Question

A projectile of mass m is fired with a velocity ${v_0}$ at angle $\theta$ with the horizontal. At the highest position in its flight, it explodes into two fragments of masses ${m_1} = \dfrac{m}{3}$ and ${m_2} = \dfrac{{2m}}{3}$. The fragment of mass ${m_1}$ falls vertically with zero initial speed.
(a) Find the initial velocity of the fragment ${m_2}$.
(b) Find the distance d at which the fragment ${m_2}$ will land with respect to ${m_1}$.

Answer 1

After the projectile breaks up, since one of the masses has zero velocity, all of the initial velocity will be applied to the second mass. Use the law of conservation of momentum to find the velocity of the second fragment, and then use the equations of kinematics for linear motion to find the distance travelled by the fragment.

Complete step by step answer:
The formula for the highest point that a projectile reaches is given as follows:
$H = \dfrac{{{v_0}^2{{\sin }^2}\theta }}{{2g}}$
Where, ${v_0}$ is the initial velocity of the projectile, g is the acceleration due to gravity and$\theta$ is the angle that the projectile makes with the horizontal.

(a) Now, according to the law of conservation of momentum, the initial and the final momentum is conserved. The initial momentum of the particle before exploding is given as:
${p_1} = m{v_0}\cos \theta$
The final momentum of the particle after exploding is given by:
${p_2} = \dfrac{2}{3}mv$
Here, ${p_1}$ and ${p_2}$ are the initial and final momentum. Here, the final momentum of the particle is only due to the second mass, as for the first mass fragment, the initial velocity is zero. Here, the velocity of the second fragment is represented as v. Thus applying the law of conservation of linear momentum, we compare the two equations as above:

{p_1} = {p_2} \\\ \Rightarrow m{v_0}\cos \theta = \dfrac{2}{3}mv \\\ \therefore v = \dfrac{{3{v_0}\cos \theta }}{2} \\\ $$ **Thus the initial velocity of the second fragment is given as above.** (b) Now to find the distance covered by the second fragment, let it take time to reach the ground. Thus, $$H = \dfrac{1}{2}g{t^2}$$ The time it takes to reach the ground is;

\dfrac{{{v_0}^2{{\sin }^2}\theta }}{{2g}} = \dfrac{1}{2}g{t^2} \\
\Rightarrow t = \dfrac{{{v_0}\sin \theta }}{g} \\ $$
Thus, the distance travelled by the second fragment would be given as follows:

D=\dfrac{3{{v}_{0}}\cos \theta }{2}\centerdot \dfrac{{{v}_{0}}\sin \theta }{g} \Rightarrow D=\dfrac{3{{v}_{0}}\cos \theta \sin \theta }{2g} \\\ \therefore D=\dfrac{3{{v}_{0}}\sin 2\theta }{4g} \\\ $$ **Above is the distance travelled by the second fragment with respect to the first fragment.** **Note:** While applying the conservation of the momentum, we must ensure that we take into account the initial momentum of the system which is non zero, since the projectile has non zero velocity at its highest point. Here, we have assumed that there is negligible air resistance when the projectile is in motion.