Centre of Mass 3 Question 5
6. An $\alpha$-particle of mass $m$ suffers one-dimensional elastic collision with a nucleus at rest of unknown mass. It is scattered directly backwards losing $64 %$ of its initial kinetic energy. The mass of the nucleus is $\quad$ (2019 Main, 12 Jan II)
(a) $1.5 m$
(b) $4 m$
(c) $3.5 m$
(d) $2 m$
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Answer:
Correct Answer: 6. (b)
Solution:
- We have following collision, where mass of $\alpha$ particle $=m$ and mass of nucleus $=M$
Let $\alpha$ particle rebounds with velocity $v _1$, then Given; final energy of $\alpha=36 %$ of initial energy
$$ \begin{array}{rlrl} \Rightarrow & & \frac{1}{2} m v _1^{2} & =0.36 \times \frac{1}{2} m v^{2} \\ \Rightarrow & v _1 & =0.6 v \end{array} $$
As unknown nucleus gained $64 %$ of energy of $\alpha$, we have
$$ \begin{aligned} \frac{1}{2} M v _2^{2} & =0.64 \times \frac{1}{2} m v^{2} \\ v _2 & =\sqrt{\frac{m}{M}} \times 0.8 v \end{aligned} $$
From momentum conservation, we have
$$ m v=M v _2-m v _1 $$
Substituting values of $v _1$ and $v _2$ from Eqs. (i) and (ii), we have
$$ \begin{array}{rlrl} & & m v & =M \sqrt{\frac{m}{M}} \times 0.8 v-m \times 0.6 v \\ \Rightarrow & 1.6 m v & =\sqrt{m M} \times 0.8 v \\ \Rightarrow & 2 m & =\sqrt{m M} \\ \Rightarrow & 4 m^{2} & =m M \Rightarrow M=4 m \end{array} $$
7 According to the given condition in the question, after collision the mass of combined system is doubled. Also, this system would be displaced from its circular orbit.
So, the combined system revolves around centre of mass of ‘system + earth’ under action of a central force.
Hence, orbit must be elliptical.
8 Pendulum’s velocity at lowest point just before striking mass $m$ is found by equating it’s initial potential energy (PE) with final kinetic energy (KE).
Initially, when pendulum is released from angle $\theta _0$ as shown in the figure below,
We have,
Here,
$$ \begin{aligned} m g h & =\frac{1}{2} m v^{2} \\ h & =l-l \cos \theta _0 \\ v & =\sqrt{2 g l\left(1-\cos \theta _0\right)} \end{aligned} $$
$$ \text { So, } $$
With velocity $v$, bob of pendulum collides with block. After collision, let $v _1$ and $v _2$ are final velocities of masses $m$ and $M$ respectively as shown
Then if pendulum is deflected back upto angle $\theta _1$, then
$$ v _1=\sqrt{2 g l\left(1-\cos \theta _1\right)} $$
Using definition of coefficient of restitution to get
$$ e=\frac{\mid \text { velocity of separation } \mid}{\mid \text { velocity of approach } \mid} $$
$$ 1=\frac{v _2-\left(-v _1\right)}{v-0} \Rightarrow v=v _2+v _1 $$
From Eqs. (i), (ii) and (iii), we get
$$ \begin{aligned} & \Rightarrow \sqrt{2 g l\left(1-\cos \theta _0\right)}=v _2+\sqrt{2 g l\left(1-\cos \theta _1\right)} \\ & \Rightarrow \quad v _2=\sqrt{2 g l}\left(\sqrt{1-\cos \theta _0}-\sqrt{1-\cos \theta _1}\right) \end{aligned} $$
According to the momentum conservation, initial momentum of the system $=$ final momentum of the system
$$ \begin{aligned} \Rightarrow \quad m v & =M v _2-m v _1 \\ \Rightarrow M v _2 & =m\left(v+v _1\right) \\ M v _2 & =m \sqrt{2 g l}\left(\sqrt{1-\cos \theta _0}+\sqrt{1-\cos \theta _1}\right) \end{aligned} $$
Dividing Eq. (v) and Eq. (iv), we get
$\Rightarrow \frac{M}{m}=\frac{\sqrt{1-\cos \theta _0}+\sqrt{1-\cos \theta _1}}{\sqrt{1-\cos \theta _0}-\sqrt{1-\cos \theta _1}}$
$$ \begin{gathered} =\frac{\sqrt{\sin ^{2} \frac{\theta _0}{2}}+\sqrt{\sin ^{2} \frac{\theta _1}{2}}}{\sqrt{\sin ^{2} \frac{\theta _0}{2}-\sqrt{\sin ^{2} \frac{\theta _1}{2}}}} \\ \frac{M}{m}=\frac{\sin \frac{\theta _0}{2}+\sin \frac{\theta _1}{2}}{\sin \frac{\theta _0}{2}-\sin \frac{\theta _1}{2}} \end{gathered} $$
For small $\theta _0$, we have
$$ \begin{aligned} \frac{M}{m} & =\frac{\frac{\theta _0}{2}+\frac{\theta _1}{2}}{\frac{\theta _0}{2}-\frac{\theta _1}{2}} \\ \text { or } \quad M & =m \frac{\theta _0+\theta _1}{\theta _0-\theta _1} \end{aligned} $$
