The Ballistic Pendulum

The ballistic pendulum is a pendulum used to measure the speed of a bullet. A bullet is fired into the pendulum and gets embedded in the pendulum. As a result the pendulum rises to a certain height. If the masses of the pendulum and the bullet are known, the speed of the bullet can be calculated using this height.

Let the mass of the bullet be mb, the speed of the bullet be vb, the mass of the pendulum be mp, their common speed after the collision be V and the height to which the pendulum rises after the collision be h. The collision is completely inelastic and the initial speed of the pendulum is zero. Therefore

mbvb = ( mb + mp )V

and

vb = ( mb + mp )V ⁄ mb

V may be expressed in terms of h by using the principle of conservation of mechanical energy. While the pendulum rises up, there are two forces acting on the pendulum: The tension in the string and gravitational force. The tension in the string does not contribute to the work done because it is perpendicular to the displacement. The only force with non-zero contribution to the work done is gravitational force which is conservative. Hence mechanical energy of the pendulum is conserved. Its mechanical energy at the bottom and at its maximum height are equal. Assuming the origin of the coordinate system is fixed at its initial location, yi = 0, yf = h, vi = V and vf = 0.

( mb + mp )V² = (mb + mp )|g|h

and

V = √( 2|g|h )

Substituting the V in the expression for vb with this expression, the following expression for the speed of the bullet is obtained in terms of the masses of the bullet and the pendulum and the height to which the bullet rises after the collision.

vb = ( mb + mp )√( 2|g|h ) ⁄ mb

Example: A bullet of mass 0.05 kg is fired into a pendulum of mass 10 kg. The bullet is embedded in the pendulum and the pendulum rises to a height of 2 m. Calculate the speed of the bullet.

Solution: mb = 0.05 kg ; mp = 10 kg ; h = 2 m ; vb ?

vb = ( mb + mp )√( 2|g|h ) ⁄ mb = { ( 0.05 + 10 )√( 2 * 9.8 * 2 ) ⁄ 0.05 } m ⁄ s = m ⁄ s

Completely Elastic Collision

A completely elastic collision is a collision where not only momentum but also kinetic energy is conserved. For a completely elastic collision, the following two equations from the conservation of momentum and kinetic energy hold.

m1v1i + m2v2i = m1v1f + m2v2f

m1v1i 2 ⁄ 2 + m2v2i 2 ⁄ 2 = m1v1f 2 ⁄ 2 + m2v2f 2 ⁄ 2

The second equation is not in a desirable form because it containes velocities raised to the power of 2. It can be simplified though with little mathematical manupulation. Collecting terms containing m1 on one side and terms containing m2 on the other side, ( and factorizing the resulting differences between squares for the second equation) the following equations are obtained.

m1 ( v1i - v1f ) = m2 ( v2f - v2i )

m1 ( v1i - v1f )(v1i + v1f ) = m2 ( v2f - v2i )(v2f + v2i )

After dividing the second equation by the first equation and collecting the initial velocities of the resulting equation on one side and the final velocities on the other side, the following equation is obtained.

v1i - v2i = -( v1f - v2f )

Using this simplified equation, the following two equations can be used for a completely elastic collision.

m1v1i + m2v2i = m1v1f + m2v2f

v1i - v2i = -( v1f - v2f )

Example: A 4 kg object moving to the right with a speed of 5 m ⁄ s collides with a 2 kg object moving to the right with a speed of 3 m ⁄ s. If the collision is completely elastic, calculate the speeds of both objects after collision.

