401. A particle moves in a circle with constant radius. At an instant, its tangential acceleration is zero, but its radial acceleration is non-zero. The motion at that instant is best described as:
ⓐ. changing speed with fixed direction
ⓑ. no change in velocity at all
ⓒ. constant speed, changing direction
ⓓ. straight-line motion at that instant
Correct Answer: constant speed, changing direction
Explanation: Tangential acceleration changes the magnitude of velocity, which is speed. If tangential acceleration is zero, the speed is not changing at that instant. Radial acceleration changes the direction of velocity and is needed for circular motion. A non-zero radial acceleration means the velocity direction is still changing. Thus, the velocity vector changes even without a change in speed.
402. The dimensional check for the same-level range formula \(R=\frac{u^2\sin2\theta}{g}\) gives:
ⓐ. \([L]\)
ⓑ. \([LT^{-1}]\)
ⓒ. \([LT^{-2}]\)
ⓓ. \([T]\)
Correct Answer: \([L]\)
Explanation: \( \textbf{Formula:} \)
\[
R=\frac{u^2\sin2\theta}{g}
\]
\( \textbf{Dimension of speed:} \)
\[
[u]=[LT^{-1}]
\]
\( \textbf{Dimension of \(u^2\):} \)
\[
[u^2]=[L^2T^{-2}]
\]
\( \textbf{Dimension of \(g\):} \)
\[
[g]=[LT^{-2}]
\]
\( \textbf{Sine factor:} \) \(\sin2\theta\) is dimensionless.
\( \textbf{Divide dimensions:} \)
\[
[R]=\frac{[L^2T^{-2}]}{[LT^{-2}]}=[L]
\]
\( \textbf{Final answer:} \) The formula has the dimension of length.
403. A circular-motion note says \(\omega=2\pi T\). The correction is:
ⓐ. \(\omega=\frac{2\pi}{T}\), angular displacement per time
ⓑ. \(\omega=\frac{T}{2\pi}\), since period is treated as an angle
ⓒ. \(\omega=T+2\pi\), since time and angle are added directly
ⓓ. \(\omega=2\pi rT\), since radius must appear in every circular formula
Correct Answer: \(\omega=\frac{2\pi}{T}\), angular displacement per time
Explanation: In one complete revolution, the angular displacement is \(2\pi\,\text{rad}\). The time for one revolution is the period \(T\). Angular speed is angular displacement divided by time, so \(\omega=\frac{2\pi}{T}\). Multiplying by \(T\) would make angular speed increase with period, which is opposite to the physical meaning. A longer period means slower rotation, not faster rotation.
404. A projectile is fired from level ground with speed \(u\). If \(\theta=0^\circ\) and air resistance is neglected, the same-level time of flight from \(T=\frac{2u\sin\theta}{g}\) becomes:
ⓐ. \(0\)
ⓑ. \(\frac{u}{g}\)
ⓒ. \(\frac{2u}{g}\)
ⓓ. \(\frac{u^2}{g}\)
Correct Answer: \(0\)
Explanation: \( \textbf{Same-level time formula:} \)
\[
T=\frac{2u\sin\theta}{g}
\]
\( \textbf{For horizontal projection from level ground:} \)
\[
\theta=0^\circ
\]
\( \textbf{Use \(\sin0^\circ=0\):} \)
\[
T=\frac{2u(0)}{g}=0
\]
\( \textbf{Interpretation:} \) If the projectile is launched horizontally from ground level and must return to the same ground level immediately, the mathematical flight time is zero. A non-zero flight time for horizontal projection requires the launch point to be above the landing level.
\( \textbf{Final answer:} \) The same-level formula gives \(T=0\).
