/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 A particle of mass \(m\) moving ... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 16

A particle of mass \(m\) moving with a speed \(v\) hits elastically another identical and stationary particle inside a smooth horizontal circular tube of radius \(r\). The time in which the next collision will take place is equal to (A) \(\frac{2 \pi r}{v}\) (B) \(\frac{4 \pi r}{v}\) (C) \(\frac{3 \pi r}{2 v}\) (D) \(\frac{\pi r}{v}\)

Short Answer

Expert verified
The time in which the next collision will take place is equal to \(\frac{2 \pi r}{v}\).

Step by step solution

01

Understand the initial conditions

Initially one particle of mass \(m\) moves with speed \(v\) inside the smooth horizontal circular tube of radius \(r\). Another identical stationary particle is inside the tube too. The collision is elastic.
02

Conservation of momentum

For an elastic collision, both momentum and the kinetic energy of the system are conserved. Let the initial velocity of the first (moving) particle be \(v_1 = v\), and the initial velocity of the second (stationary) particle be \(v_2 = 0\). As they have the same mass, we get the conservation of momentum equation as: \[m(v_1 - u_1) = m(u_2 - v_2)\]
03

Conservation of kinetic energy

The conservation of kinetic energy equation is: \[\frac{1}{2}m(v_1^2 - u_1^2) = \frac{1}{2}m(u_2^2 - v_2^2)\]
04

Solve for final velocities

Solve the system of equations formed by conservation of momentum and conservation of kinetic energy to find the final velocities \(u_1\) and \(u_2\) of the two particles: \[u_1 = \frac{v_1-v_2}{2} \quad \text{and} \quad u_2 = \frac{v_1 + v_2}{2}\] Now plug in the initial velocities \(v_1\) and \(v_2\) to get: \[u_1 = \frac{v-0}{2} =\frac{v}{2} \quad \text{and} \quad u_2 = \frac{v+0}{2}= \frac{v}{2}\]
05

Find the time for the next collision

Both particles now move with the same speed \(u_1 = u_2 = \frac{v}{2}\). To find out when the next collision will occur, we need to know the distance they need to travel in order to collide. Since the particles are identical in mass, upon collision, they will exchange their velocities, which means that for the stationary particle to hit the initially moving particle, it has to complete half of the circular path. The distance they need to travel is half the circumference of the circular tube, which is given by \(\frac{1}{2}(2\pi r) = \pi r\). Using the distance, speed, and time relationship: \[t = \frac{\text{distance}}{\text{speed}}\]
06

Calculate the time for the next collision

Plug the values for distance and speed into the above equation to find the time of the next collision: \[t = \frac{\pi r}{\frac{v}{2}} = \frac{2\pi r}{v}\] So, the time in which the next collision will take place is equal to \(\frac{2 \pi r}{v}\). Therefore, the correct answer is choice (A).

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Most popular questions from this chapter

A bomb of mass \(3 m \mathrm{~kg}\) explodes into two pieces of mass \(m \mathrm{~kg}\) and \(2 \mathrm{~m} \mathrm{~kg}\). If the velocity of \(\mathrm{m} \mathrm{kg}\) mass is \(16 \mathrm{~m} / \mathrm{s}\), the total energy released in the explosion is (A) \(192 \mathrm{~m} \mathrm{~J}\) (B) \(96 \mathrm{~m} \mathrm{~J}\) (C) \(384 \mathrm{~m} \mathrm{~J}\) (D) \(768 \mathrm{~m} \mathrm{~J}\)

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Consider a block \(P\) of mass \(2 M\) placed on another block \(Q\) of mass \(4 M\) lying on a fixed rough horizontal surface of coefficient of friction \(\mu\) between this surface and the block \(Q\). Surfaces of \(P\) and \(Q\) interacting with each other are smooth. \(A\) mass \(M\) moving in the horizontal direction along a line passing through the centre of mass of block \(Q\) is normal to the face. Speed of mass \(M\) is \(v_{0}\) when it collides elastically with the block \(Q\) at a height \(s\) from the rough surface. Both the blocks have same length \(4 s\). Velocity \(v_{0}\) is enough to make the block \(P\) topple. When mass \(M\) collides with block \(Q\) with velocity \(v_{0}\) the block \(P\). (A) Does not move (B) Moves forward (C) Moves backward (D) Either (B) or (C)
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