501. A block moves without slipping on a horizontal rotating platform of radius \(r\). At its present speed, the required static friction is exactly one-half of the maximum available static friction. The block is then moved to radius \(2r\), the coefficient of static friction is reduced to one-fourth of its original value, and the speed is kept unchanged. The normal reaction remains \(mg\). What is the new condition?
ⓐ. The required friction is half the new limiting friction, so the block remains comfortably safe.
ⓑ. The required friction equals the new limiting static friction, so the block is exactly at the no-slip threshold.
ⓒ. The required friction is twice the new limiting friction, so slipping is unavoidable.
ⓓ. The required friction is unchanged while the limiting friction becomes one-fourth, so the required-to-available ratio becomes \(2\).
Correct Answer: The required friction equals the new limiting static friction, so the block is exactly at the no-slip threshold.
Explanation: Initially let the maximum available static friction be
\[
f_{\max}=\mu_smg.
\]
The required friction is stated to be
\[
f_{\text{req}}=\frac12\mu_smg.
\]
Since
\[
f_{\text{req}}=\frac{mv^2}{r},
\]
we have
\[
\frac{mv^2}{r}=\frac12\mu_smg.
\]
At the new radius \(2r\), with the same speed,
\[
f_{\text{req}}'=\frac{mv^2}{2r}
=\frac12f_{\text{req}}
=\frac14\mu_smg.
\]
The new coefficient is \(\mu_s/4\), so the new limiting friction is
\[
f_{\max}'=\frac{\mu_s}{4}mg
=\frac14\mu_smg.
\]
Therefore,
\[
f_{\text{req}}'=f_{\max}'.
\]
The block is exactly at limiting static friction: no slipping is required yet, but any increase in speed or further reduction in available friction would exceed the threshold.
502. A \(2\,\text{kg}\) block moves without slipping in a horizontal circle of radius \(4\,\text{m}\) with speed \(6\,\text{m s}^{-1}\). It is attached to the centre by a string that pulls radially inward with tension \(22\,\text{N}\). The horizontal surface can exert static friction in either radial direction, with maximum magnitude \(6\,\text{N}\). What is the actual frictional force?
ⓐ. \(4\,\text{N}\) radially outward, and no slipping occurs
ⓑ. \(4\,\text{N}\) radially inward, and no slipping occurs
ⓒ. \(6\,\text{N}\) radially outward, at limiting equilibrium
ⓓ. \(22\,\text{N}\) radially outward, so slipping must occur
Correct Answer: \(4\,\text{N}\) radially outward, and no slipping occurs
Explanation: The required inward resultant for uniform circular motion is
\[
F_c=\frac{mv^2}{r}.
\]
Substituting,
\[
F_c=\frac{2(6^2)}{4}=18\,\text{N}.
\]
The string alone supplies \(22\,\text{N}\) inward, which is \(4\,\text{N}\) more than the required inward resultant. Therefore another horizontal force must act radially outward so that the net inward force becomes \(18\,\text{N}\). Hence,
\[
22-f=18.
\]
Thus,
\[
f=4\,\text{N}
\]
outward. The maximum available static friction is \(6\,\text{N}\), so the required \(4\,\text{N}\) is physically available. Static friction therefore need not point inward merely because the motion is circular; its direction is fixed by the force balance required to prevent relative slipping.
503. In Trial P, a particle of mass \(m\) moves in a horizontal circle of radius \(3\,\text{m}\), and the largest inward force available from the physical force provider is \(12\,\text{N}\). In Trial Q, the same particle moves in a circle of radius \(6\,\text{m}\), while the largest available inward force is \(18\,\text{N}\). What is the ratio of the maximum permitted speeds \(v_Q/v_P\)?
ⓐ. \(\sqrt{\dfrac32}\)
ⓑ. \(\sqrt3\)
ⓒ. \(\dfrac32\)
ⓓ. \(3\)
Correct Answer: \(\sqrt3\)
Explanation: At the maximum permitted speed, the available inward force equals the centripetal-force requirement:
\[
F_{\max}=\frac{mv_{\max}^2}{r}.
