Explanation:

Velocity of disc '\(A\)' in tangent direction
\(\begin{equation*}v_{1, t}=u \sin 45^{\circ}=\dfrac{u}{\sqrt{2}} \tag{1}\end{equation*}\)
The disc ' \(B\) ' will move along common normal direction only.

And according to equation both the discs move perpendicular to each other. It means disc ' \(A\) ' should move along common tangent direction.
Hence \(v_{1}=\dfrac{u}{\sqrt{2}}\)
along common tangent direction.
Now conserving linear momentum along common normal direction \(m \cdot \frac{u}{{\sqrt 2 }} + 0 = m \times 0 + 2m{v_{2,n}}\)
or \({v_{2,n}} = \frac{u}{{\sqrt 2 }}{\rm{ }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\quad {\rm{(2)}}\)
or velocity of disc \(B^{\prime}\) is \(\frac{u}{{2\sqrt 2 }}\) along common normal.
Now coefficient of restitution
\(e=\dfrac{\text { Relative velocity of separation }}{\text { Relative velocity of approach }}\)\(=\dfrac{v_{2, n}-0}{u \cos 45^{\circ}}=\dfrac{u / 2 \sqrt{2}}{u / \sqrt{2}}=\dfrac{1}{2}\)

Explanation:

After collision the horizontal component of velocity remains same and vertical component becomes e times the initial value.
\(\tan \alpha=\dfrac{v \sin \theta}{e v \cos \theta}=\dfrac{\tan \theta}{e}\)

or \(\alpha=\tan ^{-1}\left(\dfrac{\tan \theta}{e}\right)\)

Explanation:

From impulse-momentum theorem.
\(\begin{align*}& \int N d t=m(v+5 \cos \theta) \tag{1}\\& \int f d t=m 5 \sin \theta\end{align*}\)
\(\int N dt = m5\sin \theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)\)
\(\Rightarrow \quad \mu m(1+5 \cos \theta)=m 5 \sin \theta\)

According to Newton's law of restitution,\(\quad v = e\,\,5\cos \theta \)
Solve to get \(\mu=1\)

Explanation:

Explanation:

Kinetic energy of ball when reaching the ground
\( = mgh = mg \times 10\)
Kinetic energy after the impact
\( = \frac{{60}}{{100}} \times mg \times 10 = 6\,mg\)
If the ball rises to a height \(h\), then \(mgh = 6\,mg\).
Hence, \(h = 6\;m.\)

Explanation:

Velocity of disc '\(A\)' in tangent direction
\(\begin{equation*}v_{1, t}=u \sin 45^{\circ}=\dfrac{u}{\sqrt{2}} \tag{1}\end{equation*}\)
The disc ' \(B\) ' will move along common normal direction only.

And according to equation both the discs move perpendicular to each other. It means disc ' \(A\) ' should move along common tangent direction.
Hence \(v_{1}=\dfrac{u}{\sqrt{2}}\)
along common tangent direction.
Now conserving linear momentum along common normal direction \(m \cdot \frac{u}{{\sqrt 2 }} + 0 = m \times 0 + 2m{v_{2,n}}\)
or \({v_{2,n}} = \frac{u}{{\sqrt 2 }}{\rm{ }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\quad {\rm{(2)}}\)
or velocity of disc \(B^{\prime}\) is \(\frac{u}{{2\sqrt 2 }}\) along common normal.
Now coefficient of restitution
\(e=\dfrac{\text { Relative velocity of separation }}{\text { Relative velocity of approach }}\)\(=\dfrac{v_{2, n}-0}{u \cos 45^{\circ}}=\dfrac{u / 2 \sqrt{2}}{u / \sqrt{2}}=\dfrac{1}{2}\)

Explanation:

After collision the horizontal component of velocity remains same and vertical component becomes e times the initial value.
\(\tan \alpha=\dfrac{v \sin \theta}{e v \cos \theta}=\dfrac{\tan \theta}{e}\)

or \(\alpha=\tan ^{-1}\left(\dfrac{\tan \theta}{e}\right)\)

Explanation:

