A bullet buries itself in a pendulum. Determine the horizontal component of the pendulum's displacement.?

Question

The bullet has a mass of 0.014 kg and is traveling at 220 m/s, the pendulum is at rest and has a mass of 3.1 kg. The pendulum is hanging from a string that is 3.5 m long.

Answer 1

Using conservation of momentum find the velocity V of bullet+pendulum after the collision:

m*v = (M + v)V, V = m*v/(M + m), V = 0.014*220/(3.1 + 0.014)

V = 0.989 m/s or V = 1 m/s (approx)

Using the conservation of the mechanical energy find the height that the pendulum+ bullet goes:

(M +m)*V^2/2 = (M + m)*g*h, h = V^2/2*g, h = 1/2*10, h = 0.05 m.

You can determine the horizontal displacement x using the Pythagorean theorem:

x = sqrt(L^2 - (L - h)^2)

x = sqrt(3.5^2 - 3.45^2), x = 0.59 m

Answer 2

This is done by energy balance. First, call the mass of the bullet m and the mass of the pendulum M. The bullet initially have kinetic energy 0.5*m*v^2, where you were given v. Then bullet will enter the pendulum, and we assume that the bullet's kinetic energy is completely transformed into potential energy of the pendulum plus bullet system. The gravitational potential energy of the new mass will be (M+m)gh, where g is the acceleration due to gravity. So we'll have 0.5mv^2 = (M+m)gh, where h is your only unknown, and you can solve for it algebraically. Then, use the Pythagorean theorem to determine the horizontal distance that corresponds to that height, remembering that the hypotenuse of the triangle is the length L of the pendulum, which you were also given. L^2 = x^2 + y^2, where y will be the h you calculated, and x is your only unknown, which you can now calculate algebraically as well.

Answer 3

Using conservation of energy the vertical displacement of the pendulum will be y
the kinetic energy of the bullet is
1/2*m*v^2
which will get transferred to the vertical displacement of the resultant mass:

.5*.014*220^2=3.114*9.8*y
y=(.5*.014*220^2)/(3.114*9.8)
=11.1m

This is greater than twice the length of the string, so the resultant mass will spin around through the top of the circle.

This answer is incorrect, sorry.
The collision is inelastic, so kinetic energy is not conserved. The conservation of momentum is the right approach as taken by Dimos.

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