Centre of Mass 3 Question 31
34. An object of mass $5 kg$ is projected with a velocity of $20 m / s$ at an angle of $60^{\circ}$ to the horizontal. At the highest point of its path, the projectile explodes and breaks up into two fragments of masses $1 kg$ and $4 kg$. The fragments separate horizontally after the explosion. The explosion releases internal energy such that the kinetic energy of the system at the highest point is doubled. Calculate the separation between the two fragments when they reach the ground.
(1990, 8M)
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Answer:
Correct Answer: 34. $44.25 m$
Solution:
- Let $v _1$ and $v _2$ be the velocities after explosion in the directions shown in figure. From conservation of linear momentum, we have
or
$$ \begin{gathered} 5\left(20 \cos 60^{\circ}\right)=4 v _1-1 \times v _2 \\ 4 v _1-v _2=50 \\ 5 kg \end{gathered} $$
Just before explosion
$$ v _2 \stackrel{1 kg}{O^{\longleftrightarrow}} \stackrel{4 kg}{\longrightarrow} v _1 $$
Just after explosion
Further, it is given that, kinetic energy after explosion becomes two times.
Therefore,
$$ \begin{aligned} & \frac{1}{2} \times 4 \times v _1^{2}+\frac{1}{2} \times 1 \times v _2^{2}=2 \frac{1}{2} \times 5 \times\left(20 \cos 60^{\circ}\right)^{2} \\ & \text { or } \quad 4 v _1^{2}+v _2^{2}=1000 \end{aligned} $$
Solving Eqs. (i) and (ii), we have
$$ \text { or } \quad v _1=5 m / s \text { and } v _2=-30 m / s \text {. } $$
In both the cases relative velocity of separation in horizontal direction is $25 m / s$.
$\therefore x=25 t=$ distance between them when they strike the ground.
$$ \begin{aligned} & \text { Here, } \quad t=\frac{T}{2} \quad(T=\text { time of flight of projectile }) \\ & =\frac{u \sin \theta}{g}=\frac{20 \sin 60^{\circ}}{9.8}=1.77 s \\ & \therefore \quad x=25 \times 1.77 m=44.25 m \end{aligned} $$
