A 20.0 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) (9.81))) / 20.9 kg)

v = 4.96 m/s

Question

Physics

Asia

4 November, 07:09

A 20.0 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) (9.81))) / 20.9 kg)

v = 4.96 m/s

+4

Answers (1)

Know the Answer?

Kennedi Guerra · Answer 1

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.