\]
Therefore,
\[
v_{\max}^2=\frac{F_{\max}r}{m}.
\]
For the same particle, the mass cancels in a ratio:
\[
\left(\frac{v_Q}{v_P}\right)^2
=\frac{F_Qr_Q}{F_Pr_P}.
\]
Substituting,
\[
\left(\frac{v_Q}{v_P}\right)^2
=\frac{18\times6}{12\times3}.
\]
Hence,
\[
\left(\frac{v_Q}{v_P}\right)^2=3.
\]
Therefore,
\[
\frac{v_Q}{v_P}=\sqrt3.
\]
The result depends on the product \(F_{\max}r\), not on either the force limit or radius alone.
504. A particle moves with constant speed in a circle. At one instant its momentum is
\[
\vec p_i=8\hat i\,\text{kg m s}^{-1}.
\]
After completing one-quarter of the circle, its momentum is
\[
\vec p_f=8\hat j\,\text{kg m s}^{-1}.
\]
The quarter-circle motion takes \(0.40\,\text{s}\). What is the average net force over this interval?
ⓐ. \((-40\hat i+40\hat j)\,\text{N}\), magnitude \(40\sqrt2\,\text{N}\)
ⓑ. \((20\hat i+20\hat j)\,\text{N}\), magnitude \(20\sqrt2\,\text{N}\)
ⓒ. \(20\hat j\,\text{N}\), magnitude \(20\,\text{N}\)
ⓓ. \((-20\hat i+20\hat j)\,\text{N}\), magnitude \(20\sqrt2\,\text{N}\)
Correct Answer: \((-20\hat i+20\hat j)\,\text{N}\), magnitude \(20\sqrt2\,\text{N}\)
Explanation: The average net force is related to the momentum change by
\[
\vec F_{\text{avg}}=\frac{\Delta\vec p}{\Delta t}.
\]
The momentum change is
\[
\Delta\vec p=\vec p_f-\vec p_i.
\]
Thus,
\[
\Delta\vec p=8\hat j-8\hat i
=-8\hat i+8\hat j.
\]
Dividing by \(0.40\,\text{s}\),
\[
\vec F_{\text{avg}}
=\frac{-8\hat i+8\hat j}{0.40}
=-20\hat i+20\hat j\,\text{N}.
\]
Its magnitude is
\[
F_{\text{avg}}=\sqrt{20^2+20^2}=20\sqrt2\,\text{N}.
\]
Although the momentum magnitude remains \(8\,\text{kg m s}^{-1}\), its direction changes, so the momentum change and average force are non-zero. The average-force direction over the finite quarter-circle interval need not coincide with the instantaneous radial direction at either endpoint.
505. Consider two experiments using the same particle.
Graph P plots required centripetal force \(F_c\) against radius \(r\) while angular speed is kept constant; it is a straight line through the origin. Graph Q plots \(F_c\) against \(r\) while linear speed is kept constant; the curve decreases as \(r\) increases.
Which statement correctly explains both graphs?
ⓐ. P follows \(F_c=m\omega^2r\), while Q follows \(F_c=mv^2/r\).
ⓑ. P follows \(F_c=mv^2/r\), while Q follows \(F_c=m\omega^2r\).
ⓒ. Both should be straight lines through the origin because force is always proportional to radius.
ⓓ. Both should decrease with radius because centripetal force is always inversely proportional to radius.
Correct Answer: P follows \(F_c=m\omega^2r\), while Q follows \(F_c=mv^2/r\).
Explanation: The dependence on radius depends on which kinematic quantity is held fixed. At constant angular speed,
\[
F_c=m\omega^2r.
\]
Thus,
\[
F_c\propto r,
\]
so an \(F_c\)-versus-\(r\) graph is a straight line through the origin. At constant linear speed,
\[
F_c=\frac{mv^2}{r}.
\]
Therefore,
\[
F_c\propto\frac1r,
\]
so the force decreases as radius increases. There is no contradiction between the two graphs because constant \(\omega\) and constant \(v\) are different experimental conditions. At constant \(\omega\), increasing \(r\) also increases the linear speed through \(v=\omega r\). Hence option A correctly accounts for both representations.
