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My work here is: First, I use conservation of energy: (taking the plane as zero potential) $$\frac{1}{2}m(\frac{2ga}{5})+2amg=\frac{1}{2}mv^2+a(1+\cosθ)mg$$ $$\frac{1}{5}ga+2ga=\frac{1}{2}v^2+ga(1+\cosθ)$$ $$v^2=(\frac{12}{5}-2\cosθ)ga \tag1$$ Now resolving forces radially at the point of falling ($R=0$) $$mg\cosθ-\frac{mv^2}{r}-R=0$$ $$mg\cosθ=\frac{mv^2}{a}$$ $$v^2=ag\cosθ\tag2$$ Equating $(1)$ and $(2)$ $$\cosθ=\frac{4}{5}$$ As required. I am having problems in the second part. Let $v=v_1$, the velocity of arrival at the plane, $v_2$, and the velocity of rebound $v_3$.

I considered the velocity of arrival equating energies. $$\frac{1}{2}mv_1^2 + mga(1+\cosθ)=\frac{1}{2}mv_2^2$$ $v_1=\sqrt{\frac{4ag}{5}}$ $$v_2^2=\frac{22ag}{5}$$ Using Newton's Law of Restitution: $$e=\frac{-v_3}{v_2}=\frac{5}{9}$$ $$v_3=\frac{5v_2}{9}=\frac{5\sqrt{\frac{22ag}5}}{9}$$ Using the dynamics equation $v^2=u^2+2as$ $$(v_3\sinθ)^2=2gh$$ From earlier, $\sinθ=\frac{3}{5}$ $$h=\frac{(\frac{3v_3}{5})^2}{2g}$$ But this gives a wrong result. I am unsure what I did wrong and, am I wrong in assuming that the particle will bounce of with the same angle $θ$ that it came out of the ball?


1 Answer 1


The problem is that you have considered the velocity $v_2$ directly in the expression of $e$ whereas you should have taken only the vertical component(or the component perpendicular to the plane) as the horizontal component reamins unaffected because the plane is smooth.

Note: I used R in place of a
At the time of breaking from sphere $v_i=2\sqrt\frac{gR}5$.
Now, $$\large v_{\perp,i}=v\sin \theta = \frac65 \sqrt\frac{gR}5$$ $$\large v_{\parallel,i}=v\sin \theta = \frac85 \sqrt\frac{gR}5$$

Using $v^2=u^2+2as$

$$\large v_{\perp,f} = \frac95 \sqrt\frac{6gR}5$$ $$\large v_{\parallel,f}= \frac85 \sqrt\frac{gR}5$$

So after collision $v_f = e*v_i$:

$$\large v_{\perp,f2} = \frac59 * \frac95 \sqrt\frac{6gR}5 =\sqrt\frac{6gR}5 $$ $$\large v_{\parallel,f2} = \frac85 \sqrt\frac{gR}5$$

Height raised $H = \frac{v^2}{2g}$

$$\LARGE H = \frac{3R}5$$


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