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A cannot ball is fired with a velocity 2...

A cannot ball is fired with a velocity 200m / 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

Text Solution

Verified by Experts

The correct Answer is:
B
A highest point `v=u cos60^@ =200 xx1/2 = 100 m//s`
Law of conservation of momentum
`mv(hati)=(m/3)v_1(hatj)+m/3v_2(-hatj)+(m/3)v_3`
`implies v_3 =3v = 300 m//s(hati)`
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