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401. A car of mass 1000 kg moves with a constant speed of 20 m/s in a circle of radius 50 m. What is the required centripetal force?
ⓐ. 200 N
ⓑ. 4000 N
ⓒ. 8000 N
ⓓ. 10,000 N
Correct Answer: 8000 N
Explanation: Centripetal force is given by $F_c = \frac{mv^2}{r}$. Substituting values: $F_c = \frac{1000 \times 400}{50} = \frac{400,000}{50} = 8000 \, N$. This inward force could be supplied by friction, tension, or another real physical force.
402. A stone tied to a string of length 2 m is whirled in a horizontal circle at a speed of 6 m/s. What is the centripetal acceleration?
ⓐ. 9 m/s²
ⓑ. 12 m/s²
ⓒ. 15 m/s²
ⓓ. 18 m/s²
Correct Answer: 18 m/s²
Explanation: Centripetal acceleration is $a_c = \frac{v^2}{r} = \frac{36}{2} = 18 \, m/s^2$. The acceleration is directed towards the center, continuously changing the direction of the velocity vector.
403. Which of the following is true about centripetal force?
ⓐ. It acts tangential to the circular path
ⓑ. It is an outward fictitious force
ⓒ. It acts towards the center and provides the necessary acceleration
ⓓ. It is zero in uniform circular motion
Correct Answer: It acts towards the center and provides the necessary acceleration
Explanation: Centripetal force is the real force that acts inward towards the center of the circle. It maintains circular motion by continuously changing the direction of velocity.
404. A satellite of mass m revolves around Earth in a circular orbit of radius r with speed v. The centripetal force is provided by gravity. Which equation represents this balance?
ⓐ. $\frac{mv^2}{r} = \frac{GMm}{r^2}$
ⓑ. $\frac{mv}{r} = \frac{GM}{r^2}$
ⓒ. $\frac{m}{r^2} = \frac{GM}{v^2}$
ⓓ. $v = \sqrt{GM}$
Correct Answer: $\frac{mv^2}{r} = \frac{GMm}{r^2}$
Explanation: For orbital motion, gravitational pull $F = \frac{GMm}{r^2}$ supplies the centripetal force required to keep the satellite in orbit. Hence, centripetal force and gravitational force are equal in magnitude.
405. A 0.5 kg ball moves in a circle of radius 2 m with angular speed 4 rad/s. What is the centripetal force?
ⓐ. 2 N
ⓑ. 4.4 N
ⓒ. 8 N
ⓓ. 16 N
Correct Answer: 16 N
Explanation: Centripetal force $F = m\omega^2 r = 0.5 \times (16) \times 2 = 16 \, N$. Correction: actual calculation gives $F = 0.5 \times 16 \times 2 = 16 \, N$,
406. Which situation does not involve centripetal force?
ⓐ. A planet orbiting the sun
ⓑ. A car turning on a curved road
ⓒ. A stone resting on the ground
ⓓ. An electron moving around the nucleus
Correct Answer: A stone resting on the ground
Explanation: Centripetal force exists only when an object moves in a circular path. A stone at rest experiences only gravitational and normal forces, not centripetal force.
407. Why is centripetal force called “not a new force”?
ⓐ. Because it is imaginary
ⓑ. Because it is just the name given to the resultant of actual forces causing circular motion
ⓒ. Because it always equals zero
ⓓ. Because it has no effect on velocity
Correct Answer: Because it is just the name given to the resultant of actual forces causing circular motion
Explanation: Centripetal force is not an independent force. It is the net inward force (e.g., tension, gravity, friction) that provides the necessary centripetal acceleration.
408. A bike turns a circular path of radius 25 m at speed 10 m/s. What is its centripetal acceleration?
ⓐ. 2 m/s²
ⓑ. 3 m/s²
ⓒ. 4 m/s²
ⓓ. 5 m/s²
Correct Answer: 4 m/s²
Explanation: $a_c = \frac{v^2}{r} = \frac{100}{25} = 4 \, m/s^2$. This inward acceleration must be supplied by static friction between tires and road.
409. If the string breaks while whirling a stone in a circle, what path does the stone follow?
ⓐ. It continues in the circular path
ⓑ. It moves radially inward
ⓒ. It moves radially outward
ⓓ. It moves tangentially to the circle at that instant
Correct Answer: It moves tangentially to the circle at that instant
Explanation: Once centripetal force is removed, inertia carries the stone along a straight line tangential to the circle, because no inward force is acting anymore.
410. A coin of mass 0.05 kg lies on a turntable rotating at 2 revolutions per second. The radius is 0.1 m. What is the minimum friction force needed to keep the coin from sliding?
ⓐ. 0.08 N
ⓑ. 0.16 N
ⓒ. 0.20 N
ⓓ. 0.25 N
Correct Answer: 0.16 N
Explanation: Angular velocity $\omega = 2 \times 2\pi = 4\pi \, rad/s$. Centripetal force = $m\omega^2 r = 0.05 \times (4\pi)^2 \times 0.1$. = $0.05 \times 157.9 \times 0.1 \approx 0.16 \, N$. Static friction provides this inward force.
411. A body completes 40 revolutions in 20 seconds. What is its frequency?
ⓐ. 1 Hz
ⓑ. 2 Hz
ⓒ. 3 Hz
ⓓ. 4 Hz
Correct Answer: 2 Hz
Explanation: Frequency is the number of revolutions per second. Here, $f = \frac{40}{20} = 2 \,\text{Hz}$. This means the object completes 2 revolutions every second.
412. The time taken for one complete revolution in circular motion is called:
ⓐ. Frequency
ⓑ. Period
ⓒ. Angular velocity
ⓓ. Linear speed
Correct Answer: Period
Explanation: The period $T$ is the reciprocal of frequency, representing the time to complete one revolution. If $f = 2 \,\text{Hz}$, then $T = 1/f = 0.5 \,s$.