Solution: m1 = 4 kg ; vi1 = 5 m ⁄ s ; m2 = 2 kg ; vi2 = 3 m ⁄ s ; v1f = ? ; v2f = ?

m1v1i + m2v2i = m1v1f + m2v2f

m1v1f = m1v1i + m2v2i - m2v2f = ( 4 * 5 + 2 * 3) kg m ⁄ s - 2 kg v2f = 26 kg m ⁄ s - 2 kg v2f

v1f = ( 26 kg m ⁄ s - 2 kg v2f ) ⁄ m1 = ( 26 kg m ⁄ s - 2 kg v2f ) ⁄ 4 kg = 6.5 m ⁄ s - 0.5v2f

v1i - v2i = -( v1f - v2f )

v2f = v1i - v2i + v1f = (5 - 3) m ⁄ s + v1f = 2 m ⁄ s + v1f

and substituting the expression for v1f in terms of v2f

v2f = 2 m ⁄ s + 6.5 m ⁄ s - 0.5v2f

1.5v2f = 8.5 m ⁄ s

v2f = 8.5 ⁄ 1.5 m ⁄ s = m ⁄ s

and

v1f = 6.5 m ⁄ s - 0.5v2f = ( 6.5 - 0.5 * 5.7) m ⁄ s = m ⁄ s

Two Dimensional Collisions

Two dimensional collision is collision in a plane where more than one different directions for the velocities are involved. The principle of conservation of momentum applies in vector form.

m1v1i + m2v2i = m1v1f + m2v2f

If two vectors are equal, then their components also must be equal. Thus, this vector equation can be decomposed into two algebraic component equations: an equation for the x-components and an equation for the y-components.

m1v1ix + m2v2ix = m1v1fx + m2v2fx

m1v1iy + m2v2iy = m1v1fy + m2v2fy

Quite often, a velocity is given in the form of magnitude ( |v| ) and direction ( θ ). In that case, first the components should be calculated. If θ is a default angle ( angle measured with respect to the positive x-axis), then vx = |v| cos θ and and vy = |v| sin θ.

Example: An object of mass 6 kg going north with a speed of 10 m ⁄ s collides with a 2 kg object going east with a speed of 20 m ⁄ s. After collision the 6 kg object moves with a speed of 15 m ⁄ s making an angle of 40° with the horizontal (east).

  1. Calculate the components of the velocity of the 2 kg object after collision.

    Solution: m1 = 6 kg ; |v1i| = 10 m ⁄ s ; θ1i = 90° ; |v1f| = 15 m ⁄ s ; θ1f = 40° ; m2 = 2 kg ; |v2i| = 20 m ⁄ s ; θ2i = 0° ; v2fx = ? ; v2fy = ?

    v1ix = |v1i| cos θ1i = ( 10 * cos 90° ) m ⁄ s = 0

    v1iy = |v1i| sin θ1i = ( 10 * sin 90° ) m ⁄ s = 10 m ⁄ s

    v2ix = |v2i| cos θ2i = ( 20 * cos 0° ) m ⁄ s = 20 m ⁄ s

    v2iy = |v2i| sin θ2i = ( 20 * sin 0° ) m ⁄ s = 0

    v1fx = |v1f| cos θ1f = ( 15 * cos 40° ) m ⁄ s = m ⁄ s

    v1fy = |v1f| sin θ1f = ( 15 * sin 40° ) m ⁄ s = m ⁄ s

    m1v1ix + m2v2ix = m1v1fx + m2v2fx

    m2v2fx = m1v1ix + m2v2ix - m1v1fx = ( 6 * 0 + 2 * 20 - 6 * 11.5) kg m ⁄ s = -29 kg m ⁄ s

    v2fx = -29 ⁄ m2 kg m ⁄ s = -29 ⁄ 2 m ⁄ s = -14.5 m ⁄ s

    m1v1iy + m2v2iy = m1v1fy + m2v2fy

    m2v2fy = m1v1iy + m2v2iy - m1v1fy = ( 6 * 10 + 2 * 0 - 6 * 9.6) kg m ⁄ s = 57.6 kg m ⁄ s

    v2fx = 57.6 ⁄ m2 kg m ⁄ s = 57.6 ⁄ 2 m ⁄ s = 28.8 m ⁄ s

  2. Calculate the magnitude and the direction of the velocity of the 2 kg object after collision.

    Solution: |v2f| = ? ; θ2f = ?

    |v2f| = √( v2fx² + v2fy² ) = √{(-14.5)² + 28.8² } = m ⁄ s

    θ2f = arctan ( v2fy ⁄ v2fx ) + 180° = arctan ( 28.8 ⁄ -14.5 ) + 180° = °