Question

A body projected with a speed $u$ at an angle of ${{60}^{\circ }}$ with the horizontal explodes in two equal pieces at a point where its velocity makes an angle of ${{30}^{\circ }}$ with the horizontal for 1st time. One piece starts moving vertically upward with a speed of $\dfrac{u}{2\sqrt{3}}$ after explosion. What is the velocity of one piece with respect to another in the vertical direction.

Answer 1

First calculate the initial horizontal velocity of the projected body. Then with the help of fact that the horizontal velocity of a projectile is constant, find the speed of the body at that point where its velocity makes an angle of ${{30}^{\circ }}$ with the horizontal for 1st time. Then use the law of conservation of momentum and find the speeds of the two pieces.

Formula used:
${{u}_{x}}=u\cos \theta$
${{u}_{y}}=u\sin \theta$
Where ${{u}_{y}}$ and ${{u}_{x}}$ are the initial velocities of the projectile in the vertical and the horizontal directions, u is initial velocity and $\theta$ is angle of projection.
$P=mv$
where P is momentum of a body of mass m and moving with velocity v.

Complete step by step answer:
It is given that the initial speed of the body is u and the angle of projection is $\theta ={{60}^{\circ }}$.Therefore, the initial horizontal velocity of the body is,
${{u}_{x}}=u\cos {{60}^{\circ }}=u\left( \dfrac{1}{2} \right)=\dfrac{u}{2}$
And the initial vertical velocity of the body is,
${{u}_{y}}=u\sin {{60}^{\circ }}=u\left( \dfrac{\sqrt{3}}{2} \right)=\dfrac{\sqrt{3}u}{2}$
Then, it is said that the body explodes at a point where its velocity makes an angle of ${{30}^{\circ }}$ with the horizontal for the 1st time. So, let this velocity be v.
Therefore, the horizontal velocity of the body is,
${{v}_{x}}=v\cos {{30}^{\circ }}=v\left( \dfrac{\sqrt{3}}{2} \right)=\dfrac{\sqrt{3}v}{2}$ ….. (i)
And the vertical velocity of the body is,
${{v}_{y}}=v\sin {{30}^{\circ }}=v\left( \dfrac{1}{2} \right)=\dfrac{v}{2}$ …. (ii)

In a projectile motion (without air resistance), the velocity of the body in the horizontal direction remains constant, i.e. ${{v}_{x}}={{u}_{x}}$, since there is no force acting in that direction. Therefore, the horizontal speed of the body at this point is,
${{v}_{x}}={{u}_{x}}=\dfrac{u}{2}$ …. (iii).
Now, equate (i) and (iii).
$\dfrac{\sqrt{3}v}{2}=\dfrac{u}{2}$
$\Rightarrow v=\dfrac{u}{\sqrt{3}}$
Substitute the value v in (ii).Then,
${{v}_{y}}=\dfrac{1}{2}\left( \dfrac{u}{\sqrt{3}} \right)=\dfrac{u}{2\sqrt{3}}$
This means that at the speed of the body in the upward direction just before the explosion is $\dfrac{u}{2\sqrt{3}}$. Therefore, its momentum in this direction is,
${{P}_{i}}=m{{v}_{y}}=m\dfrac{u}{2\sqrt{3}}$
Explosion is due to internal forces and the net force due to internal force is always zero. Hence, the momentum of the system will be conserved. This means that the momentum in the vertical direction is equal to momentum in the same after the explosion.

It is given that the body splits into two equal pieces (both of mass $\dfrac{m}{2}$) . It is said that one piece starts moving vertically upward with a speed of ${{v}_{1}}=\dfrac{u}{2\sqrt{3}}$ after explosion.
Therefore, the its momentum is,
${{P}_{1}}=\left( \dfrac{m}{2} \right)\left( \dfrac{u}{2\sqrt{3}} \right)=\dfrac{mu}{4\sqrt{3}}$
Let the speed of the other piece in the upward direction be ${{v}_{2}}$.Then the momentum of the other piece in the same direction be ${{P}_{2}}=\dfrac{m}{2}{{v}_{2}}$.
Then, we know that ${{P}_{i}}={{P}_{1}}+{{P}_{2}}$
$\Rightarrow \dfrac{mu}{2\sqrt{3}}=\dfrac{mu}{4\sqrt{3}}+\dfrac{m}{2}{{v}_{2}}$
$\Rightarrow {{v}_{2}}=2\left( \dfrac{u}{2\sqrt{3}}-\dfrac{u}{4\sqrt{3}} \right)=\dfrac{u}{2\sqrt{3}}$
This means that the speed of both the pieces in the upward direction is $\dfrac{u}{2\sqrt{3}}$.

Since the velocities are equal, the relative velocity of one piece with respect to another is zero.

Note: The students do know that during the explosion, the momentum of the system is conserved but most of them do not know the why is the momentum of the system conserved.Explosion of a body (e.g. a bomb) take places due to some chemical reaction within the body. These chemical reactions create some internal force with the body, which push the body apart. Since internal forces of a system create zero net force, the momentum of the system is conserved.