Question

A shell is fired from a cannon with a speed of $100m{s}^{-1}$ at an angle ${30}^{\circ }$ with the vertical (y-direction). At the highest point of trajectory, the shell explodes into two fragments of masses in the ratio $1:2$. The lighter fragments move vertically upwards with an initial speed of $200m/s$. What is the speed of the heavier fragments at the time of the explosion?

Answer 1

$u_x=u\cos {60}^o=100\cos {60}^o$

or $u_x=50m/s$

At top must point the center of mass will have only horizontal component of velocity.

( & this will be $u_x=50m/s$)

If shell was of $3m$

then the fragment $m$ will move upward with $200m/s$

so to make Net velocity of center of mass zero in vertical direction $mx200+2mx{v}_y=0$

or $v_y=-100m/s$ i.e downward.

only the bigger part is possessing the horizontal component of velocity so $\frac{mx0+2m{v}_x}{m+2m}=u_x$

$\Rightarrow V_x=\frac{3}{2}{u}_x=75m/s$

$speed=\sqrt{u_x^2+v_y^2}=\sqrt{{75}^2+(-100{)}^2}=125m/s$