Questions: Motion in a Plane
Motion in a Plane24 min readLesson 6 of 7
Q1. Distinguish between a scalar and a vector with two examples of each.
Answer: A scalar is described completely by its magnitude and unit, while a vector needs a direction as well. Mass, time and temperature are scalars; displacement, velocity, acceleration and force are vectors. Two vectors are equal only when their magnitudes and directions both match, so moving a vector without rotating it produces an equal vector. A scalar can be added to a scalar of the same kind, but never to a vector.
Hinglish: Scalar sirf magnitude aur unit se poora ho jaata hai, vector ko direction bhi chahiye. Mass, time, temperature scalar hain; displacement, velocity, force vectors hain. Do vectors barabar tabhi hote hain jab magnitude aur direction dono same ho. Scalar ko vector mein add nahi kar sakte.
Q2. State the triangle law of vector addition and use it to find the resultant of two perpendicular vectors.
Answer: If two vectors are placed tail to head, the vector drawn from the free tail to the free head is the resultant. For two perpendicular vectors of magnitudes 3 and 4, the triangle is right angled, so the magnitude is the square root of 9 plus 16, which is 5. The direction makes an angle with the 3 unit vector whose tangent is 4 over 3, about 53 degrees. The same result follows from the law of cosines with the angle equal to 90 degrees, which reduces the cross term to zero.
Hinglish: Triangle law mein pehle vector ke head par doosra rakh ke tail se head tak resultant milta hai. 3 aur 4 perpendicular vectors ke liye magnitude root(9+16) = 5 hoti hai, aur direction ka tangent 4/3 hai, matlab lagbhag 53 degree. Law of cosines se bhi 90 degree par wahi answer aata hai.
Q3. A vector of magnitude 10 makes an angle of 30 degrees with the x-axis. Find its components.
Answer: The x component is 10 cos 30, which is 8.66, and the y component is 10 sin 30, which is 5. The magnitude check is the square root of 8.66 squared plus 5 squared, and that returns 10. The signs of the components depend on the quadrant: in the second quadrant the x component is negative while the y component is positive, even though the same angle of 30 degrees from the x-axis is quoted.
Hinglish: x component 10 cos 30 = 8.66 aur y component 10 sin 30 = 5. Magnitude check root(8.66^2 + 5^2) = 10 aata hai. Signs quadrant se aate hain; doosre quadrant mein x negative aur y positive hota hai.
Q4. Prove that the horizontal range of a projectile is maximum at 45 degrees.
Answer: The range is u squared sin 2 theta divided by g. For a fixed launch speed, u squared and g are constant, so the range is largest when sin 2 theta is largest. The sine of an angle cannot exceed 1, and it equals 1 when 2 theta equals 90 degrees, that is when theta is 45 degrees. The maximum range is then u squared over g. Two angles that add to 90 degrees give the same sine of the double angle, which is why they share a range.
Hinglish: Range R = u^2 sin 2 theta / g hoti hai. Fixed speed ke liye u^2 aur g constant hain, isliye range tab maximum hoti hai jab sin 2 theta maximum ho. Sin ki maximum value 1 hai, jo 2 theta = 90 degree par aati hai, matlab theta = 45 degree. Tab max range u^2/g hoti hai. 90 degree ke complementary angles ki sin 2 theta same hoti hai, isliye range bhi same.
Q5. A projectile is launched horizontally from a height. Show that its time of flight depends only on the height.
Answer: In a horizontal launch the initial vertical velocity is zero, so the vertical motion is free fall from rest. The height follows from h = half g t squared, so the time is the square root of 2h over g. The horizontal speed does not appear in this equation, which means two balls released horizontally at different speeds from the same height land at the same time. The faster one lands farther away, because its horizontal distance is its speed multiplied by the common time.
Hinglish: Horizontal launch mein initial vertical velocity zero hoti hai, isliye vertical motion rest se free fall hai. h = 1/2 g t^2 se time root(2h/g) milta hai, jismein horizontal speed aati hi nahi. Isliye same height se alag speeds par chhodi gayi do balls ek saath zameen par pahunchti hain; tez wali zyada door girti hai.
Q6. Why is the acceleration in uniform circular motion directed toward the centre?
