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The problem is as follows:

The diagram from below shows a sphere labeled $A$ and is moving with a horizontal speed of $v=4\sqrt{10}\,\frac{m}{s}$ over a frictionless table. After the collision the cable makes an angle of $53^{\circ}$ with the vertical and the sphere $A$ ends at rest. Find the $COR$ (coefficient of restitution) and the relationship between the masses.

Sketch of the problem

The alternatives given are as follows:

$\begin{array}{ll} 1.&\frac{1}{2},\, \frac{1}{2}\\ 2.&\frac{1}{3},\, \frac{1}{6}\\ 3.&\frac{1}{2},\, \frac{1}{3}\\ 4.&\frac{1}{8},\, \frac{1}{2}\\ 5.&\frac{1}{2},\, \frac{1}{6}\\ \end{array}$

When it mentions the COR for the collision I'm assuming that it is an inelastic collision. Therefore if treated as such then

$p_1=p_2$

$m_1 v_1=m_2 v_2$

$v_2=4\sqrt{10} \frac{m_1}{m_2}$

But that's how far I went. I'm stuck with this problem, can someone help me here?.

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  • $\begingroup$ How did you get the last line of your working? It doesn't look right. If I understand your notation, $v_{1}$ is the velocity of $A$ immediately before the collision, and $v_{2}$ is the velocity of $B$ immediately after. If so, then $v_{1} = 4\sqrt{10}$ is given. Perhaps you mean $v_{2} = 4\sqrt{10} \frac{m_{1}}{m_{2}}$? $\endgroup$ – preferred_anon Dec 8 '19 at 21:10
  • $\begingroup$ The angle $53^{\circ}$ is when sphere $B$ is at rest following the collision. That means it has gained some height, and thus gained some potential energy. Since it's at rest, it's lost some kinetic energy. You should try writing down some formulas for these energies. $\endgroup$ – preferred_anon Dec 8 '19 at 21:12
  • $\begingroup$ @preferred_anon Sorry it was a typo. I'll try to write the formula as the conditions you mentioned. But at the beginning it will have kinetic energy and at the end it will be potential energy. Am I getting your idea right? $\endgroup$ – Chris Steinbeck Bell Dec 8 '19 at 21:32
  • $\begingroup$ To get exactly what they have, you must use $\cos(53)=\frac 35$ $\endgroup$ – WW1 Dec 8 '19 at 22:05
  • $\begingroup$ @WW1 Perhaps can you offer some answer to guiding me on what should I do?. $\endgroup$ – Chris Steinbeck Bell Dec 8 '19 at 22:06
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Initial momentum of $A$ is $4\sqrt{10}m_A$. This is equal to the momentum of $B$ after the collision. The kinetic energy of $B$ after the collision is then $p^{2}/2m_B = 80m_{A}^{2}\big/m_B$.

As has been noted, $B$ is raised by $2\text{m}$ by the collision, giving it a potential energy of $20m_{B} = \frac{1}{2}m_{B}v_{B}^{2}$. Note therefore that $v_{B} = \sqrt{40} = 2\sqrt{10}$.

Therefore $80m_{A}^{2} = 20m_{B}^{2}$, meaning $m_{A}/m_{B} = 1/2$.

The COR is then $$\frac{v_{B}}{v_A} = \frac{2\sqrt{10}}{4\sqrt{10}} = 1/2.$$

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Total mechanical energy using $g=10$ and $\cos(53)=\frac 35$ ... $$E = \frac 12 m_a (4\sqrt {10})^2=80m_a \\E'= m_bg[5(1-\cos(53))] = 20 m_b$$ $E'$must be equal to the total Kinetic energy immediately after collision so . $$C_r =\sqrt{ \frac{KE'}{KE}}= \frac 14 \frac {m_b}{m_a} $$ The only option that could possibly make sense would be the one in which $C_r=\frac 12$ which must have $m_b = m_a$ It is not entirely clear what they mean by "the relationship between the masses" , it is possible that they mean $\frac{m_b}{m_a+m_b}$, in which case you should choose the option with both numbers equal to $\frac 12$

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  • $\begingroup$ How come $C_r=\sqrt{ \frac{KE'}{KE}}$? Isn't this only true when $m_a=m_b$? And isn't $\sqrt{ \frac{KE'}{KE}}=\sqrt{\frac 14 \frac {m_b}{m_a}}$? $\endgroup$ – Shubham Johri Dec 10 '19 at 6:45

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