506. Four particles execute uniform circular motion. Their data are shown below.
| Case | Mass | Speed | Radius |
|---|
| P | \(1\,\text{kg}\) | \(4\,\text{m s}^{-1}\) | \(2\,\text{m}\) |
| Q | \(2\,\text{kg}\) | \(3\,\text{m s}^{-1}\) | \(3\,\text{m}\) |
| R | \(3\,\text{kg}\) | \(2\,\text{m s}^{-1}\) | \(1\,\text{m}\) |
| S | \(0.5\,\text{kg}\) | \(6\,\text{m s}^{-1}\) | \(3\,\text{m}\) |
What is the correct order of required centripetal-force magnitudes?
ⓐ. \(Q=S\gt P\gt R\)
ⓑ. \(P\gt R\gt Q\gt S\)
ⓒ. \(R\gt Q\gt P\gt S\)
ⓓ. \(R\gt P\gt Q=S\)
Correct Answer: \(R\gt P\gt Q=S\)
Explanation: For each case,
\[
F_c=\frac{mv^2}{r}.
\]
For P,
\[
F_P=\frac{1(4^2)}2=8\,\text{N}.
\]
For Q,
\[
F_Q=\frac{2(3^2)}3=6\,\text{N}.
\]
For R,
\[
F_R=\frac{3(2^2)}1=12\,\text{N}.
\]
For S,
\[
F_S=\frac{0.5(6^2)}3=6\,\text{N}.
\]
Therefore,
\[
F_R\gt F_P\gt F_Q=F_S.
\]
The ranking cannot be made reliably from speed alone because mass and radius also change between the cases.
507. A \(2\,\text{kg}\) block moves without slipping in a horizontal circle of radius \(0.50\,\text{m}\) on a rotating platform. It is also attached to the centre by a string that exerts a constant \(6\,\text{N}\) inward tension. The coefficient of static friction is \(0.50\), and \(g=10\,\text{m s}^{-2}\). What is the maximum speed for no slipping, and what happens if the string is cut when the block is moving at that speed?
ⓐ. \(\sqrt5\,\text{m s}^{-1}\); after cutting the string, no friction is required
ⓑ. \(2\,\text{m s}^{-1}\); after cutting the string, the block remains exactly at limiting friction
ⓒ. \(2\,\text{m s}^{-1}\); after cutting the string, static friction alone is insufficient
ⓓ. \(4\,\text{m s}^{-1}\); after cutting the string, kinetic friction immediately supplies the same inward force
Correct Answer: \(2\,\text{m s}^{-1}\); after cutting the string, static friction alone is insufficient
Explanation: The maximum available static friction is
\[
f_{s,\max}=\mu_smg
=0.50(2)(10)
=10\,\text{N}.
\]
At the maximum no-slip speed, the string and maximum inward static friction together provide
\[
F_{\max}=6+10=16\,\text{N}.
\]
Thus,
\[
\frac{mv_{\max}^2}{r}=16.
\]
Substituting \(m=2\,\text{kg}\) and \(r=0.50\,\text{m}\),
\[
\frac{2v_{\max}^2}{0.50}=16.
\]
Hence,
\[
4v_{\max}^2=16,
\qquad
v_{\max}=2\,\text{m s}^{-1}.
\]
If the string is cut at this speed, the same circular path would still require \(16\,\text{N}\) inward. Static friction alone can provide at most \(10\,\text{N}\), so it cannot maintain the original no-slip circular motion.
508. A \(1\,\text{kg}\) block moves without slipping on a horizontal rotating platform in a circle of radius \(2\,\text{m}\). A string attached to the centre pulls inward with constant tension \(5\,\text{N}\). At a certain instant the block speed is \(2\,\text{m s}^{-1}\). The coefficient of static friction is \(0.40\), and \(g=10\,\text{m s}^{-2}\). Which statement is correct?
ⓐ. Friction is zero at \(2\,\text{m s}^{-1}\) because the string itself provides a centripetal force.