413. If the frequency of a wheel is 10 Hz, what is its period?
ⓐ. 0.01 s
ⓑ. 0.05 s
ⓒ. 0.1 s
ⓓ. 1 s
Correct Answer: 0.1 s
Explanation: Period $T = 1/f = 1/10 = 0.1 \,s$. Higher frequency means shorter time per revolution.
414. A particle moves in a circle of radius 2 m with frequency 0.5 Hz. What is its angular velocity?
ⓐ. 0.5 rad/s
ⓑ. 1 rad/s
ⓒ. $\pi$ rad/s
ⓓ. $2\pi$ rad/s
Correct Answer: $\pi$ rad/s
Explanation: Angular velocity is $\omega = 2\pi f = 2\pi \times 0.5 = \pi \,\text{rad/s}$. The angular velocity shows how fast the angle changes at the center.
415. If a fan rotates at 120 revolutions per minute (rpm), what is its frequency in Hz?
ⓐ. 1 Hz
ⓑ. 2 Hz
ⓒ. 3 Hz
ⓓ. 4 Hz
Correct Answer: 2 Hz
Explanation: 120 rpm = 120/60 = 2 revolutions per second. Therefore frequency = 2 Hz.
416. A particle completes 1 revolution in 0.25 s. What is its angular velocity?
ⓐ. 2π rad/s
ⓑ. 4π rad/s
ⓒ. 6π rad/s
ⓓ. 8π rad/s
Correct Answer: 8π rad/s
Explanation: Period $T = 0.25 \,s$. Angular velocity $\omega = \frac{2\pi}{T} = \frac{2\pi}{0.25} = 8\pi \,\text{rad/s}$.
417. Which relation connects period T, frequency f, and angular velocity ω?
ⓐ. $f = \frac{1}{T}, \, \omega = 2\pi f$
ⓑ. $f = T, \, \omega = \pi f$
ⓒ. $f = \omega, \, T = 2\pi f$
ⓓ. $f = \frac{T}{2\pi}, \, \omega = f$
Correct Answer: $f = \frac{1}{T}, \, \omega = 2\pi f$
Explanation: Frequency is reciprocal of period. Angular velocity is the rate of angular displacement per time, and in circular motion is related to frequency as $\omega = 2\pi f$.
418. A rotating disc makes 600 revolutions per minute. What is its angular velocity?
ⓐ. $10\pi \, rad/s$
ⓑ. $20\pi \, rad/s$
ⓒ. $40\pi \, rad/s$
ⓓ. $60\pi \, rad/s$
Correct Answer: $20\pi \, rad/s$
Explanation: 600 rpm = 600/60 = 10 revolutions per second. Angular velocity $\omega = 2\pi f = 2\pi \times 10 = 20\pi \, rad/s$.
419. A particle moves with angular velocity $\omega = 4 \, rad/s$. How much angle (in radians) does it sweep in 3 seconds?
ⓐ. 4 rad
ⓑ. 8 rad
ⓒ. 10 rad
ⓓ. 12 rad
Correct Answer: 12 rad
Explanation: Angular displacement = $\theta = \omega t = 4 \times 3 = 12 \, rad$. Since $2\pi \, rad$ is one revolution, 12 rad ≈ 1.91 revolutions.
420. A wheel rotates at 5 Hz. How many revolutions will it make in 2 minutes?
ⓐ. 300
ⓑ. 400
ⓒ. 500
ⓓ. 600
Correct Answer: 600
Explanation: Frequency = 5 revolutions per second. In 2 minutes = 120 seconds, total revolutions = $5 \times 120 = 600$.
421. A satellite of mass 1000 kg orbits Earth at a height where the radius of orbit is $7 \times 10^6 \, m$. If gravitational constant $G = 6.67 \times 10^{-11}\, Nm^2/kg^2$ and Earth’s mass $M = 6 \times 10^{24} \, kg$, what is the orbital speed?
ⓐ. 5.6 km/s
ⓑ. 6.4 km/s
ⓒ. 7.5 km/s
ⓓ. 8.9 km/s
Correct Answer: 7.5 km/s
Explanation: Orbital speed formula: $v = \sqrt{\frac{GM}{r}}$. Substituting: $v = \sqrt{\frac{6.67 \times 10^{-11} \times 6 \times 10^{24}}{7 \times 10^6}} \approx 7.5 \times 10^3 \, m/s$.
422. Why do satellites remain in orbit without falling to Earth?
ⓐ. Because no force acts on them
ⓑ. Because gravity balances centrifugal force
ⓒ. Because velocity is tangential and gravity provides centripetal force
ⓓ. Because of zero gravity in space
Correct Answer: Because velocity is tangential and gravity provides centripetal force
Explanation: A satellite’s tangential velocity prevents it from falling straight down. Gravity acts inward as centripetal force, curving its path to stay in orbit.
423. Which law explains planetary orbits around the Sun?
ⓐ. Newton’s first law
ⓑ. Kepler’s laws of planetary motion
ⓒ. Archimedes’ principle
ⓓ. Bernoulli’s theorem
Correct Answer: Kepler’s laws of planetary motion
Explanation: Kepler’s three laws explain elliptical orbits, equal area sweeps in equal time, and the relation between square of period and cube of semi-major axis. They are consequences of Newtonian gravitation.
424. A geostationary satellite revolves around Earth once every 24 hours. What is its orbital period in seconds?
ⓐ. 21,600 s
ⓑ. 43,200 s
ⓒ. 57,600 s
ⓓ. 86,400 s
Correct Answer: 86,400 s
Explanation: 24 hours = $24 \times 3600 = 86,400 \, s$. This is the period required to remain fixed relative to a point on Earth’s surface.
425. Which condition is essential for a satellite to be geostationary?
ⓐ. Orbit must be elliptical with 12-hour period
ⓑ. Orbit must be polar with 24-hour period
ⓒ. Orbit must be circular above equator with 24-hour period
ⓓ. Orbit must be circular above poles with 12-hour period
Correct Answer: Orbit must be circular above equator with 24-hour period
Explanation: A geostationary satellite orbits in the equatorial plane with period equal to Earth’s rotation (24 h). This makes it appear stationary to observers on Earth.
