Centre of Mass 3 Question 10
13. A particle of mass $m$ moving in the $x$-direction with speed $2 v$ is hit by another particle of mass $2 m$ moving in the $y$-direction with speed $v$. If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to
(2015 Main)
(a) $50 %$
(b) $56 %$
(c) $62 %$
(d) $44 %$
Show Answer
Answer:
Correct Answer: 13. (b)
Solution:
- In all type of collisions, momentum of the system always remains constant. In perfectly inelastic collision, particles stick together and move with a common velocity.
Let this velocity is $\mathbf{v} _c$. Then,
initial momentum of system $=$ final momentum of system
or $\quad m(2 v) \hat{\mathbf{i}}+2 m(v) \hat{\mathbf{j}}=(m+2 m) \mathbf{v} _c$
$\therefore \quad \mathbf{v} _c=\frac{2}{3}(v \hat{\mathbf{i}}+v \hat{\mathbf{j}})$
$\left|\mathbf{v} _c\right|$ or $v _c$ or speed $=\sqrt{\frac{2}{3} v^{2}+\frac{2}{3} v^{2}}=\frac{2 \sqrt{2}}{3} v$
Initial kinetic energy
$$ K _i=\frac{1}{2}(m)(2 v)^{2}+\frac{1}{2}(2 m)(v)^{2}=3 m v^{2} $$
Final kinetic energy
$$ K _f=\frac{1}{2}(3 m) \frac{2 \sqrt{2}}{3} v^{2}=\frac{4}{3} m v^{2} $$
Fractional loss $=\frac{K _i-K _f}{K _i} \times 100$
$$ =\frac{\left(3 m v^{2}\right)-\frac{4}{3} m v^{2}}{\left(3 m v^{2}\right)} \times 100=56 % $$