405. If \(g\) were zero for a projectile launched with speed \(u\) at angle \(\theta\), the path would be:
ⓐ. downward-opening parabola under gravity
ⓑ. straight line along the launch direction
ⓒ. circular arc with changing radius
ⓓ. vertical line through the launch point
Correct Answer: straight line along the launch direction
Explanation: Projectile curvature is caused by the downward acceleration due to gravity. If \(g=0\), there is no acceleration in the vertical direction. The horizontal and vertical components of velocity would both remain constant. A body with constant velocity moves in a straight line. The parabolic trajectory appears only when the vertical component changes with time due to gravity.
406. A vector formula list contains the following claims:
I. \(a_c=\frac{v^2}{r}\)
II. \(a_c=\omega^2r\)
III. \(a_c=v\omega\)
For uniform circular motion, the supported claims are:
ⓐ. I only
ⓑ. I and II only
ⓒ. II and III only
ⓓ. I, II, and III
Correct Answer: I, II, and III
Explanation: Claim I is the standard speed-radius form of centripetal acceleration. Claim II follows by substituting \(v=\omega r\) into \(a_c=\frac{v^2}{r}\). Claim III also follows from \(a_c=\frac{v^2}{r}\) because \(\frac{v}{r}=\omega\). These formulas are equivalent when \(v\), \(\omega\), and \(r\) describe the same circular motion. Choosing among them depends on the quantities given in the problem.
407. A particle moves uniformly in a circle. Over a very small time interval \(\Delta t\), the arc length covered is \(\Delta s\), and the speed is \(v\). In the derivation of centripetal acceleration, the step \(\frac{\Delta s}{\Delta t}=v\) is used because:
ⓐ. speed is the rate of motion along the path
ⓑ. speed is always equal to radius
ⓒ. angular speed is always equal to arc length
ⓓ. acceleration is zero over a small interval
Correct Answer: speed is the rate of motion along the path
Explanation: Speed measures how fast distance along the path is covered. In circular motion, the distance covered over a small interval is the arc length \(\Delta s\). Therefore, \(\frac{\Delta s}{\Delta t}\) represents the speed along the circular path. In uniform circular motion, this speed remains constant. This step connects the geometry of the small arc with the rate of change of velocity.
408. A graph of \(v_y\) against \(t\) for an ideal projectile is a straight line with slope \(-g\). If the line crosses the time axis at \(t=3\,\text{s}\), the initial vertical component \(u_y\) is:
ⓐ. \(3g\)
ⓑ. \(\frac{g}{3}\)
ⓒ. \(6g\)
ⓓ. \(0\)
Correct Answer: \(3g\)
Explanation: \( \textbf{Vertical velocity relation:} \)
\[
v_y=u_y-gt
\]
\( \textbf{Meaning of time-axis crossing:} \) At the crossing, \(v_y=0\).
\( \textbf{Given crossing time:} \)
\[
t=3\,\text{s}
\]
\( \textbf{Apply condition:} \)
\[
0=u_y-g(3\,\text{s})
\]
\( \textbf{Solve:} \)
\[
u_y=3g
\]
\( \textbf{Interpretation:} \) The crossing represents the highest point, where the vertical velocity becomes zero.
\( \textbf{Final answer:} \) The initial vertical component is \(3g\).
409. A table compares formula conditions in projectile motion.
| Formula | Condition for direct use |
|---|
| P. \(T=\frac{2u\sin\theta}{g}\) | same launch and landing level |
| Q. \(R=\frac{u^2\sin2\theta}{g}\) | same launch and landing level |
| R. \(H=\frac{u^2\sin^2\theta}{2g}\) | height above point of projection |
| S. \(x=u_xt\) | only when \(g=0\) |
The entry that needs correction is:
ⓐ. P
ⓑ. Q
ⓒ. R
ⓓ. S
Correct Answer: S
Explanation: The time-of-flight formula in P is directly for same-level landing. The range formula in Q also assumes same launch and landing level. The height formula in R gives maximum height measured above the projection point. Entry S is wrong because \(x=u_xt\) follows from \(a_x=0\), not from \(g=0\). In ideal projectile motion near Earth, \(g\) is not zero, but the horizontal acceleration is zero, so \(x=u_xt\) is still valid.