426. The centripetal force keeping planets in orbit around the Sun is provided by:
ⓐ. Their inertia
ⓑ. Gravitational attraction of the Sun
ⓒ. Solar radiation pressure
ⓓ. Magnetic fields
Correct Answer: Gravitational attraction of the Sun
Explanation: Newton showed that Sun’s gravity provides the centripetal force necessary for planetary circular or elliptical motion.
427. A satellite is in circular orbit of radius r with orbital speed v. What happens if speed is suddenly doubled?
ⓐ. It continues in the same orbit
ⓑ. It enters elliptical orbit with higher apogee
ⓒ. It crashes into Earth
ⓓ. It escapes Earth’s gravity
Correct Answer: It escapes Earth’s gravity
Explanation: Escape velocity $v_e = \sqrt{2GM/r}$. If speed equals or exceeds this, the satellite will no longer remain bound in orbit and escape Earth’s gravity.
428. Which statement is correct about energy of a satellite in circular orbit?
ⓐ. Total energy is zero
ⓑ. Total energy is positive
ⓒ. Total energy is negative
ⓓ. Kinetic energy equals zero
Correct Answer: Total energy is negative
Explanation: For bound circular orbits, kinetic energy = $\frac{GMm}{2r}$, potential energy = $-\frac{GMm}{r}$. Total energy = $-\frac{GMm}{2r}$, which is negative, confirming the satellite is gravitationally bound.
429. A planet of mass m revolves around the Sun at distance r with orbital period T. Which relation holds true?
ⓐ. $T^2 \propto r$
ⓑ. $T^2 \propto r^2$
ⓒ. $T^2 \propto r^3$
ⓓ. $T^2 \propto r^4$
Correct Answer: $T^2 \propto r^3$
Explanation: Kepler’s third law: square of orbital period is directly proportional to cube of semi-major axis. This arises from balancing gravitational and centripetal forces.
430. Why do astronauts in satellites experience weightlessness?
ⓐ. Because there is no gravity in space
ⓑ. Because the satellite moves away from Earth
ⓒ. Because both satellite and astronauts are in free fall together
ⓓ. Because they are outside Earth’s gravitational field
Correct Answer: Because both satellite and astronauts are in free fall together
Explanation: Gravity still acts strongly on astronauts, but since they and the satellite fall toward Earth at the same rate, they experience apparent weightlessness (microgravity condition).
431. A projectile is launched from a 60 m high cliff with speed 30 m/s at 40° above the horizontal. Neglecting air resistance, what is the horizontal range from the foot of the cliff? (g = 9.8 m/s²)
ⓐ. 120 m
ⓑ. 137.5 m
ⓒ. 150 m
ⓓ. 165 m
Correct Answer: 137.5 m
Explanation: Resolve velocity: $u_x = 30\cos40^\circ \approx 22.98$ m/s, $u_y = 30\sin40^\circ \approx 19.28$ m/s. Time of flight from $y(t)=60+u_y t-\tfrac12gt^2=0$ gives $4.9t^2-19.28t-60=0\Rightarrow t=\frac{19.28+\sqrt{19.28^2+4\cdot4.9\cdot60}}{9.8}\approx 5.99$ s. Range $R=u_x t\approx 22.98\times 5.99\approx 137.5$ m.
432. A shell is fired from level ground with speed 40 m/s at 30°. At the highest point it explodes into two equal fragments; one fragment drops vertically from that point. Where (measured from the launch point) does the other fragment land? (g = 9.8 m/s²)
ⓐ. 106 m
ⓑ. 141 m
ⓒ. 212 m
ⓓ. 282 m
Correct Answer: 212 m
Explanation: Original range $R=\frac{u^2\sin60^\circ}{g}=\frac{1600\cdot0.866}{9.8}\approx 141.4$ m, apex is at $R/2$. Conservation of horizontal momentum at apex (mass halves; one has zero $v_x$) gives the other fragment’s $v_x=2u\cos\theta$. Horizontal travel from apex equals $R$, so landing point is $R/2+R=\tfrac{3R}{2}\approx 212$ m.
433. A river flows east at 6 m/s. A boat can move at 8 m/s relative to water. To land directly opposite on the north bank (no drift), how long does it take to cross a 300 m wide river?
ⓐ. 37.5 s
ⓑ. 42.0 s
ⓒ. 50.0 s
ⓓ. 56.7 s
Correct Answer: 56.7 s
Explanation: Aim upstream so westward component cancels current: $8\sin\phi=6\Rightarrow \phi=\sin^{-1}(0.75)$. Then northward component $=8\cos\phi=8\sqrt{1-0.75^2}\approx 5.29$ m/s. Time $t=\frac{300}{5.29}\approx 56.7$ s.
434. A target 150 m due east moves due north at 10 m/s. A projectile is fired from the origin at speed 100 m/s (neglect gravity for the short flight). At what yaw angle to the east (towards the north) should it be aimed to hit the target?
ⓐ. 3.00°
ⓑ. 5.74°
ⓒ. 8.00°
ⓓ. 10.0°
Correct Answer: 5.74°
Explanation: For interception with constant speeds, lateral component must match target’s speed: $100\sin\phi=10\Rightarrow \sin\phi=0.1\Rightarrow \phi\approx 5.74^\circ$. Then $t=\frac{150}{100\cos\phi}\approx1.51$ s, during which target moves $10t\approx15$ m north—exactly matched by the projectile’s northward component.
435. A boat (speed 6 m/s in still water) must cross a 240 m wide river with current 3 m/s east. If the boat heads straight across (minimum-time strategy), what are the crossing time and downstream drift?
ⓐ. 40 s and 60 m
ⓑ. 40 s and 120 m
ⓒ. 80 s and 120 m
ⓓ. 80 s and 240 m
Correct Answer: 40 s and 120 m
Explanation: Minimum time occurs when heading perpendicular to the banks: northward speed $=6$ m/s, so $t=240/6=40$ s. Drift $=vt_{\text{current}}=3\times40=120$ m.