ⓑ. Friction is \(3\,\text{N}\) inward; at \(v=\sqrt{10}\,\text{m s}^{-1}\), friction reaches its maximum.
ⓒ. Friction is \(5\,\text{N}\) outward, so slipping already occurs.
ⓓ. Friction is \(3\,\text{N}\) outward; if the speed increases to \(\sqrt{10}\,\text{m s}^{-1}\), the required friction becomes zero.
Correct Answer: Friction is \(3\,\text{N}\) outward; if the speed increases to \(\sqrt{10}\,\text{m s}^{-1}\), the required friction becomes zero.
Explanation: At \(v=2\,\text{m s}^{-1}\), the required inward resultant is
\[
F_c=\frac{mv^2}{r}
=\frac{1(4)}2
=2\,\text{N}.
\]
The string already pulls inward with \(5\,\text{N}\). Therefore friction must oppose the excess inward pull:
\[
5-f=2.
\]
Hence,
\[
f=3\,\text{N}
\]
outward. This is below
\[
f_{s,\max}=0.40(1)(10)=4\,\text{N},
\]
so no slipping is required. Friction becomes zero when the string tension alone exactly equals the centripetal requirement:
\[
5=\frac{v^2}{2}.
\]
Therefore,
\[
v^2=10,
\qquad
v=\sqrt{10}\,\text{m s}^{-1}.
\]
Static friction can therefore reverse radial direction as the required centripetal force changes.
509. A particle moves uniformly in a circle. At two diametrically opposite points \(P\) and \(Q\), compare its velocity, momentum, centripetal acceleration and net force.
ⓐ. All four vectors have the same direction at both points because the speed is constant.
ⓑ. Velocity is the same at \(P\) and \(Q\), but acceleration is opposite.
ⓒ. Momentum is opposite, but net force has the same direction at both points.
ⓓ. Velocity and momentum are opposite at \(P\) and \(Q\), and the acceleration and net-force vectors are also opposite.
Correct Answer: Velocity and momentum are opposite at \(P\) and \(Q\), and the acceleration and net-force vectors are also opposite.
Explanation: At diametrically opposite points, the tangents to the circle are parallel but oppositely directed for motion in one fixed sense around the circle. Therefore the velocity vectors are opposite:
\[
\vec v_Q=-\vec v_P.
\]
For constant mass,
\[
\vec p=m\vec v,
\]
so the momentum vectors are also opposite. The centripetal acceleration at each point is directed towards the centre. From diametrically opposite positions, those inward directions are themselves opposite. Hence,
\[
\vec a_Q=-\vec a_P.
\]
By Newton's Second Law,
\[
\vec F_{\text{net}}=m\vec a,
\]
so the net-force vectors are also opposite. Their magnitudes remain unchanged if the motion is uniform with fixed radius.
510. A particle moves anticlockwise with constant speed around a horizontal circular path. At point \(A\) it is at the easternmost point of the circle. After successive quarter revolutions it reaches the northernmost point \(B\), westernmost point \(C\), and southernmost point \(D\). Which sequence correctly gives the velocity direction followed by the net-force direction at \(A,B,C,D\)?
ⓐ. \(A:\) north, west; \(B:\) west, south; \(C:\) south, east; \(D:\) east, north
ⓑ. \(A:\) west, north; \(B:\) south, west; \(C:\) east, south; \(D:\) north, east
ⓒ. \(A:\) north, east; \(B:\) west, north; \(C:\) south, west; \(D:\) east, south
ⓓ. \(A:\) south, west; \(B:\) east, south; \(C:\) north, east; \(D:\) west, north
Correct Answer: \(A:\) north, west; \(B:\) west, south; \(C:\) south, east; \(D:\) east, north
Explanation: In anticlockwise circular motion, the velocity is always tangent to the circle in the direction of motion. At the easternmost point \(A\), that tangent points north. The inward force points towards the centre, which is west of \(A\). At the northernmost point \(B\), velocity points west and the centre lies south. At \(C\), velocity points south and the inward force points east. At \(D\), velocity points east while the centre lies north. Thus the correct velocity-force pairs are
\[
A:(N,W),\quad B:(W,S),\quad C:(S,E),\quad D:(E,N).