Answer: The speed is constant, so the velocity changes only in direction. The change in velocity between two nearby points is the vector difference of two tangents, and that difference points toward the centre of the circle. Dividing the change by the small time interval gives an acceleration along the same inward direction. The magnitude is v squared over r, which grows with the square of the speed and falls with the radius. After one full revolution the average acceleration is zero because the velocity returns to its starting value.
Hinglish: Uniform circular motion mein speed constant hai, isliye velocity sirf direction mein badalti hai. Do paas ke points par velocity vectors ka difference centre ki taraf point karta hai, aur usko time se divide karne par acceleration bhi centre ki taraf milta hai. Magnitude v^2/r hoti hai. Ek poore revolution par velocity wapas shuru wali ho jaati hai, isliye average acceleration zero hota hai.
Q7. A boat can move at 5 m s^-1 in still water and the river flows at 3 m s^-1. Find the shortest time to cross a 400 m wide river and the drift in that case.
Answer: For the shortest time the boat heads straight across, so its crossing component is its full 5 m per second. The time is 400 over 5, that is 80 s. During those 80 s the current carries the boat downstream by 3 times 80, which is 240 m. The boat's ground speed is the root of 25 plus 9, about 5.83 m per second, along a direction tilted downstream from the crossing line.
Hinglish: Shortest time ke liye boat seedha across chalti hai, isliye crossing component 5 m/s. Time 400/5 = 80 s. Itni der mein current boat ko 3 x 80 = 240 m downstream le jaata hai. Ground speed root(25+9) = 5.83 m/s hoti hai, direction downstream tilt ke saath.
Q8. A body moves in a circle of radius 2 m at a constant angular speed of 3 rad s^-1. Find its speed, centripetal acceleration and period.
Answer: The speed is omega r, so 3 times 2 equals 6 m per second. The centripetal acceleration is v squared over r, which is 36 over 2, equal to 18 m per second squared, or from omega squared r, 9 times 2, the same 18. The period is 2 pi over omega, about 2.09 s. The direction of the acceleration is toward the centre and it changes continuously as the body goes around, even though its magnitude stays fixed.
Hinglish: Speed v = omega r = 3 x 2 = 6 m/s. Centripetal acceleration v^2/r = 36/2 = 18 m/s^2, ya omega^2 r = 9 x 2 = 18, same answer. Period 2 pi / omega = 2.09 s. Acceleration ki direction centre ki taraf hoti hai aur ghoomte hue badalti rehti hai, magnitude same rehta hai.
Q9. Two vectors have equal magnitudes and their resultant has the same magnitude as either vector. What is the angle between them?
Answer: Let each vector have magnitude A and let the angle be theta. The resultant squared is A squared plus A squared plus 2A squared cos theta, and setting it equal to A squared gives 1 plus 2 cos theta equal to zero, so cos theta equals minus one half. The angle is therefore 120 degrees. The vector triangle is equilateral in that case, which is a quick geometrical check on the answer.
Hinglish: Dono vectors ka magnitude A aur angle theta lo. Resultant ka square 2A^2 + 2A^2 cos theta hai, aur isko A^2 ke barabar rakhne se 1 + 2 cos theta = 0, matlab cos theta = -1/2. Angle 120 degree aata hai, aur vector triangle equilateral ban jaata hai.
Q10. A ball is thrown at 20 m s^-1 at 60 degrees to the horizontal. Find the maximum height, the time of flight and the range. Take g = 10 m s^-2.
Answer: The vertical component is 20 sin 60, about 17.32, and the horizontal component is 10. The maximum height is the square of 17.32 divided by 20, which is 15 m. The time of flight is twice 17.32 divided by 10, about 3.46 s. The range is the horizontal speed times the flight time, 10 times 3.46, about 34.6 m, which also follows from u squared sin 120 over g. The range equals that of a 30 degree throw at the same speed, because 30 and 60 are complementary.
Hinglish: Vertical component 20 sin 60 = 17.32, horizontal 10. Maximum height 17.32^2 / 20 = 15 m. Time of flight 2 x 17.32/10 = 3.46 s. Range 10 x 3.46 = 34.6 m, jo u^2 sin 120/g se bhi aata hai. 30 aur 60 complementary angles hain, isliye 30 degree wale throw ki range bhi same hogi.
Quiz below: five questions on scalars and vectors, vector components, the projectile results and circular motion. Try them without the formula sheet first, then check each wrong answer against the matching page.