Question
A shell is fired from a cannon with a velocity $$v$$ $$\left( {m/\sec .} \right)$$ at an angle $$\theta $$ with the horizontal direction. At the highest point in its path it explodes into two pieces of equal mass. One of the pieces retraces its path to the cannon and the speed (in $$m/\sec.$$ ) of the other piece immediately after the explosion is
A.
$$3v\cos \theta $$
B.
$$2v\cos \theta $$
C.
$$\frac{3}{2}v\cos \theta $$
D.
$$\sqrt {\frac{3}{2}} v\cos \theta $$
Answer :
$$3v\cos \theta $$
Solution :
As one piece retraces its path, the speed of this piece just after explosion should be $$v\cos \theta $$
Applying conservation of linear momentum at the highest point;
$$\eqalign{ & m\left( {v\cos \theta } \right) = \frac{m}{2} \times v' - \frac{m}{2} \times v\cos \theta \cr & 3v\cos \theta = v' \cr} $$
As one piece retraces its path, the speed of this piece just after explosion should be $$v\cos \theta $$
Applying conservation of linear momentum at the highest point;
$$\eqalign{ & m\left( {v\cos \theta } \right) = \frac{m}{2} \times v' - \frac{m}{2} \times v\cos \theta \cr & 3v\cos \theta = v' \cr} $$