\]
The force direction continuously changes with position even though its magnitude may remain constant.
511. A particle moves in a circle of fixed radius under a physical force whose maximum available inward magnitude is \(F_0\). At speed \(v\), the force provider is operating at only \(36\%\) of its maximum capacity. If the mass and radius remain unchanged, what is the largest possible speed before the force limit is reached?
ⓐ. \(1.36v\)
ⓑ. \(\dfrac{25}{9}v\)
ⓒ. \(\dfrac53v\)
ⓓ. \(3v\)
Correct Answer: \(\dfrac53v\)
Explanation: At the present speed,
\[
F_c=\frac{mv^2}{r}.
\]
The problem states that
\[
F_c=0.36F_0.
\]
At the maximum speed \(v_{\max}\),
\[
F_0=\frac{mv_{\max}^2}{r}.
\]
Taking the ratio,
\[
\frac{v^2}{v_{\max}^2}=0.36.
\]
Therefore,
\[
\frac{v}{v_{\max}}=0.6.
\]
Hence,
\[
v_{\max}=\frac{v}{0.6}
=\frac53v.
\]
A force reserve of \(64\%\) does not imply a \(64\%\) speed reserve because centripetal force depends on the square of speed.
512. At an instant, a \(2\,\text{kg}\) particle moving in a circle of radius \(5\,\text{m}\) has momentum
\[
\vec p=(6\hat i+8\hat j)\,\text{kg m s}^{-1}.
\]
The centre of the circle lies from the particle in the direction of the vector
\[
-4\hat i+3\hat j.
\]
What is the instantaneous net force?
ⓐ. \((-6\hat i-8\hat j)\,\text{N}\)
ⓑ. \((8\hat i-6\hat j)\,\text{N}\)
ⓒ. \((-8\hat i+6\hat j)\,\text{N}\)
ⓓ. \((6\hat i+8\hat j)\,\text{N}\)
Correct Answer: \((-8\hat i+6\hat j)\,\text{N}\)
Explanation: The momentum magnitude is
\[
p=\sqrt{6^2+8^2}=10\,\text{kg m s}^{-1}.
\]
Since
\[
p=mv,
\]
the speed is
\[
v=\frac{10}{2}=5\,\text{m s}^{-1}.
\]
The centripetal-force magnitude is
\[
F_c=\frac{mv^2}{r}
=\frac{2(25)}5
=10\,\text{N}.
\]
The given centre-direction vector is
\[
-4\hat i+3\hat j,
\]
whose magnitude is \(5\). Its unit vector is therefore
\[
-\frac45\hat i+\frac35\hat j.
\]
Hence the inward net force is
\[
\vec F=10\left(-\frac45\hat i+\frac35\hat j\right).
\]
Thus,
\[
\vec F=-8\hat i+6\hat j\,\text{N}.
\]
The force direction is determined by the centre, not simply by the direction opposite to momentum.
513. Consider a block moving without slipping on a horizontal rotating platform.
Statement I: The actual static friction need not equal \(\mu_smg\).
Statement II: For fixed \(r,\mu_s\) and \(g\), the maximum permissible angular speed is independent of the block's mass.
Statement III: Doubling the block's mass doubles both the required centripetal force at a given speed and the maximum available static friction.
Statement IV: At fixed \(\mu_s\) and \(g\), moving the block to a larger radius increases the maximum permissible angular speed.
Which option is correct?
ⓐ. I, II, III and IV
ⓑ. I and IV only
ⓒ. II, III and IV only
ⓓ. I, II and III only
Correct Answer: I, II and III only
Explanation: The actual static friction equals the inward force required:
\[
f_s=m\omega^2r
\]
as long as this does not exceed its limiting value. Therefore Statement I is correct. At the threshold,
\[
m\omega_{\max}^2r=\mu_smg.
