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A cannon ball is fired with a velocity 200m/s 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 of 100m/s, the second one falling vertically downwards with a velocity of 100m/s . What is the velocity of the third fragment?

A cannon ball is fired with a velocity 200m/s 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 of 100m/s, the second one falling vertically downwards with a velocity of 100m/s . What is the velocity of the third fragment?

Grade:11

1 Answers

Nayak Roshan
44 Points
7 years ago
Hi Anas, I am here to clear your doubt.
In this problem it is given that the flight explodes into 3 equal fragments so the mass of the fragments must also be same.
BY USING THE METHOD OF CENTER OF MASS WE GET 
V1+V2+V3/3=VELOCITY OF CENTER OF MASS 
WHERE V1= VELOCITY FRAGMENT MOVING VERTICALLY UPWARDS 
V2=VELOCITY OF FRAGMENT MOVING VERTICALLY DOWNWARDS 
V3 = VELOCITY OF UNKNOWN FRAGMENT.
HERE AS THE TWO FRAGMENTS ARE MOVING IN VERTICALLY COMPONENT THEIR HORIZONTAL COMPONENT IS ZERO.AND VERTICALLY COMPONENT OF THE FLIGHT WHEN IT EXPLODES IS ZERO AS IT IS AT THE MAX HEIGHT (I THINK YOU KNOW THE BASICS OF PROJECTILE)....
1).SO VELOCITIES ALONG VERTICALLY 
V1y+V2y+V3y/3=velocity of center of mass
Let us take take the upward coordinate axis as positive and downward as negative....
So 100-100+V3y/3=0,this is equal to zero because vertical component of the flight at max height is zero.
Therefore this gives V3y=0
2).velocity along horizontal
V1x+V2x+V3x/3=VELOCITY OF horizontal component of center of mass (here center of mass is of the flight )
V1x=V2x=0 as they are moving vertically and not horizontally or at some angle .
And velocity of center of mass = vcos (theta) ( BECAUSE AT MAX HEIGHT HORIZONTAL COMPONENT IS VCOS (THETA) , WHERE V=VELOCITY OF PROJECTION,AND THETA=ANGLE OF PROJECTION)
THEREFORE VCOS (THETA)=200 (COS60)=200×1/2=100m/s
Therefore V1x+V2x+V3x/3=100
0+0+V3x=100
Therefore V3x=100
Therefore net velocity of the third fragment is resultant of the vertices and horizontal component 
V3=root (Vx2+Vy2)
V3=rroot (10000+0)
V3=100m/s
IF YOU DO NOT KNOW THE BASICS OF PROJECTILE MOTION AND CENTER OF MASS YOU CAN'T UNDERSTAND THIS QUESTION.SO PLEASE LEARN THE BASICS FIRST AND THEN COME TO THIS SOLUTION PART.IF YOU KNOW THE BASICS THEN YOU R GOOD WELL TO GO THROUGH THIS SOLUTION. 
HOPE THIS WILL HELP YOU....
 

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