436. A particle moves in a circle of radius 0.50 m. Its speed increases uniformly from 0 to 20 m/s in 5 s. What is the magnitude of its acceleration at $t=5$ s?
ⓐ. 4.0 m/s²
ⓑ. 40 m/s²
ⓒ. 400 m/s²
ⓓ. 800 m/s²
Correct Answer: 800 m/s²
Explanation: Tangential acceleration $a_t=\frac{\Delta v}{\Delta t}=4$ m/s². At $v=20$ m/s, centripetal acceleration $a_c=\frac{v^2}{r}=\frac{400}{0.5}=800$ m/s². Total $a=\sqrt{a_c^2+a_t^2}\approx\sqrt{800^2+4^2}\approx 800$ m/s² (dominantly centripetal).
437. A projectile is fired from ground at 37° and must clear a vertical wall 10 m high located 20 m away. What is the minimum launch speed? (g = 9.8 m/s²)
ⓐ. 20.0 m/s
ⓑ. 22.4 m/s
ⓒ. 24.7 m/s
ⓓ. 27.0 m/s
Correct Answer: 24.7 m/s
Explanation: Height at $x$ is $y=x\tan\theta-\frac{g x^2}{2u^2\cos^2\theta}$. For $\theta=37^\circ$, $x=20$ m, require $y\ge10$. Compute $20\tan37^\circ\approx15.07$ and $\cos^2 37^\circ\approx0.6378$. Solve $10\le 15.07-\frac{9.8\cdot400}{2u^2\cdot0.6378}\Rightarrow u\ge 24.6$ m/s. Hence $u_{\min}\approx 24.7$ m/s.
438. A projectile is launched with speed 30 m/s at 50° up a slope inclined at 20° to the horizontal. Find the time of flight until it strikes the slope. (g = 9.8 m/s²)
ⓐ. 2.45 s
ⓑ. 3.26 s
ⓒ. 3.80 s
ⓓ. 4.10 s
Correct Answer: 3.26 s
Explanation: For a slope at $\beta$ and launch at $\alpha$, time is $T=\frac{2u\sin(\alpha-\beta)}{g\cos\beta}$. Here $\sin(\alpha-\beta)=\sin30^\circ=0.5$, $\cos20^\circ\approx0.9397$. So $T=\frac{2\cdot30\cdot0.5}{9.8\cdot0.9397}\approx 3.26$ s.
439. Using the data of Question 438, what is the range measured along the slope?
ⓐ. 54.0 m
ⓑ. 61.0 m
ⓒ. 66.9 m
ⓓ. 74.5 m
Correct Answer: 66.9 m
Explanation: Range along the incline $R=\frac{2u^2\cos\alpha\sin(\alpha-\beta)}{g\cos^2\beta}$. With $u=30$, $\cos50^\circ\approx0.6428$, $\sin30^\circ=0.5$, $\cos^2 20^\circ\approx0.8830$: $R=\frac{2\cdot900\cdot0.6428\cdot0.5}{9.8\cdot0.8830}\approx 66.9$ m.
440. Aircraft A flies east from O at 250 m/s. At $t=0$, aircraft B is 3000 m due north of O flying south at 200 m/s. What is the minimum separation between them (assume level flight and straight-line speeds)?
ⓐ. 1.50 km
ⓑ. 2.00 km
ⓒ. 2.34 km
ⓓ. 3.00 km
Correct Answer: 2.34 km
Explanation: Positions: $A(t)=(250t,0)$, $B(t)=(0,3000-200t)$. Separation squared $d^2(t)=(250t)^2+(3000-200t)^2$. Minimize by $\frac{d}{dt}d^2=0\Rightarrow t=\frac{1.2\times10^6}{2.05\times10^5}\approx 5.85$ s. Then components are $(1463,1829)$ m, so $d_{\min}\approx\sqrt{1463^2+1829^2}\approx 2.34$ km.
441. Point B is 300 m east and 200 m north of point A across a river. Current is 3 m/s east; a boat can do 5 m/s in still water. At what heading (east of north) should the boat aim to go straight to B in the shortest time, and what is that time?
ⓐ. 30°, 46 s
ⓑ. 37°, 50 s
ⓒ. 45°, 54 s
ⓓ. 53°, 60 s
Correct Answer: 37°, 50 s
Explanation: Let heading be $\phi$ east of north: boat rel. water $=(5\sin\phi,5\cos\phi)$. Ground velocity $=(3+5\sin\phi,5\cos\phi)$ must be parallel to $(300,200)$ so ratio $=\frac{3+5\sin\phi}{5\cos\phi}=1.5$. Solving gives $\phi\approx 37^\circ$. Speed magnitude $\approx\sqrt{6.00^2+4.00^2}=7.21$ m/s. Distance $|\overrightarrow{AB}|=\sqrt{300^2+200^2}\approx 360.6$ m, hence $t\approx 360.6/7.21\approx 50$ s.
442. What is the smallest launch speed needed to hit a target at $(x,y)=(120\ \text{m},40\ \text{m})$ from the origin on level ground (choose the best launch angle)? Take $g=9.8$ m/s².
ⓐ. 35.0 m/s
ⓑ. 38.0 m/s
ⓒ. 40.4 m/s
ⓓ. 43.0 m/s
Correct Answer: 40.4 m/s
Explanation: For a given $(x,y)$, real launch angles exist if the quadratic in $\tan\theta$ has nonnegative discriminant; the minimum speed occurs at discriminant zero. Setting $k=\frac{gx^2}{2u^2}$, condition $x^2=4k(k+y)\Rightarrow k=\frac{-y+\sqrt{x^2+y^2}}{2}$. Then $u_{\min}^2=\frac{gx^2}{2k}=\frac{gx^2}{-y+\sqrt{x^2+y^2}}$. With $x=120$, $y=40$: $\sqrt{x^2+y^2}=\sqrt{16000}\approx126.49$. Hence $u_{\min}^2=\frac{9.8\cdot14400}{86.49}\approx 1631\Rightarrow u_{\min}\approx 40.4$ m/s.
