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In this experiment, we were out to determine the initial velocity imparted to a projectile from a launcher. To do so, we use
a ballistic pendulum and principles of conservation of momentum and energy. In Figure 1, we have depicted the situation before the projectile launcher has been fired.
The projectile used is a small metal ball with a mass of m1 and the pendulum (including hanger) will have a mass of m2.
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 | | Figure 1 |
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| Just after the projectile launcher is released, the ball receives an instantaneous impact and continues down the muzzle of the launcher at a constant velocity of u1.
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 | | Figure 2 |
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| The projectile continues at a velocity of u1. We're neglecting air friction and therefore the ball's horizontal velocity is not diminished in any way. As a result, the horizontal acceleration of the ball is zero. However, you may recall that the ball will undergo a vertical acceleration of g=9.8 m/s2 and therefore experience a displacement in the vertical direction. Because of the short distance that the ball will travel, this vertical displacement can be ignored. You may also rememeber that in the case the the ball were able to continue, it would follow the path of a parabola.
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 | | Figure 3 |
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| Finally, the ball becomes in contact with the pendulum bob and the ball & bob system now travel off at a common velocity V.
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 | | Figure 4 |
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| Now comes the tough part, applying the momentum and energy conservation rules in order to determine the muzzle velocity of the projectile. Recall that momentum conservation can be written as follows: |
pfinal - pinitial = 0 | (1) |
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where pinitial & pfinal are the momentum of the system (ball + bob) at any two arbitrarily chosen moments in time.
It turns out that if we consider the initial time as in Figure 2, and the final time as in Figure 4, then a portion of the problem can be solved in a straight-forward manner.
With momentum generally defined as p = m v, then the momentum conservation for the ball+bob system can be written as
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m1 u1 + m2 u2 = m1 v1 + m2 v2 | (2) |
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Let's make some simplifications. The initial velocity of the bob is 0 m/s (u2=0) and immediately after the ball's collision with the bob, the ball and bob will have the same velocity because the two stick together (v1=
v2=V). With these changes in mind, equation (2) can be rewritten as:
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m1 u1 = m1 V + m2 V = (m1+ m2) V | (3) |
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Solving for u1, the initial velocity of the ball, we have
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Let's check out what we have here; u1 is what we are after, m1, m2 we know because we actually measure those in the laboratory, but V, dang!, we don't know. In order to proceed we must determine V, the velocity of the ball+bob system immediately after the ball collides with the bob.
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We've used conservation of momentum to come up with the previous relationships. What about conservation of energy? Can we somehow use it to generate another system of equations, one that hopefully might allow us to find V, so that from equation (4), we could then determine m1. As you might of guessed, the answer is yes. Let's try. Conservation of energy can be written as:
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| Efinal - Einitial = 0 | (6) |
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As before, it is up to us to select the appropriate sequences in time to consider. It behooves us to consider the instants of time that are presented in Figure 4 as the initial situation and in Figure 5 as the final situation. By doing so, and recalling that the kinetic energy can be generally written as KE = m V2/2, then equation (6) becomes (let M = m1+ m2):
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MV2/2 + Mgh1 = MV'2/2 + Mgh2 | (7)
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Followed by some simplifications, h1=0 m, and V', the velocity of the ball+bob system, at the top of the pendulum swing, just before it begins it's return swing, is also zero. Then equation (7) becomes
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Solving for V, and realizing that M cancels from both sides of the equation, we have that
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