Question

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. 100 m/s in the horizontal direction
• B. 300 m/s in the horizontal direction
• C. 300 m/s in a direction making an angle of 60° with the horizontal
• D. 200 m/s in a direction making an angle of 60° with the horizontal

Answer 1

Answer: (B)

300 m/s in the horizontal direction

Answer 2

Momentum of ball (mass m) before explosion at the highest point $= m v \hat { i } = m u \cos 60 ^ { \circ } \hat { i }$

= $m \times 200 \times \frac { 1 } { 2 } \hat { i }$=

Let the velocity of third part after explosion is V

After explosion momentum of system = $\vec { P } _ { 1 } + \vec { P } _ { 2 } + \vec { P } _ { 3 }$

= $\frac { m } { 3 } \times 100 \hat { j } - \frac { m } { 3 } \times 100 \hat { j } + \frac { m } { 3 } \times \hat { V i }$

By comparing momentum of system before and after the explosion

$\frac { m } { 3 } \times 100 \hat { j } - \frac { m } { 3 } \times 100 \hat { j } + \frac { m } { 3 } \hat { V i } = 100 m \hat { i }$ ⇒$V = 300 \mathrm {~m} / \mathrm { s }$