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I am currently studying Clssical Mechanics, fifth edition, by Kibble and Berkshire. Problem 1 of chapter 1 is as follows:

An object $A$ moving with velocity $\mathbf{v}$ collides with a stationary object $B$. After the collision, $A$ is moving with velocity $\dfrac{1}{2}\mathbf{v}$ and $B$ with velocity $\dfrac{3}{2}\mathbf{v}$. Find the ratio of their masses. If, instead of bouncing apart, the two bodies stuck together after the collision, with what velocity would they then move.

I've read over the chapter, trying to find some indication of how to do this problem, but I do not see how this is possible with the information given. The most relevant equation that I could find relates to the law of conservation of momentum:

If we allow two small bodies to collide, then during the collision the effects of more remote bodies are generally negligible in comparison with their effect on each other, and we may treat them approximately as an isolated system. (Such collisions will be discussed in detail in Chapter 2 and 7.) The mass ratio can then be determined from measurements of their velocities before and after the collision, by using (1.7) or its immediate consequence, the law of conservation of momentum, $$m_1 \mathbf{v}_1 + m_2 \mathbf{v}_2 = \text{constant}. \tag{1.8}$$

But this is all that is written on the subject, and I wonder if more information on the law of conservation of momentum is required in order to solve this problem.

The solution is said to be $m_A/m_B = 3$; $3\mathbf{v}/4$.

I would greatly appreciate it if people would please take the time to explain how

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Conservation of momentum in this case reads $$m_1\mathbf v=\tfrac 1 2m_1\mathbf v+\tfrac 3 2m_2\mathbf v$$ which is just plugging in the velocities of before and after the collision. This is enough information to solve for $m_1/m_2$.

In the second case the velocity of after the collision is unknown so let's introduce $\mathbf{v}_1',\mathbf{v}_2'$ as the velocities after the collision. We know they stick together so we can simplify it to $\mathbf v'=\mathbf{v}_1'=\mathbf{v}_2'$. What equation do you get when you plug this in for conservation of momentum?

Edit: I used $m_1,m_2$ like in the general case but you can ofcourse replace $m_1$ by $m_A$ and $m_2$ by $m_B$.

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    $\begingroup$ Thanks for the answer. Would you please explain why the textbook equation for conservation of momentum is different from yours? $\endgroup$
    – The Pointer
    Jun 8, 2020 at 23:57
  • $\begingroup$ They are the same. Your textbook states general conservation of momentum: $m_1\mathbf v_1+m_2\mathbf v_2=m_1\mathbf v_1'+m_2\mathbf v_2'$. I then plugged velocities of this specific example in. Before the collision $\mathbf v_2=0$ so this term drops out. The other velocities are described in the problem statement. $\endgroup$
    – AccidentalTaylorExpansion
    Jun 9, 2020 at 10:45
  • $\begingroup$ Interesting. So why does the textbook have a constant instead of $m_1\mathbf v_1+m_2\mathbf v_2=m_1\mathbf v_1'+m_2\mathbf v_2'$? $\endgroup$
    – The Pointer
    Jun 9, 2020 at 10:49
  • $\begingroup$ If we call the total momentum $\mathbf P(t)$ then we can express conservation of momentum as follows: $\mathbf P(t)=m_1\mathbf v_1(t)=m_2\mathbf v_2(t)=\text{constant}$. What I did is called the velocities before/after the collision different names: $\mathbf P(t_{\text{before}})=m_1\mathbf v_1=m_2\mathbf v_2$ and $\mathbf P(t_{\text{after}})=m_1\mathbf v_1'=m_2\mathbf v_2'$. I can do this because when the particles are not colliding the velocities are constant / time independent. Can you see that both expressions agree with eachother? $\endgroup$
    – AccidentalTaylorExpansion
    Jun 9, 2020 at 11:09
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    $\begingroup$ @ThePointer You're exactly right that $\mathbf C_1=\mathbf C_2$! They are equal because momentum is conserved. Why is momentum conserved? That's a bit tricky to answer in a comment (it's actually one of the fundamental assumptions in classical mechanics) but you can find more in depth answers on this site. $\endgroup$
    – AccidentalTaylorExpansion
    Jun 9, 2020 at 17:53

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