443. A projectile is fired from the ground with speed 60 m/s and must pass through the point (x, y) = (180 m, 36 m). For g = 9.8 m/s², which launch angles are possible?
ⓐ. 15.0° and 75.0°
ⓑ. 20.0° and 70.0°
ⓒ. 26.94° and 74.37°
ⓓ. 30.00° and 60.00°
Correct Answer: 26.94° and 74.37°
Explanation: Using $y = x\tan\theta – \frac{g x^2}{2u^2\cos^2\theta}$ and $\tan\theta = T$, we get $A T^2 – xT + (y+A)=0$ with $A=\frac{g x^2}{2u^2}$. For x = 180, y = 36, u = 60: $A=44.1$. Discriminant positive gives $T_{1,2}=3.5733,\,0.5083 \Rightarrow \theta_{1,2}=74.37^\circ,\,26.94^\circ$ (complements of range angles).
444. A projectile is launched from a 50 m high cliff with speed 25 m/s at 35° above horizontal. Where does it land measured horizontally from the foot of the cliff? (g = 9.8 m/s²)
ⓐ. 92 m
ⓑ. 102 m
ⓒ. 112 m
ⓓ. 122 m
Correct Answer: 102 m
Explanation: Components: $u_x = 25\cos35^\circ \approx 20.48$ m/s, $u_y = 25\sin35^\circ \approx 14.34$ m/s. Time from $y(t)=50+u_y t-\tfrac12gt^2=0$ gives $t=\frac{u_y+\sqrt{u_y^2+2g(50)}}{g}\approx 4.98$ s. Range $R=u_x t\approx 20.48\times 4.98 \approx 101.9$ m.
445. On level ground a projectile is fired at speed 50 m/s to hit a target 180 m away. What launch angles (to the nearest 0.01°) are possible? (g = 9.8 m/s²)
ⓐ. 15.00° and 75.00°
ⓑ. 22.44° and 67.56°
ⓒ. 30.00° and 60.00°
ⓓ. 37.00° and 53.00°
Correct Answer: 22.44° and 67.56°
Explanation: Range relation $R=\frac{u^2\sin 2\theta}{g} \Rightarrow \sin 2\theta = \frac{gR}{u^2}=0.7056$. Hence $2\theta= \sin^{-1}(0.7056)=44.88^\circ$ or $180^\circ-44.88^\circ$, giving $\theta=22.44^\circ$ or $67.56^\circ$.
446. A projectile must just clear the top of a hill located 60 m away and 25 m high when fired at 45°. What is the minimum launch speed? (g = 9.8 m/s²)
ⓐ. 28.0 m/s
ⓑ. 30.5 m/s
ⓒ. 31.8 m/s
ⓓ. 34.0 m/s
Correct Answer: 31.8 m/s
Explanation: Trajectory at x = 60 must satisfy $y \ge 25$: $y=x\tan\theta – \frac{g x^2}{2u^2\cos^2\theta}$. For $\theta=45^\circ$, $\cos^2\theta=0.5$. Solve for the equality case to find $u_{\min} = \sqrt{\frac{g x^2}{2\cos^2\theta (x\tan\theta – y)}} \approx 31.75$ m/s.
447. Aircraft A starts at O and flies east with $\vec{v}_A=(220,0)$ m/s. Aircraft B starts at position (0, 4000) m and flies with $\vec{v}_B=(150, -100)$ m/s relative to O. What is their minimum separation?
ⓐ. 2.00 km
ⓑ. 2.50 km
ⓒ. 3.08 km
ⓓ. 3.50 km
Correct Answer: 3.08 km
Explanation: Relative position $\vec{r}_{AB}(t) = (220t,0) – (0+150t, 4000-100t) = (70t, 100t – 4000)$. Minimize $d^2= (70t)^2 + (100t-4000)^2 \Rightarrow t^* = \frac{7000\cdot 0 + 100\cdot 4000}{70^2+100^2} \approx 10.28$ s. Then $d_{\min} \approx \sqrt{(70t^*)^2 + (100t^*-4000)^2} \approx 3.079$ km.
448. A projectile is launched at speed 35 m/s at 55° and lands on ground 20 m below the launch level. Find its time of flight. (g = 9.8 m/s²)
ⓐ. 5.60 s
ⓑ. 6.00 s
ⓒ. 6.48 s
ⓓ. 7.10 s
Correct Answer: 6.48 s
Explanation: Vertical motion: $y(t) = u_y t – \tfrac12 g t^2$ with $u_y=35\sin55^\circ \approx 28.67$ m/s and $y=-20$. Solve $\tfrac12 g t^2 – u_y t – 20=0 \Rightarrow t = \frac{u_y+\sqrt{u_y^2+2g\cdot 20}}{g} \approx 6.481$ s.
449. A projectile is fired uphill along a slope of 15° with speed 40 m/s. For maximum range along the slope, what should the launch angle be (measured from horizontal), and what is that maximum range along the slope? (g = 9.8 m/s²)
ⓐ. 45.0°, 122 m
ⓑ. 50.0°, 126 m
ⓒ. 52.5°, 130 m
ⓓ. 60.0°, 135 m
Correct Answer: 52.5°, 130 m
Explanation: On an incline of angle $\beta$, range along slope is $R=\frac{2u^2\cos\alpha\sin(\alpha-\beta)}{g\cos^2\beta}$, maximized at $\alpha=45^\circ+\beta/2 = 52.5^\circ$. Substituting gives $R_{\max}\approx 129.7$ m, i.e., about 130 m.
450. An aircraft must track exactly northeast (45°) with airspeed 200 m/s while a wind of 30 m/s blows from the west (toward east). What heading should the pilot fly, and what will be the ground speed?
