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Chapter 4: Problem 449

If linear momentum of body is increased by \(1.5 \%\), its kinetic energy increases by...... \(\%\) (A) \(0 \%\) (B) \(10 \%\) (C) \(2.25 \%\) (D) \(3 \%\)

Short Answer

Expert verified
The kinetic energy increases by approximately 3.02%. The answer is (D) \(3 \%\).

Step by step solution

01

Linear momentum is defined as the product of mass (m) and velocity (v): p = mv Kinetic energy is defined as the energy possessed by an object due to its motion, which is given as KE = \(0.5mv^2\) We are given that the linear momentum (p) increases by 1.5%. So, the new linear momentum = \(1.015p\) #Step 2: Express the new velocity in terms of old velocity #

Since we only deal with percentages, we consider a mass of 1 kg. So, the relationship between new and old linear momentum can be written as: \(1.015v = v'\), where v' is the new velocity. #Step 3: Calculate the new kinetic energy in terms of old kinetic energy #
02

The new kinetic energy (KE') can be calculated using the new velocity (v'): KE' = \(0.5m(v')^2\), here m = 1kg. Now we need to express KE' in terms of old kinetic energy (KE). Substitute \(1.015v\) for \(v'\) and use the expression for kinetic energy: KE' = \(0.5(1.015v)^2\) #Step 4: Calculate the percentage increase in kinetic energy #

The percentage increase in kinetic energy is given by: Percentage Increase = \(\frac{KE' - KE}{KE} \times 100\) Now substitute the expressions for KE and KE': Percentage Increase = \(\frac{0.5(1.015v)^2 - 0.5v^2}{0.5v^2} \times 100\) Simplify the equation: Percentage Increase = \(\frac{(1.015^2 - 1)v^2}{v^2} \times 100\) Percentage Increase = \(1.015^2 - 1\) \\ Percentage Increase = 0.030225 \\ Percentage Increase = 3.0225\% Thus, the kinetic energy increases by approximately 3.02%. The answer is (D) \(3 \%\).

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