This is a Grade 11 physics project that will break down some of your favourite Home Alone scenes to reveal the physics - or lack of physics- within the movie. Enjoy!
The Physics Behind The Spider Scene: Home Alone 1 In case you've forgotten the scene...
What we can assume... What we can take from this scene is how when Marv hits Harry over the chest, work is being done. In physics, we know that wok is equal to FdCosØ, as well as Ekfinal-Ekinitial, where Eg=mgh = (1)(9.8)(2) = 19.6 joules and that the bar comes at rest 0.03m into Harry's chest. We can also assume that the forces are in the same direction and angle so that Cos180 = -1. So we can calculate... Work = Ekfinal-Ekinitial -Fd = 0 - 19.6 -F(0.03)/-0.03 = -19.6/-0.03 F = 653.3N in magnitude
The Physics Behind The Firecracker Scene: Home Alone 1
In case you've forgotten the scene...
The physics behind firecrackers... What Kevin most likely does not know is that in lighting the firecrackers, he was lighting a fuse to open a circuit. When this short length of wire is lit, it will melt and separate since it is under excessive current. This fuse is connected in a series circuit with the other components in the firecracker to be protected from over current, so that when the fuse blows/opens, the circuit will be opened and stop current through the firecracker, thus exploding and making a great scare tactic against burglars like Marv. Physics can also explain the crackling and various pitched sounds the firecrackers made. The lower the pitch that the firecrackers made, the lower the frequency the sound waves were travelling at. On the other hand, the higher the pitch that the firecracker made, the higher the sound waves were travelling at.
The Physics Behind The Tool Chest Scene: Home Alone 2 In case you've forgotten the scene...
Vibrations and Waves in accordance to this scene... In this scene, as Marv and Harry wait by the door, listening and trying to decipher what it is that is "making that noise", physics tells us how the "noise" is a sound that is formed by some type of vibration. These vibrations formed by the movement of the tool chest falling down the stairs, in this case, are called transverse vibration since the object is moving across the rest line, even if a full cycle has not occurred. What can be assumed is that... Since frequency and period of a vibration can be defined with the equation, f = 1/t and according to the video clip, it takes the tool chest about 14 seconds to fall down the flight of stairs. Frequency can be determined by... f = 1/t = 1/14 ∴f = 0.07 Hz
The Physics Behind The BB Gun Shooting Scene: Home Alone 1
In case you've forgotten the scene...
Here... Harry is seen getting shot in the groin within a very short distance with a BB gun. What we can assume... In this scene, Kevin's gunshot is an example of projectile motion when he shoots Harry in the groin at an approximate angle of 60 degrees, only 0.02 metres away, within 0.25 seconds (0.08m/s). With these givens, we can find the height at which Kevin was shooting at. Finding the height of the gun...
x stuff: △dx = 0.02m
vx = 0.08Cos60 = 0.04m/s
t = 0.25s
y stuff:
vy1 = 0.08Sin60 = 0.07m/s
a = -9.8 m/s^2
t = 0.25s
△dy = ?
vy2 = ?
△d = v1t + 1/2at^2
= (0.07) (0.25) + (-4.9) (0.25)^2
= -0.28875
∴ h = 0.29m
Finding impact velocity...
Vimpac = √(0.04)^2 +(0.07)^2
= √0.0065
= 0.08
v2y = v1 + at
= 0.07 + (-9.8)(0.25)
= -2.38m/s
Ø = tan -1(0.07/0.04)
= 60 degrees.
∴ the impact velocity was 0,08 m/s (60 degrees above horizon)
Tuesday 17 January 2017
The Physics Behind The Fake Party Scene: Home Alone 1 In case you've forgotten the scene...
