301. The trajectory of a projectile is represented by \(y=2x-\frac{x^2}{40}\), where \(x\) and \(y\) are in \(\text{m}\). The maximum height reached is:
ⓐ. \(20\,\text{m}\)
ⓑ. \(60\,\text{m}\)
ⓒ. \(40\,\text{m}\)
ⓓ. \(80\,\text{m}\)
Correct Answer: \(40\,\text{m}\)
Explanation: \( \textbf{Given trajectory:} \)
\[
y=2x-\frac{x^2}{40}
\]
\( \textbf{At maximum height:} \) The vertex of the parabola gives the maximum value of \(y\).
\( \textbf{Write in standard form:} \) Here \(y=ax-bx^2\), with \(a=2\) and \(b=\frac{1}{40}\).
\( \textbf{x-coordinate of vertex:} \)
\[
x=\frac{a}{2b}
\]
\( \textbf{Substitution:} \)
\[
x=\frac{2}{2\left(\frac{1}{40}\right)}=40\,\text{m}
\]
\( \textbf{Find maximum \(y\):} \)
\[
y_{\max}=2(40)-\frac{(40)^2}{40}
\]
\[
y_{\max}=80-40=40\,\text{m}
\]
\( \textbf{Final answer:} \) The maximum height is \(40\,\text{m}\).
302. A projectile is observed at two points at the same height, one while going up and the other while coming down. Neglecting air resistance, the velocity components at these two points have:
ⓐ. the same \(v_x\) and opposite \(v_y\) values
ⓑ. opposite \(v_x\) and the same \(v_y\) values
ⓒ. both \(v_x\) and \(v_y\) equal to zero
ⓓ. both \(v_x\) and \(v_y\) reversed
Correct Answer: the same \(v_x\) and opposite \(v_y\) values
Explanation: In ideal projectile motion, horizontal acceleration is zero, so \(v_x\) remains constant throughout the flight. At two points at the same height, the vertical speed magnitudes are equal if air resistance is neglected. On the upward part, \(v_y\) is positive when upward is chosen positive. On the downward part at the same height, \(v_y\) has the same magnitude but negative sign. The full velocity directions are different, but their horizontal components match.
303. If the acceleration due to gravity becomes \(\frac{g}{4}\) while the same projectile is launched with the same speed and angle on level ground, its horizontal range becomes:
ⓐ. \(\frac{R}{4}\)
ⓑ. \(\frac{R}{2}\)
ⓒ. \(2R\)
ⓓ. \(4R\)
Correct Answer: \(4R\)
Explanation: The same-level range of a projectile is
\[
R=\frac{u^2\sin2\theta}{g}
\]
For the same \(u\) and \(\theta\), the numerator remains unchanged. The range is inversely proportional to \(g\). If \(g\) is replaced by \(\frac{g}{4}\), then the denominator becomes four times smaller. Hence the range becomes four times larger, while the launch speed and direction remain unchanged.
304. Uniform circular motion refers to motion in which a body:
ⓐ. moves in a straight line with constant velocity
ⓑ. moves in a circle with zero acceleration
ⓒ. moves in a circle with constantly increasing speed only
ⓓ. moves in a circle with constant speed
Correct Answer: moves in a circle with constant speed
Explanation: In uniform circular motion, the body follows a circular path with constant speed. The magnitude of velocity remains constant, but the direction of velocity changes continuously. Since velocity is a vector, this change in direction means the velocity is not constant. Therefore, acceleration is present even though the speed does not change. The word “uniform” refers to constant speed, not to constant velocity.
305. For a body moving uniformly in a circular path, the instantaneous velocity is directed:
ⓐ. along the radius toward the centre
ⓑ. tangent to the circle at that point
ⓒ. away from the centre along the radius
ⓓ. opposite to the acceleration at every point
Correct Answer: tangent to the circle at that point
Explanation: Instantaneous velocity is always along the tangent to the path. For circular motion, the tangent at any point is perpendicular to the radius at that point. The body is not moving along the radius; it is moving around the circle. The acceleration in uniform circular motion is directed toward the centre, while velocity is tangential. This perpendicular relation is why the speed can remain constant while direction keeps changing.
306. Use the arrangement described below: a particle moves anticlockwise in a circle. At the rightmost point of the circle, the centre lies to the particle’s left.
