsequently moves with constant acceleration \(-10\hat j\,\text{m s}^{-2}\). Which statement is correct?
ⓐ. Before release, \(\vec v=-6\hat i+8\hat j\) and \(\vec a_c=-16\hat i-12\hat j\,\text{m s}^{-2}\); it next reaches the same horizontal line after \(1.6\,\text{s}\), \(9.6\,\text{m}\) to the left.
ⓑ. Before release, \(\vec v=8\hat i-6\hat j\) and \(\vec a_c=-12\hat i-16\hat j\,\text{m s}^{-2}\); it next reaches the same horizontal line after \(1.2\,\text{s}\), \(9.6\,\text{m}\) to the right.
ⓒ. Before release, \(\vec v=-8\hat i+6\hat j\) and afterward \(\vec a=-10\hat j\,\text{m s}^{-2}\); it next reaches the same horizontal line after \(0.6\,\text{s}\), \(4.8\,\text{m}\) to the left.
ⓓ. Before release, \(\vec v=-8\hat i+6\hat j\) and \(\vec a_c=-12\hat i-16\hat j\,\text{m s}^{-2}\); it next reaches the same horizontal line after \(1.2\,\text{s}\), \(9.6\,\text{m}\) to the left.
Correct Answer: Before release, \(\vec v=-8\hat i+6\hat j\) and \(\vec a_c=-12\hat i-16\hat j\,\text{m s}^{-2}\); it next reaches the same horizontal line after \(1.2\,\text{s}\), \(9.6\,\text{m}\) to the left.
Explanation: The outward radial unit vector at the release point is
\[
\hat r=\frac{3\hat i+4\hat j}{5}.
\]
For counterclockwise motion, the tangent unit vector is
\[
\hat t=-\frac45\hat i+\frac35\hat j.
\]
Hence the velocity just before release is
\[
\vec v=10\hat t=-8\hat i+6\hat j.
\]
The centripetal-acceleration magnitude is
\[
a_c=\frac{10^2}{5}=20\,\text{m s}^{-2}.
\]
It points inward, so
\[
\vec a_c=-20\hat r=-12\hat i-16\hat j.
\]
After release the specified acceleration is instead \(-10\hat j\), while the initial post-release velocity remains \(-8\hat i+6\hat j\). To return to the same horizontal level,
\[
0=6t-5t^2.
\]
The non-zero solution is
\[
t=1.2\,\text{s}.
\]
The horizontal displacement during that interval is
\[
\Delta x=(-8)(1.2)=-9.6\,\text{m}.
\]
Thus it crosses the release-level line \(9.6\,\text{m}\) to the left.
519. A projectile launch velocity \(\vec u\) lies in the \(xy\)-plane. The unit vector
\[
\hat n=\frac{4\hat i+3\hat j}{5}
\]
satisfies
\[
\vec u\cdot\hat n=20\,\text{m s}^{-1},
\qquad
\vec u\times\hat n=-15\hat k\,\text{m s}^{-1}.
\]
The projectile is launched from level ground and returns to the same level under \(g=10\,\text{m s}^{-2}\). Which set of results is correct?
ⓐ. \(\vec u=20\hat i+15\hat j\,\text{m s}^{-1},\ R=60\,\text{m}\)
ⓑ. \(\vec u=24\hat i+7\hat j\,\text{m s}^{-1},\ T=1.4\,\text{s},\ R=33.6\,\text{m}\)
ⓒ. \(\vec u=7\hat i+24\hat j\,\text{m s}^{-1},\ T=4.8\,\text{s},\ R=33.6\,\text{m},\ H=28.8\,\text{m}\)
ⓓ. \(\vec u=7\hat i-24\hat j\,\text{m s}^{-1}\), so it initially moves downward.
Correct Answer: \(\vec u=7\hat i+24\hat j\,\text{m s}^{-1},\ T=4.8\,\text{s},\ R=33.6\,\text{m},\ H=28.8\,\text{m}\)
Explanation: Introduce a unit vector perpendicular to \(\hat n\):
\[
\hat m=\frac{-3\hat i+4\hat j}{5}.
\]
This choice satisfies
\[
\hat n\times\hat m=\hat k.
\]
Resolve the launch velocity as
\[
\vec u=a\hat n+b\hat m.
\]
The dot-product condition gives
\[
a=20.
\]
Also,
\[
\vec u\times\hat n
=b(\hat m\times\hat n)
=-b\hat k.
\]
Comparing with \(-15\hat k\) gives
\[
b=15.
\]
Therefore
\[
\vec u=20\hat n+15\hat m.
\]
Substitution yields
\[
\vec u=7\hat i+24\hat j\,\text{m s}^{-1}.
\]
Hence
\[
T=\frac{2(24)}{10}=4.8\,\text{s},
\]
\[
R=(7)(4.8)=33.6\,\text{m},
\]
and
\[
H=\frac{24^2}{20}=28.8\,\text{m}.
\]
The vector-product sign is essential for selecting the correct perpendicular component.
520. A particle starts from \(x=0\) and is claimed to move with one constant acceleration. Four experimental data sets are proposed:
| Set | \(v(1\,\text{s})\) | \(v(3\,\text{s})\) | \(x(4\,\text{s})\) |
|---|
| P | \(8\,\text{m s}^{-1}\) | \(4\,\text{m s}^{-1}\) | \(24\,\text{m}\) |
| Q | \(8\,\text{m s}^{-1}\) | \(2\,\text{m s}^{-1}\) | \(24\,\text{m}\) |
| R | \(6\,\text{m s}^{-1}\) | \(4\,\text{m s}^{-1}\) | \(18\,\text{m}\) |
| S | \(5\,\text{m s}^{-1}\) | \(-1\,\text{m s}^{-1}\) | \(10\,\text{m}\) |
Which data set is internally consistent with constant acceleration?
ⓐ. S only
ⓑ. Q only
ⓒ. R only
ⓓ. P only
Correct Answer: P only
Explanation: For constant acceleration, velocity varies linearly with time. Therefore the average of the velocities at \(t=1\,\text{s}\) and \(t=3\,\text{s}\) equals the velocity at the midpoint \(t=2\,\text{s}\). A more useful relation can be obtained directly for the displacement from \(0\) to \(4\,\text{s}\). If \(v_1=v(1)\) and \(v_3=v(3)\), then
\[
a=\frac{v_3-v_1}{2}.
\]
The initial velocity is
\[
u=v_1-a.
\]
Using the constant-acceleration equations, one obtains
\[
x(4)=2(v_1+v_3).
\]
For set P,
\[
2(8+4)=24\,\text{m},
\]
so P is consistent. For Q the predicted value is
\[
2(8+2)=20\,\text{m},
\]
not \(24\,\text{m}\). For R it is
\[
2(6+4)=20\,\text{m},
\]
not \(18\,\text{m}\). For S it is
\[
2(5-1)=8\,\text{m},
\]
not \(10\,\text{m}\). Thus only P can arise from one constant acceleration.