\]
Cancelling \(m\),
\[
\omega_{\max}=\sqrt{\frac{\mu_sg}{r}},
\]
so Statement II is correct. Doubling mass doubles the required force \(m\omega^2r\) and also doubles the maximum static friction \(\mu_smg\), making Statement III correct. The expression for \(\omega_{\max}\) shows
\[
\omega_{\max}\propto\frac1{\sqrt r}.
\]
Therefore increasing radius decreases, rather than increases, the maximum permissible angular speed. Statement IV is false.
514. Two identical blocks are placed on the same horizontal rotating platform, one at radius \(r\) and the other at radius \(4r\). The coefficient of static friction is the same for both. At a certain angular speed, the inner block requires exactly one-half of its maximum available static friction to remain in circular motion. What happens to the outer block at the same angular speed?
ⓐ. It requires twice its maximum available static friction and therefore cannot remain without slipping.
ⓑ. It requires exactly its maximum static friction and is at limiting equilibrium.
ⓒ. It requires one-eighth of its maximum friction because its radius is larger.
ⓓ. It requires the same friction as the inner block because their angular speeds are equal.
Correct Answer: It requires twice its maximum available static friction and therefore cannot remain without slipping.
Explanation: For a block rotating with angular speed \(\omega\),
\[
f_{\text{req}}=m\omega^2r.
\]
For identical blocks on the same horizontal platform,
\[
f_{s,\max}=\mu_smg
\]
is the same for both. The inner block requires
\[
f_{\text{inner}}=\frac12f_{s,\max}.
\]
At radius \(4r\), the required friction at the same \(\omega\) becomes four times larger:
\[
f_{\text{outer}}=4f_{\text{inner}}.
\]
Therefore,
\[
f_{\text{outer}}=4\left(\frac12f_{s,\max}\right)
=2f_{s,\max}.
\]
The required inward force is twice what static friction can supply. Hence the outer block cannot remain in the assumed no-slip circular motion.
515. For two particles A and B, graphs of required centripetal force \(F_c\) against \(v^2\) are straight lines through the origin. The slope for A is \(2\,\text{kg m}^{-1}\), while that for B is \(0.5\,\text{kg m}^{-1}\). Their masses are \(4\,\text{kg}\) and \(1\,\text{kg}\), respectively. Which statement is correct?
ⓐ. \(r_A=2\,\text{m},\ r_B=2\,\text{m}\); at equal speed, \(F_A=F_B\).
ⓑ. \(r_A=8\,\text{m},\ r_B=0.5\,\text{m}\); at equal speed, \(F_A=F_B\).
ⓒ. \(r_A=2\,\text{m},\ r_B=2\,\text{m}\); at equal speed, \(F_A=4F_B\).
ⓓ. \(r_A=8\,\text{m},\ r_B=2\,\text{m}\); at equal speed, the graph slopes should be equal.
Correct Answer: \(r_A=2\,\text{m},\ r_B=2\,\text{m}\); at equal speed, \(F_A=4F_B\).
Explanation: From
\[
F_c=\frac{m}{r}v^2,
\]
the slope of an \(F_c\)-versus-\(v^2\) graph is
\[
\frac mr.
\]
For A,
\[
\frac{4}{r_A}=2,
\]
so
\[
r_A=2\,\text{m}.
\]
For B,
\[
\frac{1}{r_B}=0.5,
\]
giving
\[
r_B=2\,\text{m}.
\]
Thus the radii are equal. At any common speed,
\[
\frac{F_A}{F_B}
=\frac{\text{slope}_A}{\text{slope}_B}
=\frac{2}{0.5}
=4.
\]
The graph therefore simultaneously reveals the equal radii and the fourfold force requirement for particle A.
516. A particle completes each revolution in the same time \(T\) while moving first in a circle of radius \(R\) and then in a circle of radius \(R/2\). Its mass is unchanged. What happens to its linear speed and required centripetal force?
ⓐ. Speed doubles and centripetal force doubles.
ⓑ. Speed becomes one-half, but centripetal force becomes one-fourth.
ⓒ. Both speed and centripetal force become one-half of their original values.
ⓓ. Speed remains unchanged while centripetal force doubles.
