A shell is fired from a cannon with a velocity `v (m//sec.)` 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 isA. `3vcos theta`B. `2vcos theta`C. `(3v)/(2)cos theta`D. `(sqrt(3v)cos theta)/(2)`

A shell is fired from a cannon with a velocity `v (m//sec.)` at an ang

1.	A shell is fired from a cannon with a velocity `v (m//sec.)` 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 isA. `3vcos theta`B. `2vcos theta`C. `(3v)/(2)cos theta`D. `(sqrt(3v)cos theta)/(2)`
Answer» Correct Answer - A In case of projectile motion as the highest point `(v)_(vertical)=0` and `(v)_(horizontal)=vcostheta` . the initial linear momentum of the system will be `mvcostheta` . Now as force of blasting is internal and force of gravity is vertical, so linear momentum of the system along horizontal is conserved, i.e., `P_(1)+P_(2)=mvcostheta` or `m_(1)v_(1)+m_(2)v_(2)=mvcostheta` . But it is given that `m_(1)=m_(2)=m//2` and as one part retraces its path, `v_(1)=-vcostheta` . `(1)/(2)m(-vcostheta)+(1)/(2)mv_(2)=mvcostheta` . Solving, we get: `v_(2)=3_(v)costheta` .

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