410. A body moves in a circle of radius \(r\) with speed \(v\). If \(r\) becomes very large while \(v\) remains finite, the centripetal acceleration tends toward:
ⓐ. \(0\)
ⓑ. \(v^2\)
ⓒ. \(r^2\)
ⓓ. infinity
Correct Answer: \(0\)
Explanation: The centripetal acceleration is
\[
a_c=\frac{v^2}{r}
\]
If \(v\) remains finite, then \(v^2\) remains finite. As \(r\) becomes very large, the denominator grows without bound. Therefore, \(\frac{v^2}{r}\) tends toward \(0\). Physically, a path with extremely large radius is nearly straight over a small region, so very little direction-changing acceleration is needed.
411. In non-uniform circular motion, the total acceleration is not generally directed toward the centre because:
ⓐ. the radial component disappears whenever speed changes
ⓑ. acceleration must always be zero in circular motion
ⓒ. a tangential component appears when speed changes
ⓓ. velocity becomes radial in non-uniform circular motion
Correct Answer: a tangential component appears when speed changes
Explanation: Circular motion always needs a radial component of acceleration when the direction of velocity changes. If the speed also changes, a tangential component appears along the tangent. The total acceleration is the vector sum of radial and tangential components. Because of the tangential part, the total acceleration is not purely centreward in general. Uniform circular motion is the special case where the tangential component is zero.
412. A particle moves in a circle with changing speed. At a certain instant, its speed is increasing. Which statement correctly describes the acceleration at that instant?
ⓐ. Only centripetal acceleration is present, directed along the tangent
ⓑ. Only tangential acceleration is present, directed toward the centre
ⓒ. Centripetal acceleration is toward the centre, and tangential acceleration is along the direction of motion
ⓓ. Centripetal acceleration is away from the centre, and tangential acceleration is zero
Correct Answer: Centripetal acceleration is toward the centre, and tangential acceleration is along the direction of motion
Explanation: When a particle moves along a circular path, its velocity direction changes continuously.
Because the velocity direction changes, there is a radial or centripetal acceleration directed toward the centre of the circle.
Here, the speed is also increasing. A change in speed produces tangential acceleration.
Since the speed is increasing, the tangential acceleration is along the direction of motion.
So, two acceleration effects are present:
\[
\text{radial acceleration toward the centre}
\]
and
\[
\text{tangential acceleration along the motion}
\]
The radial part is linked with change in direction of velocity. The tangential part is linked with change in speed.
\( \textbf{Final answer:} \) Centripetal acceleration is toward the centre, and tangential acceleration is along the direction of motion.
413. A mixed error record says: “A projectile at the highest point has zero acceleration, and a body in uniform circular motion has zero acceleration because its speed is constant.” The best diagnosis is:
ⓐ. both confuse speed with acceleration
ⓑ. both statements are correct
ⓒ. only the projectile statement is correct
ⓓ. only the circular-motion statement is correct
Correct Answer: both confuse speed with acceleration
Explanation: At the highest point of projectile motion, the vertical component of velocity is zero, but acceleration due to gravity still acts downward. In uniform circular motion, speed is constant, but velocity direction changes continuously. That change in velocity direction requires centripetal acceleration toward the centre. In both cases, a condition about velocity is being mistaken for a condition about acceleration. Acceleration depends on change of velocity, not only on whether a speed or one component is momentarily zero.