ⓐ. Heading 45.0°, ground speed 214 m/s
ⓑ. Heading 41.0°, ground speed 218 m/s
ⓒ. Heading 38.9°, ground speed 220 m/s
ⓓ. Heading 35.0°, ground speed 225 m/s
Correct Answer: Heading 38.9°, ground speed 220 m/s
Explanation: With wind-track angle $\beta=45^\circ$, drift $\delta=\sin^{-1}(\frac{V_w\sin\beta}{V_a})=\sin^{-1}(30\sin45^\circ/200)\approx 6.09^\circ$. Heading = track − drift $\approx 45^\circ-6.09^\circ=38.91^\circ$. Ground speed $\approx V_a\cos\delta + V_w\cos\beta \approx 220.1$ m/s.
451. A target initially at (200 m, 0) moves left at 10 m/s. A projectile is launched from the origin at speed 60 m/s to intercept the target at the same level (air resistance neglected). What launch angle should be used? (g = 9.8 m/s²)
ⓐ. 10.0°
ⓑ. 13.8°
ⓒ. 20.0°
ⓓ. 25.0°
Correct Answer: 13.8°
Explanation: Let time of flight $t = \frac{2u\sin\theta}{g}$. Horizontal equality: $u\cos\theta\, t = 200 – 10t$. Eliminating t gives $2u^2\sin\theta\cos\theta + 2u(10)\sin\theta – g\cdot 200 = 0$. Numerically solving in $(0,90^\circ)$ yields $\theta \approx 13.845^\circ$.
452. A particle starts from rest and moves with constant acceleration in a plane. After 5 s it is at (90 m, 120 m) and its speed is 60 m/s. What is the acceleration vector?
ⓐ. $(6.0,\, 8.0)$ m/s²
ⓑ. $(7.2,\, 9.6)$ m/s²
ⓒ. $(8.0,\, 6.0)$ m/s²
ⓓ. $(9.6,\, 7.2)$ m/s²
Correct Answer: $(7.2,\, 9.6)$ m/s²
Explanation: With rest start and constant acceleration, $\vec{r}=\tfrac12 \vec{a} t^2 \Rightarrow \vec{a}=2\vec{r}/t^2 = 2(90,120)/25 = (7.2,9.6)$ m/s². Magnitude $|\vec{a}| = 12$ m/s², giving speed $v=|\vec{a}|t=12\times 5 = 60$ m/s, consistent with the data.
453. A projectile is launched from ground with speed 40 m/s at 35° and lands on terrain 20 m lower than the launch level. Take g = 9.8 m/s^2. What is the horizontal range?
ⓐ. 162.5 m
ⓑ. 170.0 m
ⓒ. 178.1 m
ⓓ. 185.0 m
Correct Answer: 178.1 m
Explanation: Components: $u_x = 40\cos35^\circ \approx 32.76$ m/s, $u_y = 40\sin35^\circ \approx 22.96$ m/s. Vertical equation to the landing level $y=-20$: $\tfrac{1}{2}gt^2 – u_y t – 20 = 0$. Hence $t = \dfrac{u_y + \sqrt{u_y^2 + 2g\cdot 20}}{g} \approx \dfrac{22.96 + \sqrt{22.96^2 + 392}}{9.8} \approx 5.435\,\text{s}$. Range $R = u_x t \approx 32.76 \times 5.435 \approx 178.1$ m.
454. From level ground, the projectile must pass exactly through the point $(x,y) = (100\,\text{m}, 20\,\text{m})$. What is the minimum launch speed (choose the best angle)? Use $g=9.8$ m/s^2.
ⓐ. 32.0 m/s
ⓑ. 34.6 m/s
ⓒ. 36.0 m/s
ⓓ. 38.5 m/s
Correct Answer: 34.6 m/s
Explanation: The minimum speed to reach $(x,y)$ is $u_{\min}^2 = \dfrac{g x^2}{-y + \sqrt{x^2 + y^2}}$. Here $\sqrt{x^2+y^2}=\sqrt{10400}=101.98$. Thus $u_{\min}^2 = \dfrac{9.8\cdot 10000}{-20+101.98} \approx 1195.1$, so $u_{\min} \approx 34.6$ m/s.
455. A projectile is fired with speed 30 m/s at 50°. At what time does its velocity vector make 45° below the horizontal? (g = 9.8 m/s^2)
ⓐ. 3.15 s
ⓑ. 3.90 s
ⓒ. 4.31 s
ⓓ. 4.90 s
Correct Answer: 4.31 s
Explanation: Let $v_x = u\cos50^\circ \approx 19.29$ m/s (constant). The vertical component is $v_y = u\sin50^\circ – gt \approx 22.98 – 9.8t$. For 45° below horizontal, $|v_y| = v_x$ with $v_y<0$: set $22.98 – 9.8t = -19.29$ giving $t = \dfrac{22.98+19.29}{9.8} \approx 4.31$ s.
456. Two aircraft move in a plane. A starts at (0,0) with $\vec{v}_A=(30,40)$ m/s. B starts at (3000,0) m with $\vec{v}_B=(-20,60)$ m/s. What is the minimum separation between them?
ⓐ. 0.90 km
ⓑ. 1.12 km
ⓒ. 1.50 km
ⓓ. 1.80 km
Correct Answer: 1.12 km
Explanation: Relative position $\vec{r}_{AB}(t) = \vec{r}_A-\vec{r}_B = (50t-3000,\,-20t)$. Minimize $d^2(t)=(50t-3000)^2+(20t)^2$. Set derivative zero: $2[50(50t-3000)+20(20t)]=0\Rightarrow 5800t=300000\Rightarrow t=51.724$ s. Then $\vec{r}_{AB} \approx (-414,\,-1035)$ m; $d_{\min} \approx \sqrt{414^2+1035^2} \approx 1.12$ km.
457. A particle moves on a circle of radius 3.0 m with angular acceleration $\alpha=2.0$ rad/s^2 from $\omega_0=1.0$ rad/s. What is the total acceleration magnitude at $t=4.0$ s?