At that instant, the particle’s velocity is directed:
ⓐ. downward along the tangent
ⓑ. leftward toward the centre
ⓒ. rightward away from the centre
ⓓ. upward along the tangent
Correct Answer: upward along the tangent
Explanation: At the rightmost point of a circle, the radius from the centre to the particle is horizontal. The velocity must be tangential, so it must be vertical at that point. Since the motion is anticlockwise, the particle moves upward as it passes through the rightmost point. The centreward direction is leftward, but that is the direction of centripetal acceleration, not velocity. Tangent direction and centreward direction should be kept separate.
307. The centripetal acceleration of a body in uniform circular motion is:
ⓐ. \(a_c=vr\), directed along the tangent
ⓑ. \(a_c=\frac{r}{v^2}\), directed away from the centre
ⓒ. \(a_c=v^2r\), directed opposite to velocity
ⓓ. \(a_c=\frac{v^2}{r}\), directed toward the centre
Correct Answer: \(a_c=\frac{v^2}{r}\), directed toward the centre
Explanation: In uniform circular motion, acceleration is caused by the continuous change in direction of velocity. Its magnitude is \(a_c=\frac{v^2}{r}\), where \(v\) is the speed and \(r\) is the radius of the circle. This acceleration is directed toward the centre of the circular path. It is not tangential in uniform circular motion, because tangential acceleration would change the speed. The centreward direction is essential for keeping the body on the circular path.
308. A stone tied to a string moves uniformly in a circle of radius \(2\,\text{m}\) with speed \(6\,\text{m s}^{-1}\). Its centripetal acceleration is:
ⓐ. \(3\,\text{m s}^{-2}\)
ⓑ. \(12\,\text{m s}^{-2}\)
ⓒ. \(18\,\text{m s}^{-2}\)
ⓓ. \(72\,\text{m s}^{-2}\)
Correct Answer: \(18\,\text{m s}^{-2}\)
Explanation: \( \textbf{Given:} \) \(v=6\,\text{m s}^{-1}\) and \(r=2\,\text{m}\).
\( \textbf{Centripetal acceleration relation:} \)
\[
a_c=\frac{v^2}{r}
\]
\( \textbf{Substitution:} \)
\[
a_c=\frac{(6\,\text{m s}^{-1})^2}{2\,\text{m}}
\]
\( \textbf{Square the speed:} \)
\[
(6\,\text{m s}^{-1})^2=36\,\text{m}^2\text{s}^{-2}
\]
\( \textbf{Divide by radius:} \)
\[
a_c=\frac{36\,\text{m}^2\text{s}^{-2}}{2\,\text{m}}=18\,\text{m s}^{-2}
\]
\( \textbf{Direction note:} \) This acceleration is directed toward the centre of the circle.
\( \textbf{Final answer:} \) The centripetal acceleration is \(18\,\text{m s}^{-2}\).
309. The angular speed \(\omega\), period \(T\), and frequency \(f\) in uniform circular motion are related by:
ⓐ. \(\omega=2\pi T=\frac{2\pi}{f}\)
ⓑ. \(\omega=\frac{T}{2\pi}=\frac{f}{2\pi}\)
ⓒ. \(\omega=T+f\)
ⓓ. \(\omega=\frac{2\pi}{T}=2\pi f\)
Correct Answer: \(\omega=\frac{2\pi}{T}=2\pi f\)
Explanation: One complete revolution corresponds to angular displacement \(2\pi\,\text{rad}\). If the time for one revolution is \(T\), then angular speed is \(\omega=\frac{2\pi}{T}\). Frequency is the number of revolutions per second, so \(f=\frac{1}{T}\). Substituting \(f\) gives \(\omega=2\pi f\). The unit of angular speed is commonly written as \(\text{rad s}^{-1}\).
310. A wheel rotates uniformly at \(5\,\text{rev s}^{-1}\). The angular speed of the wheel is:
ⓐ. \(5\pi\,\text{rad s}^{-1}\)
ⓑ. \(10\pi\,\text{rad s}^{-1}\)
ⓒ. \(\frac{5}{2\pi}\,\text{rad s}^{-1}\)
ⓓ. \(25\pi\,\text{rad s}^{-1}\)
Correct Answer: \(10\pi\,\text{rad s}^{-1}\)
Explanation: \( \textbf{Given frequency:} \) \(f=5\,\text{rev s}^{-1}\).