414. A projectile has trajectory \(y=3x-\frac{x^2}{20}\), where \(x\) and \(y\) are in \(\text{m}\). Taking \(g=10\,\text{m s}^{-2}\), the speed of the projectile when \(x=20\,\text{m}\) is:
ⓐ. \(10\,\text{m s}^{-1}\)
ⓑ. \(10\sqrt{2}\,\text{m s}^{-1}\)
ⓒ. \(10\sqrt{3}\,\text{m s}^{-1}\)
ⓓ. \(20\,\text{m s}^{-1}\)
Correct Answer: \(10\sqrt{2}\,\text{m s}^{-1}\)
Explanation: \( \textbf{Given trajectory:} \)
\[
y=3x-\frac{x^2}{20}
\]
\( \textbf{Compare with projectile trajectory:} \)
\[
y=x\tan\theta-\frac{gx^2}{2u_x^2}
\]
\( \textbf{From the coefficient of \(x\):} \)
\[
\tan\theta=3
\]
\( \textbf{From the coefficient of \(x^2\):} \)
\[
\frac{g}{2u_x^2}=\frac{1}{20}
\]
\( \textbf{Substitute \(g=10\,\text{m s}^{-2}\):} \)
\[
\frac{10}{2u_x^2}=\frac{1}{20}
\]
\[
u_x^2=100
\]
\[
u_x=10\,\text{m s}^{-1}
\]
\( \textbf{Initial vertical component:} \)
\[
u_y=u_x\tan\theta=10(3)=30\,\text{m s}^{-1}
\]
\( \textbf{Time when \(x=20\,\text{m}\):} \)
\[
t=\frac{x}{u_x}=\frac{20\,\text{m}}{10\,\text{m s}^{-1}}=2\,\text{s}
\]
\( \textbf{Vertical velocity then:} \)
\[
v_y=u_y-gt=30\,\text{m s}^{-1}-(10\,\text{m s}^{-2})(2\,\text{s})=10\,\text{m s}^{-1}
\]
\( \textbf{Speed:} \)
\[
v=\sqrt{v_x^2+v_y^2}=\sqrt{10^2+10^2}\,\text{m s}^{-1}=10\sqrt{2}\,\text{m s}^{-1}
\]
\( \textbf{Final answer:} \) The speed is \(10\sqrt{2}\,\text{m s}^{-1}\).
415. A projectile is launched from a cliff and lands \(80\,\text{m}\) horizontally away after \(4\,\text{s}\). The landing point is \(60\,\text{m}\) below the launch point. Taking \(g=10\,\text{m s}^{-2}\), the initial velocity vector is:
ⓐ. \(20\,\text{m s}^{-1}\hat{i}+5\,\text{m s}^{-1}\hat{j}\)
ⓑ. \(5\,\text{m s}^{-1}\hat{i}+20\,\text{m s}^{-1}\hat{j}\)
ⓒ. \(20\,\text{m s}^{-1}\hat{i}-5\,\text{m s}^{-1}\hat{j}\)
ⓓ. \(80\,\text{m s}^{-1}\hat{i}+60\,\text{m s}^{-1}\hat{j}\)
Correct Answer: \(20\,\text{m s}^{-1}\hat{i}+5\,\text{m s}^{-1}\hat{j}\)
Explanation: \( \textbf{Horizontal displacement:} \) \(x=80\,\text{m}\).
\( \textbf{Time of flight:} \) \(t=4\,\text{s}\).
\( \textbf{Horizontal motion:} \)
\[
x=u_xt
\]
\[
u_x=\frac{x}{t}=\frac{80\,\text{m}}{4\,\text{s}}=20\,\text{m s}^{-1}
\]
\( \textbf{Vertical displacement:} \) \(y=-60\,\text{m}\).
\( \textbf{Vertical motion equation:} \)
\[
y=u_yt-\frac{1}{2}gt^2
\]
\( \textbf{Substitution:} \)
\[
-60=(u_y)(4)-\frac{1}{2}(10)(4)^2
\]
\[
-60=4u_y-80
\]
\( \textbf{Solve for \(u_y\):} \)
\[
4u_y=20
\]
\[
u_y=5\,\text{m s}^{-1}
\]
\( \textbf{Initial velocity vector:} \)
\[
\vec{u}=20\,\text{m s}^{-1}\hat{i}+5\,\text{m s}^{-1}\hat{j}
\]
\( \textbf{Final answer:} \) \(\vec{u}=20\,\text{m s}^{-1}\hat{i}+5\,\text{m s}^{-1}\hat{j}\).
