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Chapter 5: Problem 70

Two identical billiard balls are in contact on a table. A third identical ball strikes them symmetrically with velocity \(v\) and remains at rest after impact. The speed of balls after collision will be (A) \(\frac{v}{\sqrt{3}}\) (B) \(\frac{v}{3}\) (C) \(\frac{v}{2}\) (D) \(v\)

Short Answer

Expert verified
The speed of balls 2 and 3 after the collision is (C) \(\frac{v}{2}\).

Step by step solution

01

1. Identifying the given information and unknowns

We are given the following information: 1. Three identical billiard balls 2. The initial velocity of ball 1: \(v_1 = v\) 3. The initial velocities of ball 2 and ball 3: \(v_2 = 0 \, \text{and} \, v_3 = 0\). 4. After the collision, ball 1 comes to rest: \(v_1' = 0\) We need to find the final velocities of ball 2 and ball 3 after the collision.
02

2. Applying conservation of momentum

The total momentum before the collision is equal to the total momentum after the collision. Total momentum before collision: \(m_1v + m_2(0) + m_3(0)\) Total momentum after collision: \(m_1(0) + m_2v_2' + m_3v_3'\) Since all the balls have the same mass, we can denote the mass as "m". Conserving momentum: \(m_1v = m_2v_2' + m_3v_3'\) Since the balls are identical, we can simplify this to: \(mv = mv_2' + mv_3'\) Now, we can cancel the mass (m) from both sides: \(v = v_2' + v_3'\) (1)
03

3. Applying conservation of kinetic energy

The total kinetic energy before the collision is equal to the total kinetic energy after the collision. Total kinetic energy before collision: \(\frac{1}{2}m_1v^2\) Total kinetic energy after collision: \(\frac{1}{2}m_2{v_2'}^2 + \frac{1}{2}m_3{v_3'}^2\) Since the balls are identical, we can simplify this to: \(\frac{1}{2}mv^2 = \frac{1}{2}m({v_2'}^2 + {v_3'}^2)\) Now, we can cancel the factors \( \frac{1}{2} \) and m from both sides: \(v^2 = {v_2'}^2 + {v_3'}^2\) (2)
04

4. Combining the two equations

We have two equations (1) and (2) from the conservation of momentum and energy: \(v = v_2' + v_3'\) (1) \(v^2 = {v_2'}^2 + {v_3'}^2\) (2) Now, let's square equation (1): \(v^2 = (v_2' + v_3')^2\) \(v^2 = {v_2'}^2 + 2v_2'v_3' + {v_3'}^2\) Comparing with equation (2), we can deduce: \(2v_2'v_3' = 0\) Since both \(v_2'\) and \(v_3'\) are velocities of identical balls after collision, either one is equal to 0, and the other is equal to v, which is not possible because it means that ball 1 transferred all its energy and momentum to only one ball. Therefore, the collision must be symmetric, and balls 2 and 3 must move with the same speed: \(v_2' = v_3'\) (3)
05

5. Solving for the final speed

Now we will use equations (1) and (3) to find the final speed of balls 2 and 3. \(v = v_2' + v_3'\) \(v = 2v_2'\) (because \(v_2' = v_3'\)) Now, we will solve for \(v_2'\) (and similarly for \(v_3'\)): \(v_2' = \frac{v}{2}\) So the speed of balls 2 and 3 after the collision is \(\frac{v}{2}\). Thus, the correct answer is (C) \(\frac{v}{2}\).

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Most popular questions from this chapter

A particle of mass \(m\) moving in the \(x\)-direction with speed \(2 v\) is hit by another particle of mass \(2 m\) moving in the \(y\) direction with speed \(v\). If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to (A) \(50 \%\) (B) \(56 \%\) (C) \(62 \%\) (D) \(42 \%\)

A 20 gm bullet moving with speed \(v\) gets embedded in a \(10 \mathrm{~kg}\) block suspended from the ceiling by a massless rope. The block and the bullet swing to a height of \(45 \mathrm{~cm}\) above the equilibrium position. The initial speed of the bullet is \(\left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right)\) (A) \(1000 \mathrm{~ms}^{-1}\) (B) \(1100 \mathrm{~ms}^{-1}\) (C) \(1500 \mathrm{~ms}^{-1}\) (D) \(1503 \mathrm{~ms}^{-1}\)

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A ball of mass \(1 \mathrm{~kg}\) moving with velocity \(10 \mathrm{~m} / \mathrm{s}\) collides perpendicularly on a smooth stationary wedge of mass \(2 \mathrm{~kg}\). If the coefficient of restitution is \(e=7 / 20\) then find the velocity of ball after the collision. [in \(\mathrm{ms}^{-1}\) ].

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