Question

A bomb of \[300\;{\text{grams}}\] at rest explodes into three equal pieces, two of them found to move perpendicular to each with \[20\;{\text{m/s}}\] speed. Find the speed and the direction of \[{3^{rd}}\] piece.

Answer 1

Hint: Calculate the angle made by the third piece of exploded bomb. Adding and squaring in the equation to obtain the speed of the third piece of the exploded bomb.

Complete step by step answer:
Let us represent the first piece of the bomb by \[1\] and represent the second piece by \[2\] and represent the third piece by \[3\]. And represent the speed of pieces by \[v\].
A bomb of \[300\;{\text{grams}}\] is exploded into three equal pieces. Two of them are right angled to each. The speed of the two pieces is \[20\;{\text{m/s}}\].
Let us consider the figure,


The linear momentum is conserved before and after the explosion of the bomb.
So, the bomb is at rest before it explodes. Therefore, zero is the initial momentum of the bomb.
We consider that the piece \[1\] moves along \[x - \]axis, and the piece \[2\] along the \[y - \]axis, we assume that the third piece \[3\] moves towards the angle of \[\theta \] with the negative \[x - \]axis at speed of \[v\]
The \[x - \]axis is conserving,
\[ \Rightarrow 0 = 1\left( {20} \right) - 1\left( {v\cos \theta } \right)\]
Now we simplify the above expression.
\[ \Rightarrow v\cos \theta = 20......\left( 1 \right)\]
And the \[y - \]axis is conserving,
\[ \Rightarrow 0 = 1\left( {20} \right) - 1\left( {v\sin \theta } \right)\]
Now we simplify the above expression as,
\[ \Rightarrow v\sin \theta = 20......\left( 2 \right)\]
From equation (1) and (2) we get,
\[ \Rightarrow \tan \theta = \dfrac{{v\sin \theta }}{{v\cos \theta }}\]
Now we substitute the values as,
\[ \Rightarrow \tan \theta = \dfrac{{20}}{{20}}\]
Now, solve the above expression as,
\[ \Rightarrow \tan \theta = 1\]

Now, solve further for angle $\theta $.
\[ \Rightarrow \theta = {45^{\text{o}}}\],
So, we calculate as,
\[ \Rightarrow \phi = {180^{\text{o}}} - {45^{\text{o}}}\]
Now we solve the above expression as,
\[ \Rightarrow \phi = {135^{\text{o}}}\]
The third piece moves with an angle of \[{135^{\text{o}}}\] with \[1\].
Now, calculate the speed of the third piece and adding equation \[\left( 1 \right)\] and \[\left( 2 \right)\],
\[ \Rightarrow v\cos \theta + v\sin \theta = 20 + 20\]
By simplifying the above relation, we get,
\[ \Rightarrow v\cos \theta + v\sin \theta = 40......\left( 3 \right)\]
Now we are squaring both sides to get
\[ \Rightarrow {v^2}\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right) = {20^2} + {20^2}\]
Now, simplify the expression.
\[ \Rightarrow {v^2}\left( 1 \right) = 400 + 400\]
Now, solve the expression further.
\[ \Rightarrow {v^2} = 800\]
We solve the value of $v$ as,
\[ \Rightarrow v = 20\sqrt 2 \;{\text{m/s}}\]

Therefore, the speed of the third piece that explodes is \[20\sqrt 2 \;{\text{m/s}}\].

Note: As we know that the velocity is the vector quantity and the resultant of the velocities will be calculated by using the vector addition that is it depends on the direction of the velocities.