Question 183: Centre of Mass and Rotational Motion

Question

An alpha-particle of mass m suffers 1-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

Accepted Answer

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
$$
\begin{aligned}
\Rightarrow   \frac{1}{2} m v_1^2  =0.36 	imes \frac{1}{2} m v^2 \\
\Rightarrow  v_1  = 0.6 v .......(i)
\end{aligned}
$$
As unknown nucleus gained $64 \%$ of energy of $\alpha$, we have
$$
\begin{aligned}
& \frac{1}{2} M v_2^2=0.64 	imes \frac{1}{2} m v^2 \
& \Rightarrow v_2=\sqrt{\frac{m}{M}} 	imes 0.8 v .........(ii)
\end{aligned}
$$
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
$$
\begin{array}{rlrl}
m v =M \sqrt{\frac{m}{M}} 	imes 0.8 v-m 	imes 0.6 v \\
\Rightarrow  16 m v =\sqrt{m M} 	imes 0.8 v \\
\Rightarrow  2 m =\sqrt{m M} \\
\Rightarrow  4 m^2 =m M \Rightarrow M=4 m
\end{array}
$$

Answer

2m

Answer

4m

Answer

1.5m

Answer

3.5m

Centre of Mass and Rotational Motion

An alpha-particle of mass m suffers 1-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

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