Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Motion in A Plane question

2025 · 22 Jan · Shift 2 · Q68
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Physics
  4. /Motion in A Plane
  5. /2025 · 22 Jan · Shift 2 · Q68

Motion in A Plane question

2025 · 22 Jan · Shift 2 · Q68

JEE MainPhysicsMotion in A PlaneMCQ+4 / −1
A ball of mass 100 g is projected with velocity 20 m/s20 \mathrm{~m} / \mathrm{s}20 m/s at 60∘60^{\circ}60∘ with horizontal. The decrease in kinetic energy of the ball during the motion from point of projection to highest point is
  1. A
    20 J
  2. B
    5 J
  3. C
    15 J
  4. D
    zero
View written solutionFree

Correct answer: C

  1. Given data
  • Mass of ball: m=100 g=0.1 kgm = 100\text{ g} = 0.1\text{ kg}m=100 g=0.1 kg
  • Initial speed: u=20 m/su = 20\text{ m/s}u=20 m/s
  • Angle of projection: θ=60∘\theta = 60^\circθ=60∘

We need the decrease in kinetic energy from the point of projection to the highest point.

  1. Initial kinetic energy

At projection,

Ki=12mu2=12(0.1)(20)2K_i = \frac{1}{2}mu^2 = \frac{1}{2}(0.1)(20)^2Ki​=21​mu2=21​(0.1)(20)2 Ki=0.05×400=20 JK_i = 0.05 \times 400 = 20\text{ J}Ki​=0.05×400=20 J
  1. Velocity at the highest point

In projectile motion, at the highest point the vertical component becomes zero, while the horizontal component remains unchanged.

Horizontal component of initial velocity:

ux=ucos⁡60∘=20×12=10 m/su_x = u\cos 60^\circ = 20 \times \frac{1}{2} = 10\text{ m/s}ux​=ucos60∘=20×21​=10 m/s

So speed at highest point is 10 m/s10\text{ m/s}10 m/s.

  1. Kinetic energy at the highest point
Kh=12mv2=12(0.1)(10)2K_h = \frac{1}{2}m v^2 = \frac{1}{2}(0.1)(10)^2Kh​=21​mv2=21​(0.1)(10)2 Kh=0.05×100=5 JK_h = 0.05 \times 100 = 5\text{ J}Kh​=0.05×100=5 J
  1. Decrease in kinetic energy
ΔK=Ki−Kh=20−5=15 J\Delta K = K_i - K_h = 20 - 5 = 15\text{ J}ΔK=Ki​−Kh​=20−5=15 J
  1. Option check
  • A: 20 J20\text{ J}20 J ❌
  • B: 5 J5\text{ J}5 J ❌
  • C: 15 J15\text{ J}15 J ✅
  • D: zero ❌

Therefore, the correct answer is Option C.

PreviousNext

More from Motion in A Plane

  • An object of mass ' m ' is projected from origin in a vertical xy plane at an angle 45∘ with the x− axis with an initial velocity v0​. The magnitude and direction of the angular momentum of the object with…2025 · MCQ
  • The position vector of a moving body at any instant of time is given as r=(5t2i^−5tj^​)m. The magnitude and direction of velocity at t=2s is,2025 · MCQ
  • Two projectiles are fired with same initial speed from same point on ground at angles of (45∘−α) and (45∘+α), respectively, with the horizontal direction. The ratio of their maximum heights attained is :2025 · MCQ
  • The maximum speed of a boat in still water is 27 km/h. Now this boat is moving downstream in a river flowing at 9 km/h. A man in the boat throws a ball vertically upwards with speed of 10 m/s. Range of the ball as observed by an observer…2025 · Numerical
  • The co-ordinates of a particle moving in x-y plane are given by : x=2+4t,y=3t+8t2. The motion of the particle is :2024 · MCQ
  • The maximum height reached by a projectile is 64 m. If the initial velocity is halved, the new maximum height of the projectile is ​m.2024 · Numerical
  • The angle of projection for a projectile to have same horizontal range and maximum height is :2024 · MCQ
  • A body of mass M thrown horizontally with velocity v from the top of the tower of height H touches the ground at a distance of 100 m from the foot of the tower. A body of mass 2 M thrown at a velocity 2v​…2024 · Numerical