Answer:
(a) \(v=\frac58u\).
(b) \(\displaystyle e=\frac{25}{32k}-\frac{7}{32}=\frac{25-7k}{32k}\).
(c) \(\displaystyle \frac{25}{39}\le k\le \frac{25}{7}\).
Since \(\tan\alpha=\frac34\),
\(\sin\alpha=\frac35, \qquad \cos\alpha=\frac45.\)
Also,
\(\sin2\alpha=2\sin\alpha\cos\alpha =2\cdot\frac35\cdot\frac45 =\frac{24}{25},\)
and
\(\cos2\alpha=\cos^2\alpha-\sin^2\alpha =\frac{16}{25}-\frac{9}{25} =\frac{7}{25}.\)
(a) For smooth spheres, the component of velocity perpendicular to the line of centres is unchanged for sphere \(A\). Therefore
\(u\sin\alpha=v\sin2\alpha.\)
Substitute the values:
\(u\cdot\frac35=v\cdot\frac{24}{25}.\)
Hence
\(v=\frac{3}{5}\cdot\frac{25}{24}u =\frac58u.\)
Thus
\(v=\frac58u.\)
(b) Let \(w\) be the speed of sphere \(B\) after the collision along the line of centres.
Conservation of momentum along the line of centres gives
\(mu\cos\alpha=mv\cos2\alpha+kmw.\)
Substitute \(\cos\alpha=\frac45\), \(\cos2\alpha=\frac{7}{25}\), and \(v=\frac58u\):
\(\frac45u=\frac{7}{25}\cdot\frac58u+kw.\)
So
\(kw=\frac45u-\frac{7}{40}u =\frac{32}{40}u-\frac{7}{40}u =\frac58u.\)
Therefore
\(w=\frac{5u}{8k}.\)
Newton's law of restitution along the line of centres gives
\(eu\cos\alpha=w-v\cos2\alpha.\)
Thus
\(e\cdot\frac45u=\frac{5u}{8k}-\frac58u\cdot\frac{7}{25}.\)
Cancel \(u\):
\(\frac45e=\frac{5}{8k}-\frac{7}{40}.\)
Multiply by \(\frac54\):
\(e=\frac{25}{32k}-\frac{7}{32}.\)
So
\(e=\frac{25}{32k}-\frac{7}{32}=\frac{25-7k}{32k}.\)
(c) A coefficient of restitution must satisfy
\(0\le e\le1.\)
Therefore
\(0\le \frac{25}{32k}-\frac{7}{32}\le1.\)
Since \(k\gt 0\), multiply by \(32k\):
\(0\le 25-7k\le32k.\)
The left inequality gives
\(7k\le25, \qquad k\le\frac{25}{7}.\)
The right inequality gives
\(25-7k\le32k, \qquad 25\le39k, \qquad k\ge\frac{25}{39}.\)
Hence
\(\frac{25}{39}\le k\le\frac{25}{7}.\)
