An exploding cannonball is initially launched at a speed of 75.0 m/s at an angle

Question

An exploding cannonball is initially launched at a speed of 75.0 m/s at an angle of 30.0 o above the horizontal. At its highest point, the cannonball explodes into three pieces. The first piece, with a mass of 2.00 kg, heads straight upward at a speed of 15.0 m/s immediately after the explosion.

!!!!!!!!!!!!!!!!!! Draw free body diagram !!!!!!!!!!!!!!!!!!!!!!!!!!!!

2. An exploding cannonball is initially launched at a speed of 75.0 m/s at an angle of 30.0 degree above the horizontal. At its highest point, the cannonball explodes into three pieces. The first piece, with a mass of 2.00 kg, heads straight upward at a speed of 15.0 m/s immediately after the explosion. The second piece, with mass of 2.50 kg, initially moves in the direction 15.00 above the horizontal immediately after the explosion. The third piece, with a mass of 3.00 kg, initially moves in the direction 15.0 degree below the horizontal immediately after the explosion. Flow fast do the second and third masses move immediately after t he explosion?

Explanation / Answer

since canon exploded at top-most point; it had only x componenet of velocity, y component would be zero.
so the momentum before explosion: m*vx i.e 7.5 * (75 cos30) = 487.14 (in +x direction)

let vb and vc be velocity of second and third piece resp.

net momentum after explosion in +x direction: 2.5*vb*cos15 + 3*vc*cos15
i.e. => 2.41 vb + 2.88 vc = 487.139

net momentum after explosion in +y direction = 0
for that: 2*15 + 2.5*vb*sin15 - 3*vc*sin15 = 0
i.e. => 30 + 0.625 vb - 0.75 vc = 0

We have 2 equations with 2 variables vb and vc. solving them gives the answer:
vb= 77.3256 m/s and vc= 104.438 m/s

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