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Question

If both mass and speed of a ball are doubled, the kinetic energy becomes ______ times.

This question was previously asked in
RRB ALP 2018 CBT 2 Fitter Question Paper (21-Jan-2019) (Shift 3)
The correct answer is
8

Understanding Kinetic Energy Calculation

This question involves calculating the change in kinetic energy when both the mass and speed of an object are altered.

Kinetic Energy Formula

The formula for kinetic energy (KE) is given by:

\(KE = \frac{1}{2}mv^2\)

Where:

  • \(m\) is the mass of the object.
  • \(v\) is the speed of the object.

Analyzing the Changes

Let the initial mass be \(m_1\) and the initial speed be \(v_1\). The initial kinetic energy (\(KE_1\)) is:

\(KE_1 = \frac{1}{2}m_1v_1^2\)

The problem states that both the mass and speed are doubled. So, the new mass (\(m_2\)) and new speed (\(v_2\)) are:

  • New mass: \(m_2 = 2m_1\)
  • New speed: \(v_2 = 2v_1\)

Calculating the New Kinetic Energy

The new kinetic energy (\(KE_2\)) is calculated using the formula with the new values:

\(KE_2 = \frac{1}{2}m_2v_2^2\)

Substitute the values of \(m_2\) and \(v_2\):

\(KE_2 = \frac{1}{2}(2m_1)(2v_1)^2\)

Simplify the expression:

\(KE_2 = \frac{1}{2}(2m_1)(4v_1^2)\)

\(KE_2 = \frac{1}{2}(8m_1v_1^2)\)

Comparing Kinetic Energies

Now, relate the new kinetic energy (\(KE_2\)) to the original kinetic energy (\(KE_1\)):

\(KE_2 = 8 \times \left(\frac{1}{2}m_1v_1^2\right)\)

Since \(KE_1 = \frac{1}{2}m_1v_1^2\), we can substitute \(KE_1\) into the equation:

\(KE_2 = 8 \times KE_1\)

Therefore, when both the mass and speed of a ball are doubled, its kinetic energy becomes 8 times its original value.

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Important Questions from Kinetic Energy

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  2. In an experiment, the velocity of a non-relativistic neutron is determined by measuring the time (~50 ns) it takes to travel from the source to the detector kept at a distance L. Assume that the error in the measurement of L is negligibly small. If we want to estimate the kinetic energy T of the neutron to within 5% accuracy, i.e.,   IδT / T I ≤ 0.05, the maximum permissible error Iδt Iin measuring the time of flight is nearest to

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