ⓐ. 120 m/s^2
ⓑ. 180 m/s^2
ⓒ. 220 m/s^2
ⓓ. 243 m/s^2
Correct Answer: 243 m/s^2
Explanation: $\omega=\omega_0+\alpha t=9$ rad/s; linear speed $v=\omega r=27$ m/s. Centripetal $a_c=v^2/r=27^2/3=243$ m/s^2. Tangential $a_t=r\alpha=3\times 2=6$ m/s^2. Total $a=\sqrt{a_c^2+a_t^2}\approx \sqrt{243^2+6^2}\approx 243$ m/s^2 (tangential part is negligible compared to $a_c$).
458. A river 200 m wide flows east at 3 m/s. A swimmer can do 5 m/s relative to water and wants to land 50 m upstream (west) from the point directly opposite. Find the required heading (measured west of north) and crossing time.
ⓐ. 42.0°, 58.0 s
ⓑ. 45.0°, 60.0 s
ⓒ. 49.6°, 61.7 s
ⓓ. 53.1°, 64.0 s
Correct Answer: 49.6°, 61.7 s
Explanation: Let heading be $\phi$ west of north; ground velocity $=(3+5\sin\phi,\,5\cos\phi)$. Displacement conditions: $y=200=5\cos\phi\,t$, $x=-50=(3+5\sin\phi)t$. Eliminating $t$: $(3+5\sin\phi)\,\dfrac{200}{5\cos\phi}=-50\Rightarrow 20\sin\phi+5\cos\phi=-12$. Write $R\sin(\phi+\delta)=-12$ with $R=\sqrt{20^2+5^2}=20.618$ and $\delta=\tan^{-1}(5/20)=14.04^\circ$. Hence $\phi\approx 49.6^\circ$. Time $t=200/(5\cos\phi)\approx 61.7$ s.
459. A projectile is launched from the base of a slope ($\beta=20^\circ$) with speed $u=25$ m/s at $\alpha=50^\circ$ above horizontal. What is its speed just before striking the slope? (g = 9.8 m/s^2)
ⓐ. 15.0 m/s
ⓑ. 17.7 m/s
ⓒ. 20.0 m/s
ⓓ. 22.5 m/s
Correct Answer: 17.7 m/s
Explanation: Time of flight to the slope: $T=\dfrac{2u\sin(\alpha-\beta)}{g\cos\beta}=\dfrac{2\cdot25\sin30^\circ}{9.8\cos20^\circ}\approx 2.713$ s. Vertical rise: $y=u\sin\alpha\,T-\tfrac12 gT^2\approx 19.15\cdot2.713-4.9\cdot(2.713)^2\approx 15.88$ m. By energy, $v^2=u^2-2gy=625-2\cdot9.8\cdot15.88\approx 313.7\Rightarrow v\approx 17.7$ m/s.
460. With speed $u=40$ m/s, at what launch angle $\theta$ will the maximum height be $H=45$ m? (g = 9.8 m/s^2)
ⓐ. 41.4°
ⓑ. 45.0°
ⓒ. 47.8°
ⓓ. 51.0°
Correct Answer: 47.8°
Explanation: $H=\dfrac{u^2\sin^2\theta}{2g}\Rightarrow \sin\theta = \dfrac{\sqrt{2gH}}{u} = \dfrac{\sqrt{2\cdot 9.8\cdot 45}}{40} = \dfrac{29.70}{40}=0.7425$. Hence $\theta=\sin^{-1}(0.7425)\approx 47.8^\circ$.
461. Two projectiles are fired simultaneously with the same speed $u=50$ m/s at angles $\theta$ and $90^\circ-\theta$. A vertical wall stands at $x=120$ m. For what $\theta$ do they hit the wall at the same height? (g = 9.8 m/s^2)
ⓐ. 10.0°
ⓑ. 12.0°
ⓒ. 14.0°
ⓓ. 16.0°
Correct Answer: 14.0°
Explanation: Equal heights condition at $x$ leads to $x = K(\tan\theta + \cot\theta)$ with $K=\dfrac{g x^2}{2u^2}$. Here $K=\dfrac{9.8\cdot 120^2}{2\cdot 50^2}=28.224$. Hence $\tan\theta+\cot\theta = \dfrac{120}{28.224}=4.252$. Solve $t+1/t=4.252\Rightarrow t=\tan\theta\approx 0.249$ or $4.003$. Thus $\theta \approx 14.0^\circ$ (the other is its complement).
462. A projectile of mass $4m$ is fired from level ground with speed $u=40$ m/s at $30^\circ$. At the highest point it explodes into two fragments of masses $m$ and $3m$. The lighter fragment loses all horizontal speed and falls vertically. Where (from launch) does the heavier fragment land? (g = 9.8 m/s^2)
ⓐ. 85.0 m
ⓑ. 95.2 m
ⓒ. 102.0 m
ⓓ. 110.5 m
Correct Answer: 95.2 m
Explanation: Original range $R=\dfrac{u^2\sin 2\theta}{g}=\dfrac{1600\cdot 0.5}{9.8}=81.63$ m, apex at $R/2$. Horizontal momentum at apex: $4m\,u\cos\theta$. After explosion, light fragment has zero $v_x$, so heavy fragment (mass $3m$) must carry all horizontal momentum: $v_{xH} = \dfrac{4m\,u\cos\theta}{3m} = \dfrac{4}{3}u\cos\theta$. Time to fall from apex $T = \dfrac{u\sin\theta}{g}$. Extra horizontal distance for heavy fragment: $v_{xH}T = \dfrac{4}{3}u\cos\theta \cdot \dfrac{u\sin\theta}{g} = \dfrac{2}{3}\dfrac{u^2\sin 2\theta}{g} = \dfrac{2}{3}R$. Total distance from launch: $R/2 + 2R/3 = \tfrac{7}{6}R \approx 1.1667 \times 81.63 \approx 95.2$ m.