Physics concepts shown include... 1. Newton's second law when Kevin mans the movements of the doll people 2. Friction occurrence between the steel of the train on the steel of the train tracks What we can assume/know is how... 1. When Kevin pulls on the strings that control the movements of the doll people, we can see Newton's second law at work since the doll people move at different accelerations repeatedly in the same direction/cycles, where Newton's second law states that when Fnet does not equal to 0, the object in question will undergo acceleration (a and Fnet are in the same direction). We can assume that on one string, Kevin is pulling a 0.5kg object with a force of 8 Newtons (E), while also experiencing a frictional force of 3.5 Newtons (W). 2. The coefficient of kinetic friction of dry steel on dry steel (aka the train on the train tracks) is 0.42, the mass of the toy train + the attached cardboard cutout is 2kg, and that the normal force acting is = Fg: Fg = mg = (2)(9.8) = 19.6N What we can then determine is... 1. The acceleration of the movement of the doll person: Fnet = Fa + Ff = (8) + (3.5) = 4.5N (E) Fnet = ma 4.5/0.5 = 0.5a/0.5 9 = a ∴ a = 9m/s^2 (E) 2. The value of frictional force of the dry toy train's steel wheels on the dry steel train tracks: Ff = MnFn = (19.6)(0.42) ∴ Ff = 8.232N
The Physics Behind The Cannonball Scene: Home Alone 2 In case you've forgotten the scene... https://youtu.be/QhePJCracho?t=2296 Link to fast forward to scene^
Where's the physics?
From this scene, we can determine that Kevin demonstrates projectile motion when he launches himself into the air at a certain degree in order to cannonball effectively into the pool. We can assume the angle of his jump being 50 degrees and the length of his jump as 2 metres, with the jump lasting 1 second long. With this information we can find the depth of the pool he jumps into in metres.
Finding the depth of the pool:
x stuff: △dx = 2m
vx = 1Cos50 = 0.64m/s
t = 1s
y stuff:
vy1 = 1Sin50 = 0.77m/s
a = -9.8 m/s^2
t = 1s
△dy = ?
vy2 = ?
△d = v1t + 1/2at^2
= (0.77) (1) + (-4.9) (1)^2
= -4.13
∴ h = 4m
Now that we have a value for the depth of the pool, we can find the maximum height Kevin jumps...
△dy = ?
vy1 = 0.77m/s
vy2 = Ø
a = -9.8m/s^2
vs = v1^2 +2a△d
0 = (0.77)^2 + 2(-9.8)(△d)
= 0.5929 -4.9△d
-0.5929/-4.9 = △d-4.9/-4.9
0.121 = △d
∴ △d = 0.12m
The Physics Behind The Zip Lining Scene: Home Alone 1
Towards the end of the film, Kevin makes his great escape from Harry, Marv, and the house to the tree house via. zip line. What allows for Kevin to do so, as well as what we can find out about this scene can be explained with a few physics concepts.
What we know...
- From watching Kevin going down the zip line from start to finish, we can approximately time his velocities and total time it took him to travel (v1 = Ø, v2 = 3m/s, t = 14s)
- There are forces that act on Kevin
- Newton's first law is displayed when Kevin partially crashes through the back wall of the tree house
So what we can discover/analyze is...
1. The estimated distance Kevin traveled by using the kinematics equation: △d = (v1+v2/2)t
= (Ø+3/2)(14) = 1.5 (14)
∴△d = 21m
2.
The force of gravity, normal, friction, and applied forces are all acting on Kevin as he pushes himself off of the windowsill, down the zipline,
3. The reason why Kevin kept moving even after the zipline had come to an end is attributed to Newton's first law which states that if Fnet on an object is Ø, the object will maintain its motion and how it will either remain at rest, or continue moving uniformly (at a constant speed in a straight line). So initially, Kevin is moving uniformly with the zipline handle, so we know Fnet = 0. When the zipline handle finally stops its journey at the end of the zipline, Kevin basically still has Fnet = 0 (because only Kevin's hands are attached to the handle still, so the rest of his body moves separately from the handle). Kevin then seems to be shown flinging off of the zipline handle, crashing into the tree house wall, only because the handle on the zipline stopped so suddenly, but Kevin did not and continued to move uniformly (as predicted by Newton's 1st law).