From impulse-momentum theorem.
\(\begin{align*}& \int N d t=m(v+5 \cos \theta) \tag{1}\\& \int f d t=m 5 \sin \theta\end{align*}\)
\(\int N dt = m5\sin \theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)\)
\(\Rightarrow \quad \mu m(1+5 \cos \theta)=m 5 \sin \theta\)

According to Newton's law of restitution,\(\quad v = e\,\,5\cos \theta \)
Solve to get \(\mu=1\)

Explanation:

Explanation:

Kinetic energy of ball when reaching the ground
\( = mgh = mg \times 10\)
Kinetic energy after the impact
\( = \frac{{60}}{{100}} \times mg \times 10 = 6\,mg\)
If the ball rises to a height \(h\), then \(mgh = 6\,mg\).
Hence, \(h = 6\;m.\)

Explanation:

Velocity of disc '\(A\)' in tangent direction
\(\begin{equation*}v_{1, t}=u \sin 45^{\circ}=\dfrac{u}{\sqrt{2}} \tag{1}\end{equation*}\)
The disc ' \(B\) ' will move along common normal direction only.

And according to equation both the discs move perpendicular to each other. It means disc ' \(A\) ' should move along common tangent direction.
Hence \(v_{1}=\dfrac{u}{\sqrt{2}}\)
along common tangent direction.
Now conserving linear momentum along common normal direction \(m \cdot \frac{u}{{\sqrt 2 }} + 0 = m \times 0 + 2m{v_{2,n}}\)
or \({v_{2,n}} = \frac{u}{{\sqrt 2 }}{\rm{ }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\quad {\rm{(2)}}\)
or velocity of disc \(B^{\prime}\) is \(\frac{u}{{2\sqrt 2 }}\) along common normal.
Now coefficient of restitution
\(e=\dfrac{\text { Relative velocity of separation }}{\text { Relative velocity of approach }}\)\(=\dfrac{v_{2, n}-0}{u \cos 45^{\circ}}=\dfrac{u / 2 \sqrt{2}}{u / \sqrt{2}}=\dfrac{1}{2}\)

Explanation:

After collision the horizontal component of velocity remains same and vertical component becomes e times the initial value.
\(\tan \alpha=\dfrac{v \sin \theta}{e v \cos \theta}=\dfrac{\tan \theta}{e}\)

or \(\alpha=\tan ^{-1}\left(\dfrac{\tan \theta}{e}\right)\)

Explanation:

From impulse-momentum theorem.
\(\begin{align*}& \int N d t=m(v+5 \cos \theta) \tag{1}\\& \int f d t=m 5 \sin \theta\end{align*}\)
\(\int N dt = m5\sin \theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)\)
\(\Rightarrow \quad \mu m(1+5 \cos \theta)=m 5 \sin \theta\)

According to Newton's law of restitution,\(\quad v = e\,\,5\cos \theta \)
Solve to get \(\mu=1\)

Explanation:

Explanation:

Kinetic energy of ball when reaching the ground
\( = mgh = mg \times 10\)
Kinetic energy after the impact
\( = \frac{{60}}{{100}} \times mg \times 10 = 6\,mg\)
If the ball rises to a height \(h\), then \(mgh = 6\,mg\).
Hence, \(h = 6\;m.\)

Explanation:

Velocity of disc '\(A\)' in tangent direction
\(\begin{equation*}v_{1, t}=u \sin 45^{\circ}=\dfrac{u}{\sqrt{2}} \tag{1}\end{equation*}\)
The disc ' \(B\) ' will move along common normal direction only.

