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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

Respuesta :

MathPhys

Answer:

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.

onyebuchinnaji

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

The given parameters;

  • mass of the wood, m = 20 kg
  • length of the wire, l = 1.3 m
  • maximum tension on the wire, T = 600 N
  • mass of the projectile, = 0.9 kg

Apply the principle of conservation of linear momentum;

[tex]m_1 u_1 + m_2u_2 = v(m_1 +m_2)\\\\20(0) \ + \ 0.9u_2 = v(20 + 0.9)\\\\u_2 = \frac{20.9v}{0.9} \\\\u_2 = 23.2 v[/tex]

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

[tex]\Sigma F= 0\\\\T - g(m_1 + m_2) = (m_1+ m_2 ) \frac{v^2}{r} \\\\600 \ - \ 9.8(20 + 0.9) = (20 + 0.9) \times \frac{v^2}{1.3} \\\\395.18 = 16.08v^2\\\\v^2 = 24.58\\\\v= \sqrt{24.58} \\\\v = 4.957 \ m/s[/tex]

The largest speed the projectile can have;

[tex]u_2 = 23.2v\\\\u_2 = 23.2 \times 4.957\\\\u_2 = 115 \ m/s[/tex]

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

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