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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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Most popular questions from this chapter

A ball is allowed to fall from a height \(20 \mathrm{~m}\). If there is \(30 \%\) loss of energy due to impact, then after one impact ball will go up to (A) \(18 \mathrm{~m}\) (B) \(16 \mathrm{~m}\) (C) \(12 \mathrm{~m}\) (D) \(14 \mathrm{~m}\)

A billiard ball moving with a speed of \(8 \mathrm{~m} / \mathrm{s}\) collides with an identical ball originally at rest. If the first ball stops after collision, then the second ball will move forward with a speed of .... (elastic collision) (A) \(8 \mathrm{~m} / \mathrm{s}\) (B) \(4 \mathrm{~m} / \mathrm{s}\) (C) \(16 \mathrm{~m} / \mathrm{s}\) (D) \(1.0 \mathrm{~m} / \mathrm{s}\)

Three objects \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) are kept in a straight line on a frictionless horizontal surface. These have masses $\mathrm{m}, 2 \mathrm{~m}\( and \)\mathrm{m}\( respectively. The object \)\mathrm{A}$ moves towards \(\mathrm{B}\) with a speed \(9 \mathrm{~m} / \mathrm{s}\) and makes an elastic collision with it. Thereafter, B makes completely inelastic collision with \(\mathrm{C}\). All motion occur on the same straight line. Find final speed of the object \(\mathrm{C}\) (A) \(3 \mathrm{~m} / \mathrm{s}\) (B) \(4 \mathrm{~m} / \mathrm{s}\) (C) \(5 \mathrm{~m} / \mathrm{s}\) (D) \(1 \mathrm{~m} / \mathrm{s}\)

A sphere of mass \(\mathrm{m}\) moving the velocity \(\mathrm{v}\) enters a hanging bag of sand and stops. If the mass of the bag is \(\mathrm{M}\) and it is raised by height \(\mathrm{h}\), then the velocity of the sphere was (A) \([\\{\mathrm{m}+\mathrm{M}\\} / \mathrm{m}] \sqrt{(2 \mathrm{gh})}\) (B) \((\mathrm{M} / \mathrm{m}) \sqrt{(} 2 \mathrm{gh})\) (C) \([\mathrm{m} /\\{\mathrm{M}+\mathrm{m}\\}] \sqrt{(2 \mathrm{gh})}\) (D) \((\mathrm{m} / \mathrm{M}) \sqrt{(2 \mathrm{gh})}\)

A bomb of mass \(3.0 \mathrm{~kg}\) explodes in air into two pieces of masses \(2.0 \mathrm{~kg}\) and \(1.0 \mathrm{~kg}\). The smaller mass goes at a speed of \(80 \mathrm{~m} / \mathrm{s}\). The total energy imparted to the two fragments is (A) \(1.07 \mathrm{KJ}\) (B) \(2.14 \mathrm{KJ}\) (C) \(2.4 \mathrm{KJ}\) (D) \(4.8 \mathrm{KJ}\)
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