Centre of Mass 3 Question 36
39. A body of mass $m$ moving with a velocity $v$ in the $x$-direction collides with another body of mass $M$ moving in the $y$-direction with a velocity $V$. They coalesce into one body during collision. Find
(a) the direction and magnitude of the momentum of the composite body.
(b) the fraction of the initial kinetic energy transformed into heat during the collision.
(1978)
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Answer:
Correct Answer: 39. (a) At an angle $\tan ^{-1} \frac{M V}{m v}$ with positive $x$-axis. Magnitude is
$$ \sqrt{(m v)^{2}+(M V)^{2}} \quad \text { (b) Fraction }=\frac{M m\left(v^{2}+V^{2}\right)}{(M+m)\left(m v^{2}+M V^{2}\right)} $$
Solution:
- (a) From conservation of linear momentum, momentum of composite body
$$ \begin{aligned} \mathbf{p} & =\left(\mathbf{p} _i\right) _1+\left(\mathbf{p} _i\right) _2=(m v) \hat{\mathbf{i}}+(M V) \hat{\mathbf{j}} \\ \therefore \quad|\mathbf{p}| & =\sqrt{(m v)^{2}+(M V)^{2}} \end{aligned} $$
Let it makes an angle $\alpha$ with positive $X$-axis, then
$$ \alpha=\tan ^{-1} \frac{p _y}{p _x}=\tan ^{-1} \frac{M V}{m v} $$
(b) Fraction of initial kinetic energy transformed into heat during collision
$$ \begin{aligned} & =\frac{K _f-K _i}{K _i}=\frac{K _f}{K _i}-1=\frac{p^{2} / 2(M+m)}{\frac{1}{2} m v^{2}+\frac{1}{2} M V^{2}}-1 \\ & =\frac{(m v)^{2}+(M V)^{2}}{(M+m)\left(m v^{2}+M V^{2}\right)}-1 \\ & =\frac{M m\left(v^{2}+V^{2}\right)}{(M+m)\left(m v^{2}+M V^{2}\right)} \end{aligned} $$
