QUT EGB211: Physics Problem Task - Cart, Car, Collision Analysis

This document presents a comprehensive solution to a physics problem task, likely for an undergraduate engineering or physics course (EGB211). The assignment is divided into three exercises. Exercise A focuses on a cart pulley system, requiring the creation of a free body diagram, the identification of forces, and the derivation of equations of motion to determine the force needed to initiate movement. Exercise B delves into a car crush scenario, analyzing the speed before the crash, the maximum compression of a spring, the distance traveled backward after bouncing, and the maximum deceleration experienced. Exercise C explores a collision barrier, involving free body diagrams and the application of collision principles, potentially assuming conservation of momentum and energy. The solutions include detailed calculations, diagrams, and references to relevant physics principles and equations, such as Newton's laws of motion and equations of motion, providing a thorough analysis of each problem.

Running Head: PHYSICS PROBLEM TASK
PHYSICS PROBLEM TASK
Name
Institute of Affiliation
Date
a. Exercise A : cart pulley system

PHYSICS PROBLEM TASK 2
1. Question 1
The free body diagram
Figure 1: The free body diagram
2. Question 2
From the newtons third equation of motion
F = mg sin 15
x=x0 +v t Equation 1
But
v = 1
2 (v +v0 )
= 1
2 [ ( v0+ at )+ v0 ]
= 1
2 ( 2 v0 +at )
= v0+ 1
2 at
So now equation becomes
x=x0 +( v0 + 1
2 at ) t

PHYSICS PROBLEM TASK 3
= x0 +v0 t+ 1
2 at2 Equation 2
But a = g sinθ
Therefore equation 2 becomes
X = x0 +v0 t+ 1
2 (g sinθ )t2
3. Question 3
F = μ W
= 0.6 * ( 80+15)*9.81
=559.17 N
Exercise B: car crush
1. Question 1
Free body diagram
Figure 2: Before crash ]

PHYSICS PROBLEM TASK 4
Figure 3: After crash
2. Question 2
Speed before crash
Newtons equation of motion is as follows;
v2=u2 +2 as
Given a= g sinθ, s =110 m and θ=40
v2=02 +2 ( 9.81∗sin 40 )∗110
= 1,608.07
V = 40.10 m/s
3. Question 3
Maximum compression of the spring
The force that hits the spring
F= ma = mg sinθ
= 800 kg * 9.81sin 40
= 5,044.5971 N
The force according spring

PHYSICS PROBLEM TASK 5
F= k*x
Where k is spring constant ( 11000 N/m) and x is spring extension
5,044.5971 = 11000 N/m * x
X = 5,044.5971
11000 N m−1
= 0.4586 m
= 458.6 mm
4. Question 4
Distance travelled backward after bouncing
If no energy is lost, the final velocity before hitting the barrier is also the initial
velocity after hitting the barrier
u=40.10 m s−1
Time taken to stop
t= u
a
= 40.10
9.81 sin 40
= 6.3592 seconds
Since
v2=u2 +2 as and v =0
Then

PHYSICS PROBLEM TASK 6
S = u2
2 a
= −40.102
2∗−9.81sin 40
= 127. 5 m
5. Question 5
Maximum deceleration
v2=u2−2 as but v=0 , s =0.4586 m ( spring compression )
So , a= u2
s
a = 40.102
0.4586
= 3,506.3454 m s−2
Maximum force
F= ma
= 800 kg * 3,506.3453 m s−2
= 2,805,076.3192 N
= 2.8MN
Exercise C: collision barrier
1. Question 1
Free body diagram

PHYSICS PROBLEM TASK 7
Figure 4: The fbd of moving car
Figure 5: The fbd of stationary barrier
2. Question 2
It shall be assumed that the vehicle does do slip and the energy is not lost.
m1 v1= ( m1 +m2 ) v2
2.6 * 55 = ( 2.6+ 1.7) v2
v2 = 33.25 kmh r−1
The report
in the exercise A, 1D equations of motion are considered. The relationship between
time and velocity is simple and produces a straight line. When there is an acceleration, the
graph changes accordingly (Alrasheed, 2019). The complexity now occurs when determining
the distance travelled and the time taken when only a few components are included. The
following are three equations of motion that were written by Isaac Newton.
v2=u2 +2 as . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Equation 3

PHYSICS PROBLEM TASK 8
S =ut+ 1
2 a t2 . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Equation 4
V = u+ at . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Equation 5
Exercise B
When a body hits a spring, the springs temporary absorbs the energy and the releases
it after a very short time. The amount of energy is stored is limited to the length of the spring
and the spring constant (Kuwabara, 2013). In the case above, the car slowly moves down the
hill at and inclined angle. This translated then not the whole gravitational force is at work but
a portion of it which is proportional to the angle of inclination. The new acceleration is as
follows;
a= g sin θ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Equation 6
With that, the distance travelled, time and force are subject to the angle inclination.
Exercise C
For colliding bodies, some of the energy are lost as sound, heat or deformation of the
body itself. When assuming perfect collision, there is no energy is lost meaning no soud,
heat or deformation of the body (Boccaletti, 2016). For a perfect collision, the following
equation holds.
m1 v1+ m2 v2 +… mn vn = ( m1 +m2+ … mn ) vf
The sum of products of masses and their velocity should be equal to product of final
velocity and sum of the masses.

PHYSICS PROBLEM TASK 9
References
Boccaletti, D. (2016). Galileo and the equations of motion.
Kuwabara, T. (2013). FEM for springs. Berlin: Springer.
Alrasheed, S. (2019). Principles of mechanics: Fundamental university physics.