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Chapter 4: Problem 511

If a skater of weight \(4 \mathrm{~kg}\) has initial speed $4 \mathrm{~m} / \mathrm{s}\( and \)2^{\text {nd }}\( one of weight \)6 \mathrm{~kg}\( has \)6 \mathrm{~m} / \mathrm{s}\(. After collision, they have speed (couple) \)6 \mathrm{~m} / \mathrm{s}\(. Then the loss in \)\mathrm{K} . \mathrm{E}$. is..... (A) \(48 \mathrm{~J}\) (B) zero (C) \(96 \mathrm{~J}\) (D) None of these

Short Answer

Expert verified
The loss in Kinetic Energy is -40 J, which indicates an increase in Kinetic Energy by 40 J. Hence, the correct option is (D) None of these.

Step by step solution

01

Calculate the initial kinetic energy of each skater

We can find the kinetic energy of each skater using the formula for kinetic energy: K.E. = \(0.5 \cdot m \cdot v^2\), where m is the mass of the skater, and v is the velocity of the skater. For skater 1: Mass m1 = 4 kg Velocity v1 = 4 m/s K.E.1 = 0.5 * m1 * \(v1^2\) = 0.5 * 4 * \(4^2\) = 32 J For skater 2: Mass m2 = 6 kg Velocity v2 = 6 m/s K.E.2 = 0.5 * m2 * \(v2^2\) = 0.5 * 6 * \(6^2\) = 108 J
02

Calculate the final kinetic energy of the couple

Now let's calculate the final K.E. of the couple after the collision. We need to find their combined mass and use the couple's velocity mentioned in the problem. Total mass of the couple, M = m1 + m2 = 4 kg + 6 kg = 10 kg Velocity of the couple, V = 6 m/s Final K.E. =0.5 * M * \(V^2\) = 0.5 * 10 * \(6^2\) = 180 J
03

Calculate the loss in kinetic energy

To find the loss in kinetic energy, we subtract the final K.E. from the sum of the initial K.E.s of each skater: Loss in K.E. = (K.E.1 + K.E.2) - Final K.E. Loss in K.E. = (32 J + 108 J) - 180 J = 140 J - 180 J = -40 J Since the loss in K.E. is negative, this means that there is actually an increase in the K.E. of 40 J. Therefore, the correct answer is: (D) None of these.

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