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A rocket of initial mass 6000 kg ejects gases at a constant rate of 16 kgs-1 with constant relative speed of11 km s-1. What is the acceleration of the rocket one minute after the blast?
  • a)
    25 m s -2
  • b)
    50 m s -2
  • c)
    10 m s -2
  • d)
    35 m s -2
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
A rocket of initial mass 6000 kg ejects gases at a constant rate of 16...
Acceleration of the rocket at any instant t is

Here, M = 6000kg, dm/dt = 16kgs−1
Vr = 11kms-1 = 11km s−1 t = 1 min = 60s
∴ 
≈ 35 m s−2
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Most Upvoted Answer
A rocket of initial mass 6000 kg ejects gases at a constant rate of 16...
Acceleration of the rocket can be calculated using the principle of conservation of momentum. The momentum of the rocket before the ejection of gases is equal to the momentum of the rocket and gases after ejection.

Given:
Initial mass of the rocket (m1) = 6000 kg
Rate of ejection of gases (dm/dt) = 16 kg/s
Relative speed of ejection (v) = 11 km/s = 11000 m/s

Let the final mass of the rocket after one minute be mf and the final velocity of the rocket after one minute be vf.

Momentum before ejection = Momentum after ejection
(m1)(0) = (mf)(vf) + (dm/dt)(v)

Since the rocket starts from rest, the initial velocity is 0.

Simplifying the equation, we get:
0 = (mf)(vf) + (dm/dt)(v)

Now, we need to determine the final mass of the rocket after one minute. The rate of ejection of gases is given as 16 kg/s, so after one minute (60 seconds), the mass of the ejected gases is:
(mg) = (dm/dt)(t) = (16 kg/s)(60 s) = 960 kg

Therefore, the final mass of the rocket is:
mf = (m1) + (mg) = 6000 kg + 960 kg = 6960 kg

Substituting the values back into the equation, we have:
0 = (6960 kg)(vf) + (16 kg/s)(11000 m/s)

Simplifying the equation further:
0 = 6960vf + 176000

Rearranging the equation to solve for vf:
vf = -176000 / 6960

vf ≈ -25.29 m/s

Since the rocket is accelerating, the negative sign indicates that the velocity is in the opposite direction of the initial velocity.

Finally, we can calculate the acceleration using the equation:
Acceleration (a) = Change in velocity / Time taken

In one minute, the change in velocity is:
Δv = vf - vi = -25.29 m/s - 0 m/s = -25.29 m/s

The time taken is 60 seconds.

Therefore, the acceleration is:
a = -25.29 m/s / 60 s ≈ -0.42 m/s^2

Converting the negative sign to positive, the acceleration of the rocket one minute after the blast is approximately 0.42 m/s^2.

Since the options provided are in terms of positive values, the correct answer is option 'D' which is 35 m/s^2.
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A rocket of initial mass 6000 kg ejects gases at a constant rate of 16 kgs-1 with constant relative speed of11 km s-1. What is the acceleration of the rocket one minute after the blast?a)25 m s -2b)50 m s -2c)10 m s -2d)35 m s -2Correct answer is option 'D'. Can you explain this answer?
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A rocket of initial mass 6000 kg ejects gases at a constant rate of 16 kgs-1 with constant relative speed of11 km s-1. What is the acceleration of the rocket one minute after the blast?a)25 m s -2b)50 m s -2c)10 m s -2d)35 m s -2Correct answer is option 'D'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A rocket of initial mass 6000 kg ejects gases at a constant rate of 16 kgs-1 with constant relative speed of11 km s-1. What is the acceleration of the rocket one minute after the blast?a)25 m s -2b)50 m s -2c)10 m s -2d)35 m s -2Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A rocket of initial mass 6000 kg ejects gases at a constant rate of 16 kgs-1 with constant relative speed of11 km s-1. What is the acceleration of the rocket one minute after the blast?a)25 m s -2b)50 m s -2c)10 m s -2d)35 m s -2Correct answer is option 'D'. Can you explain this answer?.
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