416. A projectile is launched with \(u_x=15\,\text{m s}^{-1}\) and \(u_y=40\,\text{m s}^{-1}\). Taking \(g=10\,\text{m s}^{-2}\), the horizontal separation between the two points where it is at height \(30\,\text{m}\) above the launch point is:
ⓐ. \(15\sqrt{10}\,\text{m}\)
ⓑ. \(60\,\text{m}\)
ⓒ. \(120\,\text{m}\)
ⓓ. \(30\sqrt{10}\,\text{m}\)
Correct Answer: \(30\sqrt{10}\,\text{m}\)
Explanation: \( \textbf{Vertical position equation:} \)
\[
y=u_yt-\frac{1}{2}gt^2
\]
\( \textbf{Given height:} \)
\[
y=30\,\text{m}
\]
\( \textbf{Substitution:} \)
\[
30=40t-\frac{1}{2}(10)t^2
\]
\[
30=40t-5t^2
\]
\( \textbf{Rearrange:} \)
\[
5t^2-40t+30=0
\]
\( \textbf{Divide by \(5\):} \)
\[
t^2-8t+6=0
\]
\( \textbf{Solve:} \)
\[
t=\frac{8\pm\sqrt{64-24}}{2}
\]
\[
t=\frac{8\pm\sqrt{40}}{2}=4\pm\sqrt{10}
\]
\( \textbf{Time separation:} \)
\[
\Delta t=(4+\sqrt{10})-(4-\sqrt{10})=2\sqrt{10}\,\text{s}
\]
\( \textbf{Horizontal separation:} \)
\[
\Delta x=u_x\Delta t
\]
\[
\Delta x=(15\,\text{m s}^{-1})(2\sqrt{10}\,\text{s})=30\sqrt{10}\,\text{m}
\]
\( \textbf{Final answer:} \) The two points are separated horizontally by \(30\sqrt{10}\,\text{m}\).
417. For a projectile launched from level ground with speed \(u\), the horizontal range is \(4\) times the maximum height. The angle of projection is:
ⓐ. \(30^\circ\)
ⓑ. \(45^\circ\)
ⓒ. \(60^\circ\)
ⓓ. \(75^\circ\)
Correct Answer: \(45^\circ\)
Explanation: \( \textbf{Range formula:} \)
\[
R=\frac{u^2\sin2\theta}{g}
\]
\( \textbf{Maximum height formula:} \)
\[
H=\frac{u^2\sin^2\theta}{2g}
\]
\( \textbf{Given condition:} \)
\[
R=4H
\]
\( \textbf{Form the ratio:} \)
\[
\frac{R}{H}=\frac{\frac{u^2\sin2\theta}{g}}{\frac{u^2\sin^2\theta}{2g}}
\]
\( \textbf{Cancel common factors:} \)
\[
\frac{R}{H}=\frac{2\sin2\theta}{\sin^2\theta}
\]
\( \textbf{Use \(\sin2\theta=2\sin\theta\cos\theta\):} \)
\[
\frac{R}{H}=4\cot\theta
\]
\( \textbf{Apply \(R=4H\):} \)
\[
4=4\cot\theta
\]
\[
\cot\theta=1
\]
\[
\theta=45^\circ
\]
\( \textbf{Final answer:} \) The angle of projection is \(45^\circ\).
