Previous year questions: Motion in a Plane

Ten questions follow from the official JEE Main 2026 papers and the NCERT Exemplar. Each heading gives the paper and the marks. Attempt each question before reading the answer, and resolve every vector into components before substituting numbers.

Q1 (JEE Main 2026, 4 Apr Shift 2, 4 marks)

The x and y coordinates of a projectile at time t are 24t and 43.6t - 4.9t squared, with the lengths in metre. The angle made by the velocity with the horizontal at t = 2 s is

Answer: Differentiate the coordinates. The horizontal velocity is 24 m per second and stays constant. The vertical velocity is 43.6 - 9.8t, which at t = 2 s is 43.6 - 19.6 = 24 m per second. The angle with the horizontal satisfies tan theta equal to 24 over 24, that is 1, so the angle is 45 degrees. The equal components at that instant make the answer immediate.

Hinglish: Coordinates ko differentiate karo: horizontal velocity 24 m/s constant, vertical 43.6 - 9.8t, jo t = 2 s par 43.6 - 19.6 = 24 m/s. Ab tan theta = 24/24 = 1, isliye angle 45 degree. Us instant par dono components barabar hain.

Q2 (JEE Main 2026, 4 Apr Shift 1, 4 marks)

Two projectiles are thrown with the same initial speed at 15 degrees and 30 degrees to the horizontal. The ratio of their ranges is 1 : x. The value of x is

Answer: The range is proportional to sin 2 theta. For 15 degrees the double angle is 30 degrees and for 30 degrees it is 60 degrees. The ratio of the ranges is sin 30 divided by sin 60, which is 0.5 divided by root 3 over 2, that is 1 over root 3. Writing the ratio as 1 : x gives x equal to root 3.

Hinglish: Range sin 2 theta ke proportional hoti hai. 15 degree ke liye sin 30 aur 30 degree ke liye sin 60. Ratio 0.5 / (root3/2) = 1/root3, isliye x = root 3.

Q3 (JEE Main 2026, 2 Apr Shift 2, 4 marks)

A 0.5 kg mass stays in contact with the inner wall of a cylindrical drum of radius 4 m that rotates about its vertical axis. The minimum rotational speed for the mass not to fall is 5 rad s^-1. The coefficient of friction between the mass and the wall is (take g = 10 m s^-2)

Answer: The normal reaction supplies the centripetal force, so N equals m omega squared r. The friction must balance the weight, so mu N equals m g. Divide the second equation by the first: mu equals g over omega squared r. Substitute 10 over 25 times 4, which is 10 over 100, so mu equals 0.1. The mass of the block cancels, which is why the answer does not depend on the 0.5 kg.

Hinglish: Normal reaction centripetal force deta hai, N = m omega^2 r, aur friction weight balance karta hai, mu N = mg. Divide karne par mu = g/(omega^2 r) = 10/(25 x 4) = 0.1. Mass cancel ho jaata hai, isliye 0.5 kg se farak nahi padta.

Q4 (JEE Main 2026, 4 Apr Shift 2, 4 marks)

A gun mounted on the ground fires bullets in all directions with the same speed. The farthest distance the bullets reach is 6.4 m. The speed of the bullets is (take g = 10 m s^-2)

Answer: The farthest distance a projectile can reach from the launch point on level ground is the maximum range, which is u squared over g at 45 degrees. Set u squared over 10 equal to 6.4, so u squared equals 64 and u is 8 m per second. The answer is a direct use of the maximum range result, and the all directions phrase only tells us that the 45 degree launch is available.

Hinglish: Level ground par maximum range u^2/g hoti hai, 45 degree par. u^2/10 = 6.4 se u^2 = 64 aur u = 8 m/s. All directions ka matlab yahi hai ki 45 degree wala launch available hai, isliye seedha formula lagta hai.

Q5 (JEE Main 2026, 2 Apr Shift 1, 4 marks)

A particle rotates in a circular path and its angular position is theta = 5t^4/40 - t^3/3 radian. The angular acceleration at t = 10 s is

Answer: Differentiate the angular position: omega equals d theta by dt, which is t cubed over 2 minus t squared. Differentiate again: alpha equals 3t squared over 2 minus 2t. At t = 10 s this is 150 - 20, that is 130 radian per second squared. The motion is non uniform, so the angular acceleration is a function of time rather than a constant.

Hinglish: Position ko differentiate karo: omega = t^3/2 - t^2. Dobara differentiate: alpha = 3t^2/2 - 2t. t = 10 s par 150 - 20 = 130 rad/s^2. Motion non uniform hai, isliye angular acceleration time ka function hai.