And according to equation both the discs move perpendicular to each other. It means disc ' \(A\) ' should move along common tangent direction.
Hence \(v_{1}=\dfrac{u}{\sqrt{2}}\)
along common tangent direction.
Now conserving linear momentum along common normal direction \(m \cdot \frac{u}{{\sqrt 2 }} + 0 = m \times 0 + 2m{v_{2,n}}\)
or \({v_{2,n}} = \frac{u}{{\sqrt 2 }}{\rm{ }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\quad {\rm{(2)}}\)
or velocity of disc \(B^{\prime}\) is \(\frac{u}{{2\sqrt 2 }}\) along common normal.
Now coefficient of restitution
\(e=\dfrac{\text { Relative velocity of separation }}{\text { Relative velocity of approach }}\)\(=\dfrac{v_{2, n}-0}{u \cos 45^{\circ}}=\dfrac{u / 2 \sqrt{2}}{u / \sqrt{2}}=\dfrac{1}{2}\)

Explanation:

After collision the horizontal component of velocity remains same and vertical component becomes e times the initial value.
\(\tan \alpha=\dfrac{v \sin \theta}{e v \cos \theta}=\dfrac{\tan \theta}{e}\)

or \(\alpha=\tan ^{-1}\left(\dfrac{\tan \theta}{e}\right)\)

Explanation:

From impulse-momentum theorem.
\(\begin{align*}& \int N d t=m(v+5 \cos \theta) \tag{1}\\& \int f d t=m 5 \sin \theta\end{align*}\)
\(\int N dt = m5\sin \theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)\)
\(\Rightarrow \quad \mu m(1+5 \cos \theta)=m 5 \sin \theta\)

According to Newton's law of restitution,\(\quad v = e\,\,5\cos \theta \)
Solve to get \(\mu=1\)

Explanation:

Explanation:

Kinetic energy of ball when reaching the ground
\( = mgh = mg \times 10\)
Kinetic energy after the impact
\( = \frac{{60}}{{100}} \times mg \times 10 = 6\,mg\)
If the ball rises to a height \(h\), then \(mgh = 6\,mg\).
Hence, \(h = 6\;m.\)

Explanation:

Velocity of disc '\(A\)' in tangent direction
\(\begin{equation*}v_{1, t}=u \sin 45^{\circ}=\dfrac{u}{\sqrt{2}} \tag{1}\end{equation*}\)
The disc ' \(B\) ' will move along common normal direction only.

And according to equation both the discs move perpendicular to each other. It means disc ' \(A\) ' should move along common tangent direction.
Hence \(v_{1}=\dfrac{u}{\sqrt{2}}\)
along common tangent direction.
Now conserving linear momentum along common normal direction \(m \cdot \frac{u}{{\sqrt 2 }} + 0 = m \times 0 + 2m{v_{2,n}}\)
or \({v_{2,n}} = \frac{u}{{\sqrt 2 }}{\rm{ }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\quad {\rm{(2)}}\)
or velocity of disc \(B^{\prime}\) is \(\frac{u}{{2\sqrt 2 }}\) along common normal.
Now coefficient of restitution
\(e=\dfrac{\text { Relative velocity of separation }}{\text { Relative velocity of approach }}\)\(=\dfrac{v_{2, n}-0}{u \cos 45^{\circ}}=\dfrac{u / 2 \sqrt{2}}{u / \sqrt{2}}=\dfrac{1}{2}\)

Explanation:

After collision the horizontal component of velocity remains same and vertical component becomes e times the initial value.
\(\tan \alpha=\dfrac{v \sin \theta}{e v \cos \theta}=\dfrac{\tan \theta}{e}\)

or \(\alpha=\tan ^{-1}\left(\dfrac{\tan \theta}{e}\right)\)

Explanation:

From impulse-momentum theorem.
\(\begin{align*}& \int N d t=m(v+5 \cos \theta) \tag{1}\\& \int f d t=m 5 \sin \theta\end{align*}\)
\(\int N dt = m5\sin \theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)\)
\(\Rightarrow \quad \mu m(1+5 \cos \theta)=m 5 \sin \theta\)

According to Newton's law of restitution,\(\quad v = e\,\,5\cos \theta \)
Solve to get \(\mu=1\)

Explanation:

Explanation:

Kinetic energy of ball when reaching the ground
\( = mgh = mg \times 10\)
Kinetic energy after the impact
\( = \frac{{60}}{{100}} \times mg \times 10 = 6\,mg\)
If the ball rises to a height \(h\), then \(mgh = 6\,mg\).
Hence, \(h = 6\;m.\)