Objective:
The objective of this lab is to measure the initial velocity of a projectile in two independent ways. The first way is to treat it as though it is a projectile moving according to the kinematic equation. The second way is to apply the conservation of linear momentum and energy to a ballistic pendulum.
Equipment:
Theory and Equations:
If the projectile is fired horizontally with an initial velocity V_{o}, it will follow the parabolic path shown in figure 1.The horizontal range x and vertical displacement y of the projectile are:
x = vt (1)
y = 1/2 gt^{2} (2)
Eliminating the value of t from (1) and (2) we get:
V_{0}=x√(g/2y) (3)
The initial velocity V_{o }of the projectile can also be determined by using the ballistic pendulum (Figure 2). It consists of a spring gun that fires a metallic ball of mass m. This ball is caught by a catcher at the end of a pendulum of mass M. The collision between the ball and pendulum is perfectly inelastic. As a result, the combination (m + M) swings upward until it stops at the highest point by a ratchet.
From conservation of momentum,
mV_{o } = (m + M) V_{1 }(4)
where V_{1 }is the common velocity of pendulum – ball just after collision. If the combination rises through a height h, then from conservation of mechanical energy we have:
(1/2) (m + M) V_{1}^{2 } = ( m + M ) g h (5)
By combining (4) and (5) we determine the velocity V_{o} to be:
V_{0} = (1+ M/m) √2gh (6)
Therefore, we can find the speed V_{o} of the projectile in two different ways using equation (3) and (6) and compare the results.
Procedures:
Projectile Motion
Ballistic Pendulum
Data Table and Calculation:
Part 1 Projectile Motion
Shots |
Range, X (cm) |
Deviation (cm) |
1 |
169.8 |
3.8 |
2 |
173.9 |
0.3 |
3 |
175.2 |
1.6 |
4 |
172.8 |
0.8 |
5 |
176.1 |
2.5 |
Average |
173.6 |
1.8 |
Vertical distance Y __0.828 m___
Velocity V_{o} of projectile from ( 3 ) _4.22 m/s_____
Part 2 Ballistic pendulum
Mass M of pendulum __0.3825 kg_
Mass m of ball ___0.0646 kg__
Trials |
1 |
2 |
3 |
4 |
5 |
Averages |
Height of CM at its lowest point (cm) |
12.0 |
12.0 |
12.0 |
12.0 |
12.0 |
12.0 |
Height of CM at its highest point (cm |
13.5 |
13.7 |
13.5 |
13.3 |
13.4 |
13.5 |
Vertical distance h (cm) |
1.5 |
1.7 |
1.5 |
1.3 |
1.4 |
1.5 |
Velocity V_{o} of projectile from ( 6 ) ___3.76 m/s_____
Percent difference between the two values of V_{o ___}_11.5%____
Calculations
Average range X = (168.8+173.9+175.2+172.8+176.1)/5 = 173.6 cm = 1.736 m
Average deviation = (3.8+0.3+1.6+0.8+2.5)/5 = 1.8 cm
V_{o} = 1.736 m x sqrt[9.8/(2 x 0.828)] = 4.22 m/s
Average h = (1.5+1.7+1.5+1.3+1.4)/5 = 1.5 cm = 0.015 m
V_{o} = (1 + 382.8g/64.6g) x sqrt(2 x 9.8 x 0.015) = 3.76 m/s
% Difference= |1^{st} value – 2^{nd} value| / [(1^{st} value + 2^{nd} value)/2] x 100%
% Difference = |4.22 – 3.76| / [(4.22 + 3.76)/2] x 100% = 11.5%
Pre Lab Questions:
By what factor is its kinetic energy changed?
p = mv => if speed is doubled, momentum is doubled.
KE = ½ mv^{2} => if speed is double, kinetic energy is quadrupled.
Before the step the person’s momentum was zero (no velocity), but afterward that person has some momentum. Therefore, linear momentum is not conserved. This is because the system (which is this person) is not an isolated system. The external force on the system is friction, which helps the person move forward.
A perfectly inelastic collision is one in which the colliding particles stick together after the collision. In an inelastic collision, only momentum is conserved, while kinetic energy is not. Example 1: two cue balls moving toward each other with velocities opposite in direction, after the collision they have the same velocity. Example 2: An arrow is shot at a stationary apple, after the collision the arrow penetrates the apple and they moves with the same velocity.
Post Lab Questions:
According to the conservation of mechanical energy of the system after the collision, we have equation (5): (1/2) (m + M) V_{1}^{2 } = (m + M) g h
Therefore, ∆KE = KE_{f }- KE_{i}
= ½(m + M)V_{1}^{2} – ½ m(V_{o})^{2}
= (m + M) g h – ½ m(V_{o})^{2}
= (0.0646kg + 0.3828kg) x 9.8m/s^{2} x 0.015m – ½ 0.0646kg(3.76m/s)^{2}
= – 0.305 J (the negative sign implies the loss)
KE_{i} = ½ m(V_{o})^{2} = 0.457 J
Fractional loss = |∆KE |/ KE_{i} = 67%
This loss of energy may be converted into heat and sound.
While it is correct that the final potential energy of the pendulum-ball system is equal to the kinetic energy of this system just after the collision (due to conservation of mechanical energy after the collision). This is a perfectly inelastic collision, and consequently this kinetic energy of the pendulum-ball system after the collision is not the initial kinetic energy of the ball. In a perfectly inelastic collision, kinetic energy is not conserved.
Conclusion:
It was a bit difficult to accurately measure the x and y values in Part 1, but despite human error the percent difference between the two methods of calculating V_{o }was fairly low. To get an even closer approximation, we could have doubled the number of tests (from 5 to 10). One of the most important deductions from this experiment is the fact that in an inelastic collision, the kinetic energy of the system is not conserved, only the momentum is conserved.
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