A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei the products must move in opposite directions.
A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei the products must move in opposite directions.
Let $m, m_{1},$ and $m_{2}$ be the respective masses of the parent nucleus and the two daughter nuclei. The parent nucleus is at rest.
Initial momentum of the system (parent nucleus) $=0$
Let $v_{1}$ and $v_{2}$ be the respective velocities of the daughter nuclei having masses $m_{1}$ and $m_{2} .$
Total linear momentum of the system after disintegration $=m_{1} v_{1}+m_{2} v_{2}$
According to the law of conservation of momentum:
Total initial momentum $=$ Total final momentum
$0=m_{1} v_{1}+m_{2} v_{2}$
$v_{1}=\frac{-m_{2} v_{2}}{m_{1}}$
Here, the negative sign indicates that the fragments of the parent nucleus move in directions opposite to each other.
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A machine gun of mass $10\,kg$ fires $20\,g$ bullets at the rate of $180$ bullets per minute with a speed of $100\,m s ^{-1}$ each. The recoil velocity of the gun is $.............\,m/s$
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The balls, having linear momenta $\vec{p}_1=\hat{p} \hat{i}$ and $\vec{p}_2=-p \hat{i}$, undergo a collision in free space. There is no external force acting on the balls. Let $\vec{p}_1^{\prime}$ and $\vec{p}_2^{\prime}$ be their final momenta. The following option$(s)$ is (are) $NOT ALLOWED$ for any non-zero value of $\mathrm{p}, \mathrm{a}_1, \mathrm{a}_2, \mathrm{~b}_1, \mathrm{~b}_2, \mathrm{c}_1$ and $\mathrm{c}_2$.
$(A)$ $ \overrightarrow{\mathrm{p}}_1^{\prime}=\mathrm{a}_1 \hat{\mathrm{i}}+\mathrm{b}_1 \hat{\mathrm{j}}+\mathrm{c}_1 \hat{\mathrm{k}} $
$ \overrightarrow{\mathrm{p}}_2^{\prime}=\mathrm{a}_2 \hat{\mathrm{i}}+\mathrm{b}_2 \hat{\mathrm{j}}$
$(B)$ $ \overrightarrow{\mathrm{p}}_1^{\prime}=\mathrm{c}_1 \hat{\mathrm{k}} $
$ \overrightarrow{\mathrm{p}}_2^{\prime}=\mathrm{c}_2 \hat{\mathrm{k}}$
$(C)$ $ \overrightarrow{\mathrm{p}}_1^{\prime}=\mathrm{a}_1 \hat{\mathrm{i}}+\mathrm{b}_1 \hat{\mathrm{j}}+\mathrm{c}_1 \hat{\mathrm{k}} $
$ \overrightarrow{\mathrm{p}}_2=\mathrm{a}_2 \hat{\mathrm{i}}+\mathrm{b}_2 \hat{\mathrm{j}}-\mathrm{c}_1 \hat{\mathrm{k}}$
$(D)$ $ \vec{p}_1^{\prime}=a_1 \hat{i}+b_1 \hat{j} $
$ \overrightarrow{\mathrm{p}}_2^{\prime}=a_2 \hat{\mathrm{i}}+b_1 \hat{\mathrm{j}}$
The balls, having linear momenta $\vec{p}_1=\hat{p} \hat{i}$ and $\vec{p}_2=-p \hat{i}$, undergo a collision in free space. There is no external force acting on the balls. Let $\vec{p}_1^{\prime}$ and $\vec{p}_2^{\prime}$ be their final momenta. The following option$(s)$ is (are) $NOT ALLOWED$ for any non-zero value of $\mathrm{p}, \mathrm{a}_1, \mathrm{a}_2, \mathrm{~b}_1, \mathrm{~b}_2, \mathrm{c}_1$ and $\mathrm{c}_2$.
$(A)$ $ \overrightarrow{\mathrm{p}}_1^{\prime}=\mathrm{a}_1 \hat{\mathrm{i}}+\mathrm{b}_1 \hat{\mathrm{j}}+\mathrm{c}_1 \hat{\mathrm{k}} $
$ \overrightarrow{\mathrm{p}}_2^{\prime}=\mathrm{a}_2 \hat{\mathrm{i}}+\mathrm{b}_2 \hat{\mathrm{j}}$
$(B)$ $ \overrightarrow{\mathrm{p}}_1^{\prime}=\mathrm{c}_1 \hat{\mathrm{k}} $
$ \overrightarrow{\mathrm{p}}_2^{\prime}=\mathrm{c}_2 \hat{\mathrm{k}}$
$(C)$ $ \overrightarrow{\mathrm{p}}_1^{\prime}=\mathrm{a}_1 \hat{\mathrm{i}}+\mathrm{b}_1 \hat{\mathrm{j}}+\mathrm{c}_1 \hat{\mathrm{k}} $
$ \overrightarrow{\mathrm{p}}_2=\mathrm{a}_2 \hat{\mathrm{i}}+\mathrm{b}_2 \hat{\mathrm{j}}-\mathrm{c}_1 \hat{\mathrm{k}}$
$(D)$ $ \vec{p}_1^{\prime}=a_1 \hat{i}+b_1 \hat{j} $
$ \overrightarrow{\mathrm{p}}_2^{\prime}=a_2 \hat{\mathrm{i}}+b_1 \hat{\mathrm{j}}$
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Two billiard balls of mass $0.05\,kg$ each moving in opposite directions with $10\,ms ^{-1}$ collide and rebound with the same speed. If the time duration of contact is $t=0.005\,s$, then $\dots N$is the force exerted on the ball due to each other.
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