\( \textbf{One revolution:} \) \(1\,\text{rev}=2\pi\,\text{rad}\).
\( \textbf{Angular speed relation:} \)
\[
\omega=2\pi f
\]
\( \textbf{Substitution:} \)
\[
\omega=2\pi(5\,\text{s}^{-1})
\]
\( \textbf{Calculation:} \)
\[
\omega=10\pi\,\text{rad s}^{-1}
\]
\( \textbf{Unit note:} \) The factor \(2\pi\) converts revolutions into radians.
\( \textbf{Final answer:} \) The angular speed is \(10\pi\,\text{rad s}^{-1}\).
311. For a particle moving uniformly in a circle of radius \(r\) with angular speed \(\omega\), the linear speed is:
ⓐ. \(v=\omega r\)
ⓑ. \(v=\frac{\omega}{r}\)
ⓒ. \(v=\frac{r}{\omega}\)
ⓓ. \(v=\omega+r\)
Correct Answer: \(v=\omega r\)
Explanation: Linear speed is the rate at which arc length is covered. For circular motion, arc length is related to angular displacement by \(s=r\theta\). Differentiating with respect to time gives \(v=r\frac{d\theta}{dt}\). Since \(\frac{d\theta}{dt}=\omega\), the relation becomes \(v=\omega r\). For the same angular speed, a larger radius gives a larger linear speed.
312. A particle moves in a circle of radius \(0.5\,\text{m}\) with angular speed \(8\,\text{rad s}^{-1}\). Its linear 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: \( \textbf{Given:} \) \(r=0.5\,\text{m}\) and \(\omega=8\,\text{rad s}^{-1}\).
\( \textbf{Linear speed relation:} \)
\[
v=\omega r
\]
\( \textbf{Substitution:} \)
\[
v=(8\,\text{rad s}^{-1})(0.5\,\text{m})
\]
\( \textbf{Calculation:} \)
\[
v=4\,\text{m s}^{-1}
\]
\( \textbf{Unit reading:} \) The radian is dimensionless, so the final unit becomes \(\text{m s}^{-1}\).
\( \textbf{Final answer:} \) The linear speed is \(4\,\text{m s}^{-1}\).
313. Study the table for uniform circular motion.
| Row | Quantity | Statement |
|---|
| P | Speed | constant in uniform circular motion |
| Q | Velocity | changes because its direction changes |
| R | Acceleration | directed toward the centre |
| S | Acceleration | zero because speed is constant |
The row that needs correction is:
ⓐ. Row P
ⓑ. Row Q
ⓒ. Row R
ⓓ. Row S
Correct Answer: Row S
Explanation: Row P is correct because uniform circular motion has constant speed. Row Q is correct because velocity is a vector and its direction changes continuously. Row R is correct because centripetal acceleration is directed toward the centre. Row S is wrong because constant speed does not mean zero acceleration when direction changes. In circular motion, acceleration exists because the velocity vector keeps turning.
314. Assertion: A body in uniform circular motion has acceleration even though its speed is constant.
Reason: Its velocity changes continuously due to change in direction.
ⓐ. Both Assertion and Reason are true, and Reason explains Assertion
ⓑ. Both Assertion and Reason are true, but Reason does not explain 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 explains Assertion
Explanation: The assertion is true because acceleration is related to change in velocity, not only change in speed. In uniform circular motion, the magnitude of velocity remains constant. However, the direction of velocity changes at every point of the circular path. The reason correctly explains that this directional change produces acceleration. The acceleration is centreward and is called centripetal acceleration.
315. If the speed of a body in uniform circular motion is doubled while the radius remains the same, its centripetal acceleration becomes:
ⓐ. one-half as large
ⓑ. twice as large
ⓒ. four times as large
ⓓ. unchanged in magnitude
Correct Answer: four times as large
Explanation: The centripetal acceleration is
\[
a_c=\frac{v^2}{r}
\]
For fixed radius \(r\), \(a_c\propto v^2\). If the speed becomes \(2v\), then the new acceleration is
\[
a_c'=\frac{(2v)^2}{r}
\]
\[
a_c'=\frac{4v^2}{r}=4a_c
\]
The square on \(v\) makes the acceleration grow faster than the speed itself.
\( \textbf{Final answer:} \) The centripetal acceleration becomes four times.
