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MCQEasyJEE 2025Projectile Motion

JEE Physics 2025 Question with Solution

A ball of mass 100 g100 \, \text{g} is projected with velocity 20 m/s20 \, \text{m/s} at 60∘60^\circ with horizontal. The decrease in kinetic energy of the ball during the motion from point of projection to highest point is:

  • A

    Zero

  • B

    5 J5 \, \text{J}

  • C

    20 J20 \, \text{J}

  • D

    15 J15 \, \text{J}

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: Mass of ball, m=100 g=0.1 kgm = 100 \, \text{g} = 0.1 \, \text{kg}, velocity, u=20 m/su = 20 \, \text{m/s}, and angle with horizontal, θ=60∘\theta = 60^\circ.

Find: The decrease in kinetic energy from the point of projection to the highest point.

At the highest point of projectile motion, the vertical component of velocity becomes zero, while the horizontal component remains unchanged.

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

Initial kinetic energy:

Ki=12mu2=12×0.1×(20)2=20 JK_i = \frac{1}{2} m u^2 = \frac{1}{2} \times 0.1 \times (20)^2 = 20 \, \text{J}

Kinetic energy at the highest point:

Kf=12mvx2=12×0.1×(10)2=5 JK_f = \frac{1}{2} m v_x^2 = \frac{1}{2} \times 0.1 \times (10)^2 = 5 \, \text{J}

Therefore, the decrease in kinetic energy is

ΔK=Ki−Kf=20−5=15 J\Delta K = K_i - K_f = 20 - 5 = 15 \, \text{J}

Therefore, the decrease in kinetic energy is 15 J15 \, \text{J}. The solution concludes this value, although the answer key listing marks option (2), which is inconsistent with the working. Hence the correct option is D.

Energy Change from Lost Vertical Component

Given: m=0.1 kgm = 0.1 \, \text{kg}, u=20 m/su = 20 \, \text{m/s}, θ=60∘\theta = 60^\circ.

Find: The loss in kinetic energy up to the highest point.

The decrease in kinetic energy is due only to the loss of the vertical component of velocity.

uy=usin⁡60∘=20×32=103 m/su_y = u \sin 60^\circ = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3} \, \text{m/s}

So the lost kinetic energy is

ΔK=12muy2\Delta K = \frac{1}{2} m u_y^2ΔK=12×0.1×(103)2\Delta K = \frac{1}{2} \times 0.1 \times (10\sqrt{3})^2ΔK=0.05×300=15 J\Delta K = 0.05 \times 300 = 15 \, \text{J}

Therefore, the decrease in kinetic energy is 15 J15 \, \text{J}, so the correct option is D.

Common mistakes

  • Using the kinetic energy at the highest point as the required answer. 5 J5 \, \text{J} is the remaining kinetic energy, not the decrease. Subtract the final kinetic energy from the initial kinetic energy.

  • Assuming the velocity becomes zero at the highest point. Only the vertical component becomes zero there; the horizontal component remains constant.

  • Not resolving the initial velocity into horizontal and vertical components. In projectile motion, the unchanged horizontal component determines the speed at the highest point.

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