A cannon ball is fired with a velocity $200\, m/sec$ at an angle of $60° $ with the horizontal. At the highest point of its flight it explodes into $3$ equal fragments, one going vertically upwards with a velocity $100 \,m/sec$, the second one falling vertically downwards with a velocity $100\, m/sec$. The third fragment will be moving with a velocity
A cannon ball is fired with a velocity $200\, m/sec$ at an angle of $60° $ with the horizontal. At the highest point of its flight it explodes into $3$ equal fragments, one going vertically upwards with a velocity $100 \,m/sec$, the second one falling vertically downwards with a velocity $100\, m/sec$. The third fragment will be moving with a velocity
$100\, m/s$ in the horizontal direction
$300\, m/s$ in the horizontal direction
$300 \,m/s$ in a direction making an angle of $60°$ with the horizontal
$200\, m/s$ in a direction making an angle of $60°$ with the horizontal
A cannon ball is fired with a velocity $200\, m/sec$ at an angle of $60° $ with the horizontal. At the highest point of its flight it explodes into $3$ equal fragments, one going vertically upwards with a velocity $100 \,m/sec$, the second one falling vertically downwards with a velocity $100\, m/sec$. The third fragment will be moving with a velocity
Let the velocity of third part after explosion is$ V$
After explosion momentum of system = ${\vec P_1} + {\vec P_2} + {\vec P_3}$
= $\frac{m}{3} \times 100\hat j - \frac{m}{3} \times 100\hat j + \frac{m}{3} \times V\hat i$
By comparing momentum of system before and after the explosion
$\frac{m}{3} \times 100\hat j - \frac{m}{3} \times 100\hat j + \frac{m}{3}V\hat i = 100m\hat i$
==>$V = 300\,m/s$
Other Language