Correct Answer: Both speed and centripetal force become one-half of their original values.
Explanation: The angular speed is
\[
\omega=\frac{2\pi}{T}.
\]
Because the period \(T\) is unchanged, \(\omega\) is unchanged. Linear speed is
\[
v=\omega r.
\]
When the radius changes from \(R\) to \(R/2\),
\[
v'=\omega\frac R2=\frac v2.
\]
The centripetal force may be written as
\[
F_c=m\omega^2r.
\]
Since \(m\) and \(\omega\) remain unchanged,
\[
F_c\propto r.
\]
Therefore,
\[
F_c'=\frac{F_c}{2}.
\]
Using only \(F_c\propto1/r\) here would be incorrect because that proportionality assumes constant linear speed, whereas this problem keeps the period and angular speed fixed.
517. At an instant, use radial-tangential axes for a particle undergoing uniform circular motion. Positive radial direction is inward and positive tangential direction is along the velocity. One physical force has components
\[
\vec F_1=(6\hat r+4\hat t)\,\text{N}.
\]
The particle's required net force at that instant is purely radial and equals
\[
\vec F_{\text{net}}=10\hat r\,\text{N}.
\]
A second physical force \(\vec F_2\) is the only other force with radial or tangential components. What is \(\vec F_2\)?
ⓐ. \((10\hat r-4\hat t)\,\text{N}\), magnitude \(2\sqrt{29}\,\text{N}\)
ⓑ. \((4\hat r+4\hat t)\,\text{N}\), magnitude \(4\sqrt2\,\text{N}\)
ⓒ. \((-4\hat r-4\hat t)\,\text{N}\), magnitude \(4\sqrt2\,\text{N}\)
ⓓ. \((4\hat r-4\hat t)\,\text{N}\), magnitude \(4\sqrt2\,\text{N}\)
Correct Answer: \((4\hat r-4\hat t)\,\text{N}\), magnitude \(4\sqrt2\,\text{N}\)
Explanation: Uniform circular motion requires no tangential acceleration, so the net tangential force must be zero. The first force contributes
\[
+4\,\text{N}
\]
tangentially. Therefore the second force must contribute
\[
-4\,\text{N}
\]
in the tangential direction. Radially, the required net inward force is \(10\,\text{N}\). The first force already supplies \(6\,\text{N}\) inward, so the second force must supply the remaining
\[
10-6=4\,\text{N}
\]
inward. Hence,
\[
\vec F_2=4\hat r-4\hat t\,\text{N}.
\]
Its magnitude is
\[
F_2=\sqrt{4^2+(-4)^2}=4\sqrt2\,\text{N}.
\]
This illustrates that the centripetal force is the radial resultant of the actual physical forces, while their tangential components must cancel for uniform speed.
518. Two particles \(A\) and \(B\), each of mass \(2\,\text{kg}\), move along the same straight line. At a certain instant their momentum-time graphs intersect at \(p=6\,\text{kg m s}^{-1}\). The slopes of the two graphs at that instant are \(+4\,\text{N}\) for \(A\) and \(-2\,\text{N}\) for \(B\). Which statement is correct?
ⓐ. Same velocity; \(|a_B|=2|a_A|\) because \(|F_B|=2|F_A|\).
ⓑ. Opposite velocities; \(a_A=+2\,\text{m s}^{-2}\), \(a_B=-1\,\text{m s}^{-2}\).
ⓒ. Same velocity; \(a_A=a_B=+2\,\text{m s}^{-2}\).
ⓓ. Same velocity; \(a_A=+2\,\text{m s}^{-2}\), \(a_B=-1\,\text{m s}^{-2}\).
Correct Answer: Same velocity; \(a_A=+2\,\text{m s}^{-2}\), \(a_B=-1\,\text{m s}^{-2}\).
Explanation: At the intersection,
\[
p_A=p_B=6\,\text{kg m s}^{-1}.
\]
Because the masses are equal,
\[
v=\frac{p}{m}=\frac62=3\,\text{m s}^{-1}
\]
for both particles at that instant. The slope of a momentum-time graph gives the instantaneous net force:
\[
F=\frac{dp}{dt}.
