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Chapter 5: Problem 74

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) \(\left(\frac{m_{1}}{m}\right) \sqrt{g l}\)

Short Answer

Expert verified
The short answer to the question is: \(v_1 = \frac{3}{2}\left(\frac{m}{m_1}\right) \sqrt{5 g l}\) which corresponds to option (B).

Step by step solution

01

Conservation of Momentum Before Collision

Before the collision between the bullet and the wooden bob, we can apply the conservation of momentum for the given system. The total momentum before and after the collision should be the same. Initial Momentum: \(m_1 v_1\) After the collision, the bullet and the wooden bob move together: Final Momentum: \(m_1 (\frac{v_1}{3}) + m(v_1 - \frac{v_1}{3})\) By conservation of momentum, we equate the initial and final momenta: \(m_1 v_1 = m_1 (\frac{v_1}{3}) + m(v_1 - \frac{v_1}{3})\)
02

Solving for the Velocity of the Wooden Bob

Next, we need to solve for the velocity of the wooden bob after the collision. Let this velocity be denoted by \(v\). From the equation obtained in Step 1, solve for \(v\): \(m_1 v_1 = m_1 (\frac{v_1}{3}) + m(v - \frac{v_1}{3})\) \(v = \frac{2v_1}{3}(\frac{m_1}{m})\)
03

Energy Conservation

Now we need to apply energy conservation to determine the conditions for the wooden bob to complete exactly one vertical circle. At the bottom of the circle when the bob is moving: Initial Energy: Kinetic Energy + Potential Energy = \(\frac{1}{2}(m + m_1)v^2 + 0\) At the top of the circle: Final Energy: Kinetic Energy + Potential Energy = \(0 + (m + m_1)gl\) By conservation of energy, we equate the initial and final energies: \(\frac{1}{2}(m + m_1)v^2 = (m + m_1)gl\)
04

Condition for Completion of Vertical Circle

For the bob to complete exactly one vertical circle, the tension in the string must reduce to zero at the top of the circle. Thus, we can simplify the equation in Step 3 as follows: \(\frac{1}{2}v^2 = gl\) Substitute the value of \(v\) from Step 2: \(\frac{1}{2}\left(\frac{2v_1}{3}(\frac{m_1}{m})\right)^2 = gl\)
05

Solving for \(v_1\)

Finally, we can solve the equation from Step 4 for \(v_1\): \(\frac{1}{2}\left(\frac{2v_1}{3}(\frac{m_1}{m})\right)^2 = gl\) \(v_1 = \frac{3}{2}\left(\frac{m}{m_1}\right) \sqrt{5 g l}\) Comparing this solution to the given options, the correct answer is: (B) \(\frac{3}{2}\left(\frac{m}{m_1}\right) \sqrt{5 g l}\)

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Most popular questions from this chapter

A ball is dropped from a height \(h\). It rebounds from the ground a number of times. Given that the coefficient of restitution is \(e\), to what height does it go after \(n\)th rebounding? (A) \(h / e^{n}\) (B) \(h / e^{2 n}\) (C) \(h e^{n}\) (D) \(h e^{2 n}\)

A ball hits a floor and rebounds after an inelastic collision. In this case (A) the momentum of the ball just after the collision is the same as that just before the collision. (B) the mechanical energy of the ball remains the same is the collision. (C) the total momentum of the ball and the earth is conserved. (D) the total energy of the ball and the earth is conserved.

A plate remains in equilibrium in air when \(n\) bullets are fired per second on it. The mass of each bullet is \(m\) and it strikes the plate with speed \(v\). The coefficient of restitution is \(e=0.5\). The mass of the plate is (A) \(M=\frac{3 m n v}{2 g}\) (B) \(M=\frac{m n v}{g}\) (C) \(M=\frac{2 m n v}{3 g}\) (D) \(M=\frac{2 m n v}{g}\)

A ball of mass \(1 \mathrm{~kg}\) strikes a wedge of mass \(4 \mathrm{~kg}\) horizontally with a velocity of \(10 \mathrm{~m} / \mathrm{s}\). Just after collision velocity of wedge becomes \(4 \mathrm{~m} / \mathrm{s}\). Friction is absent everywhere and collision is elastic. Then (A) the speed of ball after collision is \(6 \mathrm{~m} / \mathrm{s}\). (B) the speed of ball after collision is \(8 \mathrm{~m} / \mathrm{s}\). (C) the speed of ball after collision is \(4 \mathrm{~m} / \mathrm{s}\). (D) the speed of ball after collision is \(10 \mathrm{~m} / \mathrm{s}\).

A body of mass \(m_{1}\) makes a head on collision perfectly elastic with a body of mass \(m_{2}\) initially at rest. (i) What fraction of initial energy of mass \(m_{1}\) is lost in collision? (ii) For what ratio of \(\frac{m_{2}}{m_{1}}=\eta\), the fraction of energy loss is maximum?
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