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Work Power and Energy question

2022 · 28 Jul · Shift 2 · Q44
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Work Power and Energy question

2022 · 28 Jul · Shift 2 · Q44

JEE MainPhysicsWork Power and EnergyMCQ+4 / −1
A bullet of mass 200 g200 \mathrm{~g}200 g having initial kinetic energy 90 J90 \mathrm{~J}90 J is shot inside a long swimming pool as shown in the figure. If it's kinetic energy reduces to 40 J40 \mathrm{~J}40 J within 1 s1 \mathrm{~s}1 s, the minimum length of the pool, the bullet has to travel so that it completely comes to rest is JEE Main 2022 (Online) 28th July Evening Shift Physics - Work Power & Energy Question 59 English
  1. A
    45 m
  2. B
    90 m
  3. C
    125 m
  4. D
    25 m
View written solutionFree

Correct answer: A

  1. Given data
  • Mass of bullet: m=200 g=0.2 kgm = 200\text{ g} = 0.2\text{ kg}m=200 g=0.2 kg
  • Initial kinetic energy: K0=90 JK_0 = 90\text{ J}K0​=90 J
  • Kinetic energy after 1 s1\text{ s}1 s: K1=40 JK_1 = 40\text{ J}K1​=40 J

We need the minimum length of the pool so that the bullet comes to rest.

  1. Assumption for minimum length

Inside water, resistive force is taken to be constant so that retardation is constant. This is the standard interpretation in such work-energy questions.

Then,

  • loss of kinetic energy is proportional to distance,
  • motion is uniformly decelerated.
  1. Find initial speed

Using

K0=12mv02K_0 = \frac{1}{2}mv_0^2K0​=21​mv02​

So,

90=12(0.2)v02=0.1v0290 = \frac{1}{2}(0.2)v_0^2 = 0.1v_0^290=21​(0.2)v02​=0.1v02​ v02=900v_0^2 = 900v02​=900 v0=30 m/sv_0 = 30\text{ m/s}v0​=30 m/s
  1. Find speed after 1 s

Using

K1=12mv12K_1 = \frac{1}{2}mv_1^2K1​=21​mv12​

So,

40=0.1v1240 = 0.1v_1^240=0.1v12​ v12=400v_1^2 = 400v12​=400 v1=20 m/sv_1 = 20\text{ m/s}v1​=20 m/s
  1. Find retardation

Using

v=u+atv = u + atv=u+at

with u=30u=30u=30, v=20v=20v=20, t=1t=1t=1:

20=30+a(1)20 = 30 + a(1)20=30+a(1) a=−10 m/s2a = -10\text{ m/s}^2a=−10 m/s2

So the retardation magnitude is 10 m/s210\text{ m/s}^210 m/s2.

  1. Find total distance travelled before stopping

At stopping,

v=0v=0v=0

Use

v2=u2+2asv^2 = u^2 + 2asv2=u2+2as 0=302+2(−10)s0 = 30^2 + 2(-10)s0=302+2(−10)s 0=900−20s0 = 900 - 20s0=900−20s s=90020=45 ms = \frac{900}{20} = 45\text{ m}s=20900​=45 m

Thus the bullet needs 45 m45\text{ m}45 m of water to come to rest.

  1. Check options
  • A: 45 m45\text{ m}45 m ✅
  • B: 90 m90\text{ m}90 m ❌
  • C: 125 m125\text{ m}125 m ❌
  • D: 25 m25\text{ m}25 m ❌

Hence the correct option is A.

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