418. A particle in uniform circular motion of radius \(4\,\text{m}\) has average acceleration magnitude \(\frac{8}{\pi}\,\text{m s}^{-2}\) over half a revolution. Its speed is:
ⓐ. \(2\,\text{m s}^{-1}\)
ⓑ. \(4\,\text{m s}^{-1}\)
ⓒ. \(8\,\text{m s}^{-1}\)
ⓓ. \(16\,\text{m s}^{-1}\)
Correct Answer: \(4\,\text{m s}^{-1}\)
Explanation: \( \textbf{For half revolution:} \) The velocity reverses direction.
\[
|\Delta\vec{v}|=2v
\]
\( \textbf{Time for half revolution:} \)
\[
\Delta t=\frac{T}{2}
\]
\( \textbf{Average acceleration magnitude:} \)
\[
a_{\text{avg}}=\frac{|\Delta\vec{v}|}{\Delta t}=\frac{2v}{T/2}=\frac{4v}{T}
\]
\( \textbf{Period in terms of speed:} \)
\[
T=\frac{2\pi r}{v}
\]
\( \textbf{Substitute in average acceleration:} \)
\[
a_{\text{avg}}=\frac{4v}{2\pi r/v}
\]
\[
a_{\text{avg}}=\frac{2v^2}{\pi r}
\]
\( \textbf{Given \(a_{\text{avg}}=\frac{8}{\pi}\,\text{m s}^{-2}\) and \(r=4\,\text{m}\):} \)
\[
\frac{8}{\pi}=\frac{2v^2}{\pi(4)}
\]
\[
\frac{8}{\pi}=\frac{v^2}{2\pi}
\]
\( \textbf{Solve:} \)
\[
16=v^2
\]
\[
v=4\,\text{m s}^{-1}
\]
\( \textbf{Final answer:} \) The speed is \(4\,\text{m s}^{-1}\).
419. A wheel of radius \(0.5\,\text{m}\) rotates uniformly at \(120\,\text{rev min}^{-1}\). The centripetal acceleration of a point on its rim is:
ⓐ. \(2\pi^2\,\text{m s}^{-2}\)
ⓑ. \(4\pi^2\,\text{m s}^{-2}\)
ⓒ. \(8\pi^2\,\text{m s}^{-2}\)
ⓓ. \(16\pi^2\,\text{m s}^{-2}\)
Correct Answer: \(8\pi^2\,\text{m s}^{-2}\)
Explanation: \( \textbf{Given:} \) \(r=0.5\,\text{m}\) and rotation rate \(=120\,\text{rev min}^{-1}\).
\( \textbf{Convert frequency:} \)
\[
120\,\text{rev min}^{-1}=\frac{120}{60}\,\text{rev s}^{-1}
\]
\[
f=2\,\text{s}^{-1}
\]
\( \textbf{Centripetal acceleration in frequency form:} \)
\[
a_c=4\pi^2f^2r
\]
\( \textbf{Substitution:} \)
\[
a_c=4\pi^2(2)^2(0.5)
\]
\( \textbf{Calculation:} \)
\[
a_c=4\pi^2(4)(0.5)=8\pi^2\,\text{m s}^{-2}
\]
\( \textbf{Final answer:} \) The centripetal acceleration is \(8\pi^2\,\text{m s}^{-2}\).
420. A particle moves in a circle and its speed is increasing with time. Which comparison with uniform circular motion is correct?
ⓐ. In both cases, speed changes continuously
ⓑ. In uniform circular motion, velocity direction changes; when speed also changes, tangential acceleration is also present
ⓒ. In uniform circular motion, acceleration is zero because speed is constant
ⓓ. When speed changes in circular motion, centripetal acceleration disappears
Correct Answer: In uniform circular motion, velocity direction changes; when speed also changes, tangential acceleration is also present
Explanation: In uniform circular motion, the speed remains constant, but the velocity is not constant because its direction changes continuously.
This change in velocity direction gives centripetal acceleration toward the centre.
When the speed of circular motion also changes, there is an additional tangential acceleration.
The tangential acceleration is responsible for changing the speed.
The centripetal acceleration is still present because the direction of velocity is still changing along the circular path.
So, changing speed does not remove centripetal acceleration. It adds a tangential part to the acceleration description.
\( \textbf{Final answer:} \) In uniform circular motion, velocity direction changes; when speed also changes, tangential acceleration is also present.
