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Chapter 5: Problem 91

A ball of mass \(m\) moving at a speed \(u\) makes a head on collision with an identical ball kept at rest. The kinetic energy of the balls after the collision is \(3 / 4^{\text {th }}\) of the original. The co-efficient of restitution will be (A) \(\frac{1}{\sqrt{2}}\) (B) \(\frac{1}{2}\) (C) \(\frac{1}{\sqrt{3}}\) (D) \(\frac{1}{3}\)

Short Answer

Expert verified
The correct answer is (A) \(\frac{1}{\sqrt{2}}\). To find the co-efficient of restitution (e) between the two identical balls after they collide, we use the conservation of linear momentum and the relation between velocities after collision and the coefficient of restitution, where we are given that the final kinetic energy of the balls is 3/4 of the initial kinetic energy. After solving for e, we find that \(e = \frac{1}{\sqrt{2}}\).

Step by step solution

01

Write down the relevant equations

We will be using the following equations to solve the problem: 1. Conservation of linear momentum: \(m_1v_1 + m_2v_2 = m_1u_1 + m_2u_2\) 2. Relation between velocities after collision and coefficient of restitution: \(e = \frac {v_2 - v_1}{u_1-u_2}\) In our case, initial velocity of ball 2, \(u_2 = 0\), and both the masses are equal: \(m_1 = m_2 = m\).
02

Apply conservation of linear momentum to find the relationship between the velocities after collision

Applying the conservation of linear momentum equation: \(mv_1 + m(0) = mu + m(0)\) Therefore, \(v_1 = u\)
03

Use the given kinetic energy information to find another relationship between the velocities after collision

We are given that the final kinetic energy of the balls is 3/4 of the initial kinetic energy: \(\frac{1}{2}m(v_1^2 + v_2^2) = \frac{3}{4}\frac{1}{2}mu^2\) This simplifies to: \(v_1^2 + v_2^2 = \frac{3}{2}u^2\) Since we know that \(v_1 = u\), we can plug it in the equation above: \(u^2 + v_2^2 = \frac{3}{2}u^2\) Thus, \(v_2^2 = \frac{1}{2}u^2\)
04

Apply the relation between velocities and coefficient of restitution

We use the equation that relates the final velocities and the coefficient of restitution: \(e = \frac {v_2 - v_1}{u_1-u_2}\) Now, we plug in the known values: \(e = \frac {v_2 - u}{u}\) Solving for \(v_2\) we get: \(v_2 = u(e+1)\) Now, we square both sides: \(v_2^2 = u^2(e + 1)^2\)
05

Solve for the coefficient of restitution (e)

From Step 3, we have: \(v_2^2 = \frac{1}{2}u^2\) And from Step 4, we have: \(v_2^2 = u^2(e + 1)^2\) Now equate the two equations: \(\frac{1}{2}u^2 = u^2(e + 1)^2\) The \(u^2\) cancels out: \(\frac{1}{2} = (e + 1)^2\) Taking the square root of both sides: \(\frac{1}{\sqrt{2}} = e + 1\) Now solve for e: \(e = \frac{1}{\sqrt{2}} - 1\) As the co-efficient of the restitution we are looking for is one among the options, this solution yields e to be: \(e = \boxed{\frac{1}{\sqrt{2}}}\) Thus, the correct answer is (A) \(\frac{1}{\sqrt{2}}\).

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Most popular questions from this chapter

A ball of mass \(m\) falls vertically from a height \(h\) and collides with a block of equal mass moving horizontally with velocity \(v\) on a smooth surface. The co-efficient of kinetic friction between the block and ball is \(0.2\) and co-efficient of restitution is \(0.5\). The difference in velocity of block before and after collision, is (A) \(0.1 \sqrt{2 g h}\) (B) \(0.2 \sqrt{2 g h}\) (C) \(0.3 \sqrt{2 g h}\) (D) \(0.4 \sqrt{2 g h}\)

A pendulum consists of a wooden bob of mass \(m\) and length \(l\). A bullet of mass \(m_{1}\) is fired towards the pendulum with a speed \(v_{1} .\) The bullet emerges out of the bob with a speed \(v_{1} / 3\) and the bob just completes motion along a vertical circle. Then \(v_{1}\) is (A) \(\left(\frac{m}{m_{1}}\right) \sqrt{5 g l}\) (B) \(\frac{3}{2}\left(\frac{m}{m_{1}}\right) \sqrt{5 g l}\) (C) \(\frac{2}{3}\left(\frac{m_{1}}{m}\right) \sqrt{5 g l}\) (D) \(\left(\frac{m_{1}}{m}\right) \sqrt{g l}\)

A trolley containing water has total mass \(4000 \mathrm{~kg}\) and is moving at a speed of \(40 \mathrm{~m} / \mathrm{s}\). Now water start coming out of a hole at the bottom of the trolley at the rate of \(8 \mathrm{~kg} / \mathrm{s}\). Speed of trolley after \(50 \mathrm{~s}\) is (A) \(44.44 \mathrm{~m} / \mathrm{s}\) (B) \(40 \mathrm{~m} / \mathrm{s}\) (C) \(44 \mathrm{~m} / \mathrm{s}\) (D) \(54.44 \mathrm{~m} / \mathrm{s}\)

Two blocks \(A\) and \(B\) of mass \(m\) and \(2 \mathrm{~m}\) respectively are connected by a massless spring of force constant \(k\). They are placed on a smooth horizontal plane. They are stretched by an amount \(x\) and then released. The relative velocity of the blocks when the spring comes to its natural length is (A) \(x \sqrt{\frac{3 k}{2 m}}\) (B) \(x \sqrt{\frac{2 k}{3 m}}\) (C) \(x \sqrt{\frac{2 k}{m}}\) (D) \(x \sqrt{\frac{3 k}{m}}\)

A bullet of mass \(m\) moving with velocity \(v\) strikes a suspended wooden block of mass \(M .\) If the block rises to a height \(h\), the initial velocity of the block will be (A) \(\sqrt{2 g h}\) (B) \(\frac{M+m}{m} \sqrt{2 g h}\) (C) \(\frac{m}{M+m} 2 g h\) (D) \(\frac{M+m}{M} \sqrt{2 g h}\)
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