An alpha-particle of mass m undergoes a 1-dim | vedclass

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(D) Let the initial velocity of the alpha-particle be $v_0$ and its final velocity be $-v_1$ (since it is scattered backwards).Let the mass of the nuc

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(D) Let the initial velocity of the alpha-particle be $v_0$ and its final velocity be $-v_1$ (since it is scattered backwards).Let the mass of the nucleus be $M$ and its final velocity be $v_2$.According to the conservation of linear momentum:$mv_0 = -mv_1 + Mv_2$  --- $(1)$According to the property of $1$-dimensional elastic collision:$v_0 = v_1 + v_2$  --- $(2)$From $(2)$,$v_1 = v_0 - v_2$. Substituting this into $(1)$:$mv_0 = -m(v_0 - v_2) + Mv_2$$mv_0 = -mv_0 + mv_2 + Mv_2$$2mv_0 = (m + M)v_2 \Rightarrow v_2 = \frac{2mv_0}{m + M}$The final kinetic energy of the alpha-particle is $K_f = \frac{1}{2}mv_1^2$.Given that it loses $64\%$ of its initial kinetic energy $(K_i = \frac{1}{2}mv_0^2)$,the remaining kinetic energy is $36\%$ of $K_i$:$K_f = 0.36 K_i \Rightarrow \frac{1}{2}mv_1^2 = 0.36 	imes \frac{1}{2}mv_0^2$$v_1^2 = 0.36 v_0^2 \Rightarrow v_1 = 0.6 v_0$Using the formula for final velocity in elastic collision: $v_1 = \left( \frac{m - M}{m + M} ight) v_0$ (where $v_1$ is in the direction of $v_0$).Since it is scattered backwards,$v_1 = -0.6 v_0$,so $\frac{m - M}{m + M} = -0.6$.$m - M = -0.6m - 0.6M$$1.6m = 0.4M \Rightarrow M = 4m$.

An alpha-particle of mass $m$ undergoes a $1$-dimensional elastic collision with a nucleus of unknown mass $M$ at rest. It is scattered directly backwards,losing $64\%$ of its initial kinetic energy. The mass of the nucleus is .......... $m$.

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