Is the kinetic energy conserved in the collision? If not, then what percentage of energy is lost in the collision between the steel ball and the pendulum? (See equation) what happens to remaining energy?

Physics for Scientists and Engineers: Foundations and Connections
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Author:Katz, Debora M.
Publisher:Katz, Debora M.
Chapter39: Relativity
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Is the kinetic energy conserved in the collision? If not, then what percentage of energy is lost in the collision between the steel ball and the pendulum? (See equation) what happens to remaining energy?
4 5 I II L 6
Part 1: Ballistic Pendulum Method
Total Mass (M = mb +Mp) = 0.06586 + 0.24709 = 0.31295 +0.00001 kg
Length of the pendulum (Rcm) = 0.290m
Data Table I:
Trial
Angle (e)°
h = Rcm(1-cose)
1
35
0.052
36
0.055
36
0.055
4.
36
0.055
35
0.052
<O>= 35.6 +-1
<h>=0.054+-0.011
1a) Use the average height and the average angle calculated above in part I to find the
initial speed of the ball. Include the absolute uncertainty in the initial velocity.
Delta h = delta Rcm + delta (1-cose)
= +/- (0.001+0.01) = +/- 0.011
v= sqrt {2*g*h}
= sqrt {2*9.8*0.054}
= 1.03 m/s
Error in velocity = delta v = 2 (Delta h)h^(-2)
%3D
= ½ (0.011)(0.054)^-½
= 0.0236
Initial Velocity = (1.03 +/- 0.0236) m/s
1. s the kinetic energy conserved in the collision? If not, then what percentage of energy is
lost" in the collision between the steel ball and the pendulum? (See Equation 6 in part I.)
What happens to the remaining energy?
Transcribed Image Text:4 5 I II L 6 Part 1: Ballistic Pendulum Method Total Mass (M = mb +Mp) = 0.06586 + 0.24709 = 0.31295 +0.00001 kg Length of the pendulum (Rcm) = 0.290m Data Table I: Trial Angle (e)° h = Rcm(1-cose) 1 35 0.052 36 0.055 36 0.055 4. 36 0.055 35 0.052 <O>= 35.6 +-1 <h>=0.054+-0.011 1a) Use the average height and the average angle calculated above in part I to find the initial speed of the ball. Include the absolute uncertainty in the initial velocity. Delta h = delta Rcm + delta (1-cose) = +/- (0.001+0.01) = +/- 0.011 v= sqrt {2*g*h} = sqrt {2*9.8*0.054} = 1.03 m/s Error in velocity = delta v = 2 (Delta h)h^(-2) %3D = ½ (0.011)(0.054)^-½ = 0.0236 Initial Velocity = (1.03 +/- 0.0236) m/s 1. s the kinetic energy conserved in the collision? If not, then what percentage of energy is lost" in the collision between the steel ball and the pendulum? (See Equation 6 in part I.) What happens to the remaining energy?
Loss of Mechanical Energy
If we take our reference height for potential energy the line of travel of the ball as it leaves the
gun, then the initial potential energy of the ball is zero. Thus, the total mechanical energy of the
system before the collision is just the kinetic energy of the ball:
%3D
before
After the collision, but before swinging has begun, the ball and catcher are at the same height,
so the potential energy is still zero. The total energy of the system is just the kinetic energy, but
now it is the kinetic energy of the ball and the catcher, and the speed is different:
Eafter=(m+ M)v,? = (m + M)gh = (m + M)gRem(1 – cos e)
%3D
%3D
ст
Using conservation of momentum, one can show that Eafteris less than Epefore - often very
much less. The fractional energy loss in the collision, Q, can be defined as
E
after
Q =1--
Epefore
Lab report requirements: Include in your report the following for both methods:
Transcribed Image Text:Loss of Mechanical Energy If we take our reference height for potential energy the line of travel of the ball as it leaves the gun, then the initial potential energy of the ball is zero. Thus, the total mechanical energy of the system before the collision is just the kinetic energy of the ball: %3D before After the collision, but before swinging has begun, the ball and catcher are at the same height, so the potential energy is still zero. The total energy of the system is just the kinetic energy, but now it is the kinetic energy of the ball and the catcher, and the speed is different: Eafter=(m+ M)v,? = (m + M)gh = (m + M)gRem(1 – cos e) %3D %3D ст Using conservation of momentum, one can show that Eafteris less than Epefore - often very much less. The fractional energy loss in the collision, Q, can be defined as E after Q =1-- Epefore Lab report requirements: Include in your report the following for both methods:
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