Q.32 A cannon ball is fired with a velocity 200m/secat an angle of 60�

1.	Q.32 A cannon ball is fired with a velocity 200m/secat an angle of 60° with the horizontal. At thehighest point of its flight. It explodes into 3equal fragments, one going vertically upwardswith a velocity 100m/sec, the second onefalling vertically downwards with a velocity100 m/sec. The third fragment will be movingwith a velocity(A) 100 m/sec in the horizontal direction(B) 300m/sec in the horizontal direotion(C) 300 m/sec in a direction making an angleof 60° with the horizontal(D) 200 m/sec in a direction making an angleof 60° with the horizontal
Answer» i think A option is correct answer 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

Q.32 A cannon ball is fired with a velocity 200m/secat an angle of 60° with the horizontal. At thehighest point of its flight. It explodes into 3equal fragments, one going vertically upwardswith a velocity 100m/sec, the second onefalling vertically downwards with a velocity100 m/sec. The third fragment will be movingwith a velocity(A) 100 m/sec in the horizontal direction(B) 300m/sec in the horizontal direotion(C) 300 m/sec in a direction making an angleof 60° with the horizontal(D) 200 m/sec in a direction making an angleof 60° with the horizontal

Answer»

i think A option is correct answer

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

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