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.

that would be because of both gravity and friction. so     c