A20.0 kg wood ball hangs from a 1.30 m -long wire. the maximum tension the wire can withstand without breaking is 600 n . a 0.900 kg projectile traveling horizontally hits and embeds itself in the wood ball. what is the largest speed this projectile can have without causing the cable to break?

original work:

m = 20 kg wood ball + .9 kg projectile = 20.9 kg
ft max = 600 n
1.3 m radius

f = mar = ft - fg
m(v^2)/r = ft - fg

v = sqrt((r(ft - fg))/m)
v = sqrt((1.3m(600n - (20.9 kg)(9./20.9 kg)
v = 4.96 m/s

The largest speed the projectile can have without causing the cable to break is 115 m/s.

The given parameters;

Apply the principle of conservation of linear momentum;

The net force of the wood-projectile system in the circular path is calculated as follows;

The largest speed the projectile can have;

Thus, the largest speed the projectile can have without causing the cable to break is 115 m/s.

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115 m/s

Explanation:

Momentum before = momentum after

mu = (M + m) v

u = (M + m) v / m

u = (20.0 + 0.900) v / (0.900)

u = 23.2 v

Sum of the forces in the radial direction:

∑F = ma

T − (M + m) g = (M + m) v² / r

600 − (20.0 + 0.900) (9.8) = (20.0 + 0.900) v² / 1.30

v = 4.96

u = 23.2 (4.96)

u = 115 m/s

You correctly found the speed of the ball/projectile after the collision.  You just have to use conservation of momentum to find the projectile's speed before the collision.

any one of the statements could be an accurate comparison. but the statement that would define the correct choice and seal the deal is not even on the list:

"the visibilities of vector-x and vector-y in the drawing provided with the question are identical and nonexistent."