\]
Hence,
\[
F_A=+4\,\text{N},\qquad F_B=-2\,\text{N}.
\]
Using \(F=ma\),
\[
a_A=\frac42=+2\,\text{m s}^{-2},
\qquad
a_B=\frac{-2}{2}=-1\,\text{m s}^{-2}.
\]
Thus equal instantaneous momentum does not imply equal force or acceleration; those depend on the local slope of the momentum-time graph.
519. The momentum of a particle moving along the \(x\)-axis is measured at equal time intervals.
| \(t\) (s) | 0 | 1 | 2 | 3 |
|---|
| \(p_x\) (\(\text{kg m s}^{-1}\)) | 2 | 5 | 10 | 17 |
Which conclusion is justified from these data alone?
ⓐ. Average forces over the three intervals are \(3,5,7\,\text{N}\); the instantaneous forces at \(1,2,3\,\text{s}\) are exactly the same values.
ⓑ. Average forces over the three intervals are \(3,5,7\,\text{N}\); the instantaneous force at \(1\,\text{s}\) is not fixed by these data alone.
ⓒ. Average forces over the three intervals are all equal because the sampling intervals are equal; the instantaneous force is constant.
ⓓ. The average force over \(0\!-\!2\,\text{s}\) is \(4\,\text{N}\); therefore the instantaneous force at \(1\,\text{s}\) must be \(4\,\text{N}\).
Correct Answer: Average forces over the three intervals are \(3,5,7\,\text{N}\); the instantaneous force at \(1\,\text{s}\) is not fixed by these data alone.
Explanation: Over a finite interval,
\[
F_{\text{avg}}=\frac{\Delta p}{\Delta t}.
\]
For \(0\!-\!1\,\text{s}\),
\[
F_{\text{avg}}=\frac{5-2}{1}=3\,\text{N}.
\]
For \(1\!-\!2\,\text{s}\),
\[
F_{\text{avg}}=\frac{10-5}{1}=5\,\text{N}.
\]
For \(2\!-\!3\,\text{s}\),
\[
F_{\text{avg}}=\frac{17-10}{1}=7\,\text{N}.
\]
However, the table contains only discrete momentum values. The instantaneous force requires the derivative \(dp/dt\), or equivalently the local tangent slope of a continuous \(p-t\) curve. Many different smooth curves can pass through the same tabulated points while having different slopes at \(t=1\,\text{s}\). Therefore the interval-average forces are determined, but the exact instantaneous force is not.
520. Two particles move along the \(x\)-axis with the same momentum function
\[
p(t)=6+4t
\]
in SI units. Their masses are \(m_A=2\,\text{kg}\) and \(m_B=4\,\text{kg}\). Which option is correct at every instant?
ⓐ. \(F_A=2F_B,\ v_A=v_B,\ a_A=a_B\)
ⓑ. \(F_A=F_B=4\,\text{N},\ v_A=2v_B,\ a_A=2a_B\)
ⓒ. \(F_A=F_B=2\,\text{N},\ v_A=v_B/2,\ a_A=a_B/2\)
ⓓ. \(F_A=2F_B,\ v_A=2v_B,\ a_A=2a_B\)
Correct Answer: \(F_A=F_B=4\,\text{N},\ v_A=2v_B,\ a_A=2a_B\)
Explanation: The net force is obtained directly from the momentum function:
\[
F=\frac{dp}{dt}.
\]
Since
\[
p(t)=6+4t,
\]
we have
\[
F=4\,\text{N}
\]
for both particles. Thus the same \(p(t)\) implies the same \(dp/dt\), independent of mass. Their velocities are
\[
v_A=\frac{p}{2},
\qquad
v_B=\frac{p}{4},
\]
so
\[
v_A=2v_B.
\]
Their accelerations follow from \(a=F/m\):
\[
a_A=\frac42=2\,\text{m s}^{-2},
\qquad
a_B=\frac44=1\,\text{m s}^{-2}.
\]
Hence \(a_A=2a_B\), while the net forces remain equal.
