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Work Power and Energy question

2022 · 29 Jul · Shift 1 · Q52
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  5. /2022 · 29 Jul · Shift 1 · Q52

Work Power and Energy question

2022 · 29 Jul · Shift 1 · Q52

JEE MainPhysicsWork Power and EnergyMCQ+4 / −1
A ball is projected with kinetic energy E, at an angle of 60∘60^{\circ}60∘ to the horizontal. The kinetic energy of this ball at the highest point of its flight will become :
  1. A
    Zero
  2. B
    E2\frac{E}{2}2E​
  3. C
    E4\frac{E}{4}4E​
  4. D
    E
View written solutionFree

Correct answer: C

  1. Initial kinetic energy

Let the mass of the ball be mmm and initial speed be uuu.

Then its initial kinetic energy is

E=12mu2E = \frac{1}{2}mu^2E=21​mu2
  1. Resolve the initial velocity into components

The ball is projected at an angle 60∘60^\circ60∘ to the horizontal.

So,

ux=ucos⁡60∘=u2u_x = u\cos 60^\circ = \frac{u}{2}ux​=ucos60∘=2u​ uy=usin⁡60∘=3u2u_y = u\sin 60^\circ = \frac{\sqrt{3}u}{2}uy​=usin60∘=23​u​
  1. Velocity at the highest point

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

Hence, speed at the highest point is

v=ucos⁡60∘=u2v = u\cos 60^\circ = \frac{u}{2}v=ucos60∘=2u​
  1. Kinetic energy at the highest point

Therefore,

Ktop=12m(u2)2=12m⋅u24=18mu2K_{\text{top}} = \frac{1}{2}m\left(\frac{u}{2}\right)^2 = \frac{1}{2}m\cdot \frac{u^2}{4} = \frac{1}{8}mu^2Ktop​=21​m(2u​)2=21​m⋅4u2​=81​mu2

But,

E=12mu2E = \frac{1}{2}mu^2E=21​mu2

So,

Ktop=14EK_{\text{top}} = \frac{1}{4}EKtop​=41​E
  1. Check options
  • A: Zero →\rightarrow→ Incorrect
  • B: E2\frac{E}{2}2E​ →\rightarrow→ Incorrect
  • C: E4\frac{E}{4}4E​ →\rightarrow→ Correct
  • D: EEE →\rightarrow→ Incorrect

Therefore, the kinetic energy at the highest point is

E4\boxed{\frac{E}{4}}4E​​
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