316. If the radius of circular motion is doubled while the speed remains the same, the centripetal acceleration becomes:
ⓐ. half
ⓑ. double
ⓒ. four times
ⓓ. unchanged
Correct Answer: half
Explanation: The centripetal acceleration in uniform circular motion is
\[
a_c=\frac{v^2}{r}
\]
For constant speed \(v\), the quantity \(v^2\) remains unchanged. Therefore, \(a_c\) is inversely proportional to \(r\). If the radius changes from \(r\) to \(2r\), then the new acceleration is
\[
a_c'=\frac{v^2}{2r}
\]
\[
a_c'=\frac{1}{2}\frac{v^2}{r}=\frac{a_c}{2}
\]
\( \textbf{Final answer:} \) The centripetal acceleration becomes half.
317. The centripetal acceleration can also be written in terms of angular speed as:
ⓐ. \(a_c=\frac{\omega}{r^2}\)
ⓑ. \(a_c=\omega r^2\)
ⓒ. \(a_c=\frac{r}{\omega^2}\)
ⓓ. \(a_c=\omega^2r\)
Correct Answer: \(a_c=\omega^2r\)
Explanation: The centripetal acceleration is \(a_c=\frac{v^2}{r}\). In circular motion, linear speed and angular speed are related by \(v=\omega r\). Substituting this into the centripetal acceleration formula gives \(a_c=\frac{(\omega r)^2}{r}\). This simplifies to \(a_c=\omega^2r\). The relation is useful when the motion is described by angular speed rather than linear speed.
318. A particle moves in a circle of radius \(3\,\text{m}\) with angular speed \(4\,\text{rad s}^{-1}\). Its centripetal acceleration is:
ⓐ. \(12\,\text{m s}^{-2}\)
ⓑ. \(24\,\text{m s}^{-2}\)
ⓒ. \(48\,\text{m s}^{-2}\)
ⓓ. \(64\,\text{m s}^{-2}\)
Correct Answer: \(48\,\text{m s}^{-2}\)
Explanation: \( \textbf{Given:} \) \(r=3\,\text{m}\) and \(\omega=4\,\text{rad s}^{-1}\).
\( \textbf{Required:} \) Centripetal acceleration \(a_c\).
\( \textbf{Formula in angular form:} \)
\[
a_c=\omega^2r
\]
\( \textbf{Substitution:} \)
\[
a_c=(4\,\text{rad s}^{-1})^2(3\,\text{m})
\]
\( \textbf{Square angular speed:} \)
\[
(4)^2=16
\]
\( \textbf{Calculation:} \)
\[
a_c=16(3)\,\text{m s}^{-2}=48\,\text{m s}^{-2}
\]
\( \textbf{Direction note:} \) This acceleration is directed toward the centre of the circular path.
\( \textbf{Final answer:} \) The centripetal acceleration is \(48\,\text{m s}^{-2}\).
319. A graph is described below.
For a body moving in circles of different radii at the same speed \(v\), a graph is plotted between centripetal acceleration \(a_c\) on the vertical axis and \(\frac{1}{r}\) on the horizontal axis.
The graph should be:
ⓐ. a straight line through the origin with slope \(v\)
ⓑ. a curve through the origin proportional to \(r^2\)
ⓒ. a horizontal line through \(a_c=v^2\)
ⓓ. a straight line through the origin, slope \(v^2\)
Correct Answer: a straight line through the origin, slope \(v^2\)
Explanation: The centripetal acceleration is \(a_c=\frac{v^2}{r}\). If \(v\) is constant, then \(v^2\) is constant. This relation can be written as \(a_c=v^2\left(\frac{1}{r}\right)\). Therefore, a graph of \(a_c\) against \(\frac{1}{r}\) is a straight line through the origin. Its slope is the coefficient of \(\frac{1}{r}\), which is \(v^2\).
320. In uniform circular motion, the velocity and centripetal acceleration at any point are:
ⓐ. parallel in the same direction
ⓑ. parallel in opposite directions
ⓒ. perpendicular to each other
ⓓ. both zero
Correct Answer: perpendicular to each other
Explanation: The velocity of a body in circular motion is tangential to the path. The centripetal acceleration is directed along the radius toward the centre. A tangent to a circle is perpendicular to the radius at the point of contact. Therefore, velocity and centripetal acceleration are perpendicular at every point in uniform circular motion. This perpendicular acceleration changes the direction of velocity without changing its magnitude.