Q6 (JEE Main 2026, 8 Apr Shift 2, 4 marks)

Two identical bodies are projected with the same speed at two different angles and cover the same horizontal range R. Their times of flight are 5 s and 10 s. The value of R is (take g = 10 m s^-2)

Answer: Equal ranges at the same speed need complementary angles theta and 90 minus theta. The times of flight are 2u sin theta over g and 2u cos theta over g, so u sin theta equals 25 and u cos theta equals 50. The range is 2u squared sin theta cos theta over g, which equals 2 times the product of those two quantities divided by g: 2 times 25 times 50 divided by 10, that is 250 m.

Hinglish: Same speed par same range ke liye angles complementary hote hain. Times of flight se u sin theta = 25 aur u cos theta = 50 milta hai. Range 2(u sin theta)(u cos theta)/g = 2 x 25 x 50/10 = 250 m.

Q7 (NCERT Exemplar, 1 mark)

The angle between the vectors A = i + j and B = i - j is (a) 45 degrees (b) 90 degrees (c) -45 degrees (d) 180 degrees

Answer: The dot product is the product of the coefficients of i plus the product of the coefficients of j: 1 times 1 plus 1 times minus 1, which is zero. A zero dot product between non-zero vectors means the vectors are perpendicular, so the angle is 90 degrees and option (b) is correct. The two vectors lie along the two diagonals of a square, and those diagonals cut at right angles.

Hinglish: Dot product 1x1 + 1x(-1) = 0 aata hai. Zero dot product ka matlab vectors perpendicular hain, isliye angle 90 degree aur option (b). Yeh do vectors square ke diagonals ki direction mein hain, jo right angle par milte hain.

Q8 (NCERT Exemplar, 1 mark)

Which one of the following statements is true? (a) A scalar is conserved during a process. (b) A scalar can never be negative. (c) A scalar stays the same from point to point in space. (d) A scalar reads the same for observers whose axes point in different directions.

Answer: A scalar is a single number with a unit that does not depend on the direction of the axes chosen. Options (a) and (c) confuse scalar with conserved and with uniform, and (b) is false because temperature and work can be negative. Option (d) is the definition that survives: rotating the axes leaves a scalar unchanged, while it changes the components of a vector.

Hinglish: Scalar ek number plus unit hota hai jo axes ki direction par depend nahi karta. (a) aur (c) scalar ko conserved ya uniform samajh rahe hain, aur (b) galat hai kyunki temperature aur work negative ho sakte hain. (d) sahi hai, kyunki axes ghumane par scalar nahi badalta, vector ke components badal jaate hain.

Q9 (NCERT Exemplar, 1 mark)

The horizontal range of a projectile fired at 15 degrees is 50 m. If it is fired with the same speed at 45 degrees, its range will be (a) 60 m (b) 71 m (c) 100 m (d) 141 m

Answer: The range is u squared sin 2 theta over g. At 15 degrees the double angle is 30 degrees, so 50 equals u squared over g times 0.5, which gives u squared over g equal to 100 m. At 45 degrees the double angle is 90 degrees and its sine is 1, so the range is exactly 100 m. The answer is option (c).

Hinglish: R = u^2 sin 2 theta / g. 15 degree par sin 30 = 0.5, isliye 50 = 0.5 u^2/g, matlab u^2/g = 100 m. 45 degree par sin 90 = 1, isliye range 100 m, option (c).

Q10 (NCERT Exemplar, 3 marks)

A fighter plane flies horizontally at an altitude of 1.5 km with a speed of 720 km per hour. At what angle of sight with the horizontal should the pilot release a bomb so that it hits a target?

Answer: Convert the speed: 720 km per hour is 200 m per second. The vertical fall from 1500 m takes the square root of 2 times 1500 over 9.8, about 17.5 s. In that time the bomb travels horizontally by 200 times 17.5, about 3500 m. The angle of sight below the horizontal satisfies tan theta equal to the horizontal distance over the altitude, 3500 over 1500, about 2.33, which gives theta about 66.8 degrees.

Hinglish: Speed 720 km/h = 200 m/s. 1500 m girne ka time root(2 x 1500/9.8) = 17.5 s. Itni der mein horizontal distance 200 x 17.5 = 3500 m. Angle of sight ke liye tan theta = 3500/1500 = 2.33, matlab theta lagbhag 66.8 degree below horizontal.