The problem is as follows:

A steel sphere has a weight of $40\,N$ and is anchored to a pivot in the ceiling as shown in the picture from below. The wire which connects the sphere to the pivot has a length of $5\,m$ and the system is released when the sphere is held horizontally with respect to the floor. After this, the sphere swings from that position and hits a brass block which has a weight of $200\,N$ which was initially at rest. This latter action is depicted by dashed lines in the figure. Assuming that the surface where the block lies has a negligible friction and the collision between those bodies is perfectly elastic. Find the speed of the ball and of the block inmediately after the collision. Assume $g=10\,\frac{m}{s^2}$

Sketch of the problem

The alternatives given in my book are:

$\begin{array}{ll} 1.&2.6\,\frac{m}{s}\,;\,2.5\,\frac{m}{s}\\ 2.&5.2\,\frac{m}{s}\,;\,5\,\frac{m}{s}\\ 3.&1.3\,\frac{m}{s}\,;\,1.25\,\frac{m}{s}\\ 4.&6.2\,\frac{m}{s}\,;\,5.2\,\frac{m}{s}\\ 5.&4.5\,\frac{m}{s}\,;\,3.8\,\frac{m}{s}\\ \end{array}$

What I've attempted to do was to use the conservation of mechanical energy and use the principle of conservation of momentum to solve this problem. But when I did the calculations none of the answers given appears. What could I did wrong?.

In this scenario there is conservation of mechanical energy at the beginning and in the end unlike other kinds of collisions and there is also conservation of momentum.

Thus: The little letters are $\textrm{b=block}$ and $\textrm{s=sphere}$





Since I'm given some number I can begin plugin these in:

$4\cdot 10 \cdot 5 = \frac{1}{2}\cdot 4 u^2_s+\frac{1}{2}\cdot 20 u^2_{b}$

But the part where I'm stuck is at how to use the fact from the perfectly ellastic collision?. Would it mean to use the $COR=1$?

If this is the case then:



Initially $v_b=0$

But how to find $v_s=?$

What I did was to use:




But from then on. I couldn't get a clear answer. Can someone help me to clear these doubts?.

My book tells that the answer is the first option. But I am not sure how did the author get there?.

  • $\begingroup$ Yes you are correct. I think the alternatives given are incorrect. $\endgroup$ Jul 7 '20 at 3:53
  • 1
    $\begingroup$ Perfectly elastic collision means that the energy is not transformed into heat (or plastic deformation energy). $\endgroup$
    – Andrei
    Jul 7 '20 at 4:36

Going through the problem, there are two stages to it -

  1. Sphere kinematics to reach the bottom most point - Since you are releasing it at the end of a wire (which I'm assuming is rigid), the initial energy must be equal to the final energy of the ball just before collision

$$m_sgh = \frac{1}{2}mv_s^2$$ $$\implies v_s = \sqrt{2gl}$$

  1. Collision mechanics - to determine final speeds of sphere and block, you would need to conserve horizontal momentum as well as total energy (this is where you use COR = 1, otherwise energy is not conserved).

For this, you have initial sphere velocity is $v_{si} = \sqrt{2gl}$, $v_{bi} = 0$. Hence

$$m_{s}v_{si} + m_bv_{bi} = m_sv_{sf} + m_bv_{bf}$$ $$\frac{1}{2}m_sv_{si}^2 = \frac{1}{2}m_sv_{sf}^2 + \frac{1}{2}m_bv_{bf}^2$$

If you express it as a quadratic and solve, you should be able to find the answer. In general, for elastic collisions, it is usually dependant on the mass ratio of the two objects, so try to divide by any one mass and simplify


"Perfectly elastic" means that both kinetic energy and momentum are conserved. Let $u_1$,$u_2$ be the speed of the ball before and after the collision, and let $v$ be the speed of the block after the collision. The mass of the ball is $$\frac{40\text{N}}{10\text{m}\cdot\text{s}^{-2}}=4\text{kg}=m$$ Likewise the mass of the block is $$\frac{200\text{N}}{10\text{m}\cdot\text{s}^{-2}}=20\text{kg}=M$$ We can find $u_1$ by using conservation of energy. $$\frac{1}{2}m{u_1}^2=mg(5\text{ m})$$ $${u_1}=\sqrt{2(10\text{m}\cdot\text{s}^{-2})(5\text{m})}=\sqrt{100}\sqrt{\text{m}^2\text{s}^{-2}}=10\frac{\text{m}}{\text{s}}$$ The total momentum of the system is $mu_1=40\frac{\text{kg}\cdot\text{m}}{\text{s}}$ and the total energy is $\frac{1}{2}m{u_1}^2=200 \frac{\text{kg}\cdot\text{m}^2}{\text{s}^2}$. Therefore, conservation of momentum requires $$mu_1=40=mu_2+Mv=4u_2+20v$$ and conservation of energy requires $$\frac{1}{2}m{u_1}^2=200 =\frac{1}{2}m{u_2}^2+\frac{1}{2}Mv^2=2{u_2}^2+10v^2$$ Solve the first equation for $u_2$ to get $u_2=10-5v$ then plug into the second equation- $$200=2(10-5v)^2+10v^2$$ $$40=2(2-v)^2+2v^2$$ $$40=4-2v+v^2+2v^2$$ $$3v^2-2v-36=0\implies v\approx 3.813\frac{\text{m}}{\text{s}}$$ Now substituting back into the conservation of momentum equation, $$40=4u_2+20v$$ $$10=u_2+5v$$ $$u_2=5(2-3.813)\implies u_2\approx -9.067 \frac{\text{m}}{\text{s}}$$ So right after the collision, the block is going 3.813 meters per second to the right, and the ball is going 9.067 meters per second to the left. This makes intuitive sense as the block is 5 times heavier than the ball, so the ball should sort of "bounce" off.

  • $\begingroup$ In the figure you have weight of the ball is $40N$, but you used $10$ in your first equation $\endgroup$
    – Andrei
    Jul 7 '20 at 4:37
  • $\begingroup$ I've just noticed I've put in completely the wrong numbers for the masses of the ball and block, somehow. I'll fix it. $\endgroup$
    – K.defaoite
    Jul 7 '20 at 4:37

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