Physics lab 4 - lab PDF

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Name: Caroline Gerges Physics 1100 Lab #4: One and Two Dimensional Motions With Constant Acceleration

The purpose of this lab is to be able to identify free fall motion as well as be able to use the equations of kinematics. In this lab we will also be studying two dimensional motion with constant acceleration. There are 4 kinematic equations that we will be using. In the first part of the experiment we will be doing an experiment of an object in free fall motion from rest. In the second part of the experiment we will be testing an object that is thrown by someone with initial velocity. In the third part of the experiment we will be using the kinematic equations in order to test two dimensional projectile motion and then using a stimulator to verify our answers. Data sheet Introduction exercise: Answer for i) Red car velocity graph: Horizontal Answer for ii) Blue car velocity graph: Straight line with +ve slope. Answer for iii) Intersection point: The intersection point is when both cars had the same velocity (5m/s) at 4seconds. Answer for iv) Red car position graph: ) Straight line with +ve slope. Answer for v) Blue car position graph: A part of a parabola Part 1: Free fall from rest Step 2) a at t = 2 s a at t = 3 s -9.82 m/s^2

-9.74 m/s^2

Step 3) vo

Is a constant? Yes

Reason for value of

value of

a

-9.80 m/s^2 value of slope

vo +0.09 m/s

The object started at rest

Is the motion free fall?

-9.8

Step 4) Answer for a)upward opening parabola

Yes slope relation to acceleration a The value of the slope = acceleration rate

Answer for b) because the +y is considered to be downward so the acceleration is -9.8 m/s^2

Part 2: Free fall with initial v Step 2) Is the whole motion considered free fall? No

Reason Because the ball was held in the hand, so it is not just under the influence of gravity

Is a constant? Yes

value of

a -9.8 m/s^2

Step 3) value of

Reason for value of

vo 39.1 m/s

Step 4) xo x

vo Because it is not free fall and the object was thrown by someone at

v at

v at maximum height 0.0 m/s

compare

v

at final

v as a function of t

v

at t = 2.5 s

V= v o +at

V= 39.1+ -9.8*2.5= 14.6m/s

Is the position

In this case, how

0m

vo

final time

final time

time to

0m

-39.1 m/s

The magnitude is the same but final velocity is negative and initial velocity is positive

f) Expression for x

graph symmetrical?

Yes

1 2 as a function of t : x = x o + v o t + a t 2

2.5 s: 67.1m h) answer: yes, on the graph it is also 67.1m Step 5) a) v as a function of t : v =1.9+( −2) t 1 2 x=3+1.9 t+ (−2)t 2 Equations for v & x displayed by simulator: Calculated value of t at Measured value of t maximum height 1s

g) Calculate x

x

at

maximum height 1s

does the time of maximum height relate to the final time? The time of maximum height is half the time of final time

at t =

as a function of t : :

Was your calculated value correct? yes

Part 3: Two-dimensional projectile motion o θo =45 Step 4) a) Calculate t final: 4.1s b) Calculate v y final: -20.0 m/s c) Calculate the time t for reaching maximum height: 2.0s d) Calculate the range for the motion up to y final ¿ 0 m : 82.0m Step 6) a) Is the position graph symmetrical? Yes measured t final measured range R measured v y final measured t for 4.08s

-19.9 m/s

maximum height 2.1s

81.9m

o

Step 7) θo =30 a) Calculate t final: 2.89s b) Calculate v y final: -14.2m/s c) Calculate the time t for reaching maximum height: 1.45s d) Calculate the range for the motion up to y final ¿ 0 m : 70.8m o For θo =30 : Repeated step 6) a) Is the position graph symmetrical? Yes measured t final measured range R measured v y final measured t for 2.892

-14.18m/s o

maximum height 1.50s

70.78m

θo =60 a) Calculate t final: 5.0s b) Calculate v y final: -24.5m/s c) Calculate the time t for reaching maximum height: 2.5s d) Calculate the range for the motion up to y final ¿ 0 m :70.9m

For θo =60o : Repeated step 6) a) Is the position graph symmetrical? measured final 5.0s

t

measured v y -24.51m/s

final

measured t for maximum height 2.5s

measured range

R

70.85m

Step 8) Max Range Rmax is for θo =¿ 45 degrees...