Question

A body of mass m explodes into three fragments of with masses in the ratio 2:2:6. if the two similar masses move of perpendicular to each other with the speed of 10m/s each,find the velocity of the third particle and its direction relative to the two other bodies?

Answer 1

Assume the body is initially at rest. Let the mass of each fragment be $\displaystyle\sf {2x,\ 2x}$ and $\displaystyle\sf {6x}$ respectively, so that the total mass is $\displaystyle\sf {10x=m.}$

$\displaystyle\Longrightarrow\sf {x=\dfrac {m}{10}}$

The magnitude of the linear momentum of the first fragment equals that of the second one, i.e.,

• $\displaystyle\sf {p_1=p_2=20x=2m}$

And that of third one is taken as,

• $\displaystyle\sf {p_3=6xv_3=\dfrac {3mv_3}{5}}$

By conservation of linear momentum,

$\displaystyle\longrightarrow\sf {\vec {p_1}+\vec {p_2}+\vec {p_3}=0}$

$\displaystyle\longrightarrow\sf {-\vec {p_3}=\vec {p_1}+\vec {p_2}}$

$\displaystyle\longrightarrow\sf {(-\vec {p_3})^2=(\vec {p_1}+\vec {p_2})^2}$

$\displaystyle\longrightarrow\sf {(p_3)^2=(p_1)^2+(p_2)^2+2(\vec {p_1}\cdot\vec{p_2})}$

$\displaystyle\longrightarrow\sf {(p_3)^2=(p_1)^2+(p_2)^2+2p_1p_2\cos 90^{\circ}}$

$\displaystyle\longrightarrow\sf {(p_3)^2=(p_1)^2+(p_2)^2}$

Since the two similar masses are moving perpendicular to each other.

Then,

$\displaystyle\longrightarrow\sf {(p_3)^2=(2m)^2+(2m)^2}$

$\displaystyle\longrightarrow\sf {(p_3)^2=8m^2}$

$\displaystyle\longrightarrow\sf {p_3=-2m\sqrt2}$

Because this momentum should act opposite to the resultant of the other two of the similar masses, since the initial linear momentum of the system is zero.

$\displaystyle\longrightarrow\sf {\dfrac {3mv_3}{5}=-2m\sqrt2}$

$\displaystyle\longrightarrow\sf {\vec {v_3}=-\dfrac {10}{3}\sqrt2\ \hat i}$

The resultant velocity of the other two similar masses should be,

$\displaystyle\longrightarrow\sf {\vec {v_1}+\vec {v_2}=10\sqrt2\ \hat i}$

since each has a magnitude of $\displaystyle\sf {10\ m\ s^{-1}}$ and are perpendicular to each other.

Now, the velocity of the third particle relative to the other two is,

$\displaystyle\longrightarrow\sf {\vec {v_3}-(\vec {v_1}+\vec {v_2})=\left (-\dfrac {10\sqrt2}{3}-10\sqrt2\right)\ \hat i}$

$\displaystyle\longrightarrow\sf {\underline {\underline {\vec {v_3}-(\vec {v_1}+\vec {v_2})=\left (-\dfrac {40\sqrt2}{3}\ \hat i\right)\ m\ s^{-1}}}}$