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Inelastic collision between two balls

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1) Two balls ($A$ and $B$) of masses $m_a$ and $m_b$ collide in an oblique way in a table. Initially, $A$ goes towards $B$ with an initial velocity $V_o$, while $B$ stays at rest. After the inelastic collision $A$ has a velocity $\frac{V_o}{2}$ and moves with a right angle (regards the initial direction of the movement).
The exercise didn't provide any draw; the image shown is what I interpreted from the statement.

a) Determine the angle formed by the ball $B$ and the horizontal after the collision.
b) Determine the velocity of $B$ after the collision.
c) Compare and study the difference between kinetic energy before and after the collision (assess change in kinetic energy) and deduce the rate ma/mb so as to have an inelastic collision.

These are my solutions

As there are not external forces we can assure momentum is conserved. Therefore, we know the following equations are valid:

$m_aV_o, _a + m_bV_o, _b = m_aV_f, _a + m_bV_f, _b$

a) In the x axis case:

$m_aV_o, _a = m_bV_f, _bcos(\alpha)$ ---> EQ 1

In the y axis case:

$m_a\frac{V_o}{2} - m_bV_f, _bsin(\alpha) = 0$ ---> EQ2

With these two EQs we can get $tg(\alpha)=1/2$; therefore $\alpha=26.56º$

b) Using the two previous EQs I get $V_f, _b= \frac{m_aV_o}{m_bcos(\alpha)}=\frac{m_aV_o}{2sin(\alpha)m_b}$ but I am pretty sure there is something I am missing, as I do not obtain a clear result for $V_f, _b$. What I mean is that I am supposed to obtain a numeric outcome, but as I just acquired the angle as a result, I do not see how I could combine both EQ1 and EQ2 so as to obtain a number.

c) First of all I regarded the both masses as point ones. As we have an inelastic collision, there is a loss in kinetic energy. I analysed it as follows:

$KE_o= 1/2m_aV_o^2$ ---> EQ3
$KE_f= 1/2m_a(\frac{V_o}{2})^2 + 1/2m_bV_f^2$ ---> EQ4
$KE_f < KE_o$ ---> EQ5

From these EQs I got:

$\frac{ma}{mb}>(\frac{V_fcos(\alpha)}{V_o})^2 AND \frac{ma}{mb}<(\frac{2V_fsin(\alpha)}{V_o})^2$
But again I got a result slightly conclusive, what makes me doubtful. The reason why I am unsure about my result is that I am supposed to get a numeric outcome in the rate $\frac{m_a}{m_b}\$. I could obtain it if I had a numeric result from b). Notice I am using EQ3, EQ4 and EQ5 at c). As far as I am concerned, my setback here is not physics but algebra. Please let me know if it is backwards.

I am looking for an accurate solution (as much as possible).


asked Apr 11, 2018 in Physics Problems by Jorge Daniel (716 points)
edited Apr 14, 2018 by Jorge Daniel
In (b) what do you mean by "I don't obtain a clear result for Vfb"? In (c) what do you mean by "I got a result slightly conclusive"? Why are you "doubtful"?
I will make everything clear after changing equtions with MathJax. Thanks for the link
Thank you for using MathJax. It is much easier to ready now.

1 Answer

1 vote
Best answer

a) I agree with your result that $\tan\alpha = \frac12$. (Incidentally, a no-head-on collision is usually called an oblique collision.)

Momentum is conserved in the x and y directions so the x and y components of momentum of B are
$p_y=m_a V_0/2$

b) The momentum of B after collision is
$p=\sqrt{p_x^2+p_y^2}=m_aV_0 \sqrt{1^2+(\frac12)^2}=m_aV_0 \frac{\sqrt5}{2}$
so the final velocity of B is $\frac{p}{m_b}=\frac{m_a}{m_b}V_0\frac{\sqrt5}{2}$.

c) Initial and final kinetic energies are
$K_0=\frac12 m_a V_0^2$
$K_1=\frac12 m_a (V_0/2)^2+\frac12 m_b (\frac{m_a}{m_b}V_0\frac{\sqrt5}{2})^2=\frac12 m_aV_0^2.\frac14(1+5 \frac{m_a}{m_b})$
You are correct to assume that A and B are point particles, so they cannot have any rotational KE, nor any elastic PE (internal energy).

For an inelastic collision $K_1 \lt K_0$ so
$\frac14(1+5\frac{m_a}{m_b}) \lt 1$
$\frac{m_a}{m_b} \lt \frac35$.
For an elastic collision we require $K_1=K_0$. Then $\frac{m_a}{m_b}=\frac35$.

Comment :

It looks as though we could set $\frac{m_a}{m_b} \gt \frac35$ then we would have $K_1 \gt K_0$ - ie final KE is greater than initial KE. Surely this is not possible?

Actually $K_1 \gt K_0$ is possible (if A and B are not point particles). This is called a super-elastic collision. It means that there is some internal energy which is converted into KE during the collision. EG there could be an explosion which converts chemical energy into KE, or a pre-stretched elastic band could be released by the collision, the elastic PE being converted into KE. Alternatively there could be some rotational KE which we did not include in $K_0$ and $K_1$.

The conservation of energy in any collision does not require that KE cannot increase, only that total energy of all kinds cannot increase or decrease. Not only can KE be transformed into heat and sound energy, reducing KE (so $K_1 \lt K_0$), also chemical and elastic energy can be transformed into KE, increasing KE (so $K_1 \gt K_0$).

If we are told that $\frac{m_a}{m_b} \gt \frac35$ and that KE does not increase in the collision then we can deduce that the velocity of A cannot be reduced by half and turned through $90^{\circ}$ as stated in the question. This is not possible with the given conditions .

answered Apr 12, 2018 by sammy gerbil (28,896 points)
edited Apr 15, 2018 by sammy gerbil
Great answer. My mistake was I did not consider total linear momentum of the system (ball $A$ and ball $B$ ) after the collision:

$p=\sqrt{p_x^2+p_y^2}=m_aV_0 \sqrt{1^2+(\frac12)^2}=m_aV_0 \frac{\sqrt5}{2}$

Then $V_b, _f = \frac{3\sqrt{5}V_o}{10}$ would be a bad result, as this is an inelastic collision.
Could we characterise $V_b$ as follows?

$V_b, _f = \frac{3\sqrt{5}V_o}{(a5)2}$ being a>1, so as to obtain a number smaller than $\frac{3}{5}$
Why not leave the factor $\frac{m_a}{m_b}$ in the equation for $V_b$? I don't see any advantage in removing it. If you replace this factor by $a$ then you have to explain how $a$ relates to the quantities in the problem.
I thought we could plugg in it as we already knew it.  Then I will leave it in funcion of  $\frac{m_a}{m_b}$
Your answer for part b) was correct, but you should have inserted a value for $\cos\alpha=\frac{2}{\sqrt5}$.
To obtain$ \frac{2}{\sqrt5}$ factor before I have to get the modulus of linear momentum after de colision Don't I? As I did not calculate it, I could not plugg in
$\cos\alpha=\frac{2}{\sqrt5}$ follows from $\tan\alpha=\frac12
$ which you did get.
Mm I got it,. My calculator provides me the number 0.894466... not the fraction $\frac{2}{\sqrt5}$. What should I do, trial and error until I get the desired fraction?
Draw a right-angled triangle in which $\tan\alpha=\frac12$.