463. A projectile must clear a vertical wall 15 m high located 60 m from the launch point and then land back on level ground at x = 180 m. What is the minimum launch speed required? (Take g = 9.8 m/s^2)
ⓐ. 40.0 m/s
ⓑ. 41.0 m/s
ⓒ. 42.0 m/s
ⓓ. 44.0 m/s
Correct Answer: 42.0 m/s
Explanation: The landing at x = 180 m fixes $u^2\sin 2\theta = gR \Rightarrow u^2 = \frac{gR}{\sin 2\theta}$. The height at x = 60 m is $y = x\tan\theta – \frac{gx^2}{2u^2\cos^2\theta}$. Eliminating $u$ using $u^2$ from the range relation gives $y = \tan\theta\big(x – \frac{x^2}{R}\big)$. With $x=60$ m and $R=180$ m, $x – x^2/R = 60 – 3600/180 = 40$. To clear the wall, $\tan\theta \ge 15/40 = 0.375 \Rightarrow \theta \ge 20.56^\circ$. The minimum $u$ occurs when $\sin 2\theta$ is maximized subject to this, i.e. at $\theta = 45^\circ$. Then $u_{\min} = \sqrt{gR} = \sqrt{9.8 \times 180} = \sqrt{1764} = 42.0 \,\text{m/s}$.
464. A projectile is launched with speed 30 m/s at angle 35° above the horizontal on a downward slope of angle 15° (slope falls by 15° from the horizontal). Find the time of flight until it strikes the slope. (g = 9.8 m/s^2)
ⓐ. 4.35 s
ⓑ. 4.56 s
ⓒ. 4.86 s
ⓓ. 5.10 s
Correct Answer: 4.86 s
Explanation: For an incline with angle $\beta$ (negative for downward), the time to hit the slope is $T = \frac{2u\sin(\alpha-\beta)}{g\cos\beta}$, where $\alpha$ is the launch angle from horizontal. With $\alpha=35^\circ, \beta=-15^\circ$, $\alpha-\beta=50^\circ$ and $\cos\beta=\cos 15^\circ$. Thus $T = \frac{2 \times 30 \times \sin 50^\circ}{9.8 \cos 15^\circ} \approx \frac{60 \times 0.7660}{9.8 \times 0.9659} \approx 4.86 \,\text{s}$. This result uses the standard incline-impact kinematics derived by resolving motion along and perpendicular to the slope.
465. A target is 500 m due east of the launcher and moves due north at 30 m/s. A projectile is fired at speed 120 m/s, and the short flight allows us to neglect gravity. At what angle north of east should the gun be aimed to hit the target?
ⓐ. 12.0°
ⓑ. 14.5°
ⓒ. 18.0°
ⓓ. 22.5°
Correct Answer: 14.5°
Explanation: For interception with constant speeds (gravity neglected), the projectile’s northward component must match the target’s northward speed: $120 \sin\phi = 30 \Rightarrow \sin\phi = 0.25 \Rightarrow \phi \approx 14.48^\circ$. The flight time is then $t = \frac{500}{120\cos\phi} \approx 4.30 \,\text{s}$, during which the target moves $30t$ north, exactly matched by the projectile’s north component.
466. A particle moves in a circle of radius 2.0 m. Its speed is 25 m/s and its tangential acceleration is 30 m/s^2 at an instant. What is the angle between its total acceleration vector and its instantaneous velocity vector?
ⓐ. 78.7°
ⓑ. 80.0°
ⓒ. 84.5°
ⓓ. 87.5°
Correct Answer: 84.5°
Explanation: The total acceleration has perpendicular components: centripetal $a_c = v^2/r = 25^2/2 = 312.5 \,\text{m/s}^2$ (radially inward) and tangential $a_t = 30 \,\text{m/s}^2$ (along velocity). The angle $\psi$ between acceleration and velocity (tangent) satisfies $\tan\psi = a_c/a_t$. Thus $\psi = \tan^{-1}(312.5/30) \approx 84.52^\circ$. The acceleration is almost radial because $a_c \gg a_t$.
467. Two ships move in a plane. Ship A starts at the origin and sails with velocity $\vec{v}_A = (15,\,9)$ m/s. Ship B starts at $(1000,\,1500)$ m and sails with $\vec{v}_B = (-5,\,-3)$ m/s. What is their minimum separation distance?
ⓐ. 0.62 km
ⓑ. 0.77 km
ⓒ. 0.95 km
ⓓ. 1.20 km
Correct Answer: 0.77 km
Explanation: Relative position $\vec{r}_{AB}(t) = \vec{r}_A – \vec{r}_B = (15t,\,9t) – (1000-5t,\,1500-3t) = (20t – 1000,\,12t – 1500)$. Minimizing $d^2(t) = (20t-1000)^2 + (12t-1500)^2$ gives $\frac{d}{dt}d^2 = 2[20(20t-1000) + 12(12t-1500)] = 0 \Rightarrow 1088t = 76000 \Rightarrow t^* \approx 69.85 \,\text{s}$. Then $\vec{r}_{AB}(t^*) \approx (397.1,\,-661.8)$ m, so $d_{\min} \approx \sqrt{397.1^2 + 661.8^2} \approx 772 \,\text{m} = 0.77 \,\text{km}$.
This is the final section of Class 11 Physics MCQs – Chapter 4: Motion in a Plane (Part 5). The complete chapter provides a total of 467 MCQs with solutions, divided into 5 parts for structured learning and exam practice. Covering important subtopics like projectile motion on inclined planes, relative velocity problems, centripetal force, uniform circular motion, and applications in satellite dynamics, this chapter is vital for success in board exams and competitive exams like JEE, NEET, IIT, and state-level tests. Here in Part 5, you will attempt the last 67 MCQs with detailed answers, perfect for final revision and confidence building. 🎉 Congratulations on completing the entire chapter! Now explore other chapters and subjects using the navigation bar above to continue your preparation.
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