# Physics mcq questions based on mass,displacement,position.

Physics mcq questions based on mass,displacement,position.

1.
A 50 gramme bullet traveling at 182 m/s buries itself in a 5.4 kg pendulum hanging on
a 2.06 m length of string, which makes the pendulum swing upward in an arc.
Determine the horizontal component of the displacement of the pendulum.
A) 0.752m
B) 1.046 m
C) 1376.598
D) 0.511 m
E) 0.057 m
conservation of momentum
m1v1 = (m1+m2)v2
Where
m1 = mass of bullet
m2 = mass of pendulum
v1 = bullet velocity
we solve for v2, which is the velocity
of the pendulum after the bullet strikes
v2 = (m1v1)/(m2+m1)
= (.05kg)(182)/(5.4+.05)
= 9.1/5.45
= 1.669
V=1.29 m/s
So then we use conservation of energy,
to find the max height the pendulum can rise.
mgh = ½mv²
where m cancels out
g = 9.8 m/s²
v = v2 = 1.66 m/s
we want to solve for h
h = ½v²/g
h = ½(1.66)/(9.8)
h = .084 m
so now that we have the vertical displacement,
we can use the length of the string to
find the horizontal displacement.
The horizontal displacement is given by
x = [L² – (L – h)²]
where
L = length of string = 2.06 m
h = .140 m
x = [2.06² – (2.06 – .084)²]
x = [2.66² – 2.07²]
x = 1.157
x = 1.046 m
2.
You are trying to balance a tray of dishes to make it easy to carry. At one end of the 0.8m
tray are four plates with total mass 1.4kg with their centers of mass exactly above the left
edge of the tray. How far from the left edge should you place a 2.1kg jug of milk so that
the centre of mass of the system is at the centre of the tray?
A) 0.58 m
B) 0.67 m
C) 0.72 m
D) 0.53 m
E) 0.78 m

Let the Jug is kept at a meter from the center of the tray, Thus by taking the
momentum of at the center of the tray:-
=>1.4 x 0.4 = a x 2.1
=>a = 0.4/2.1 = 0.26 m
=>The distance from the left edge = 0.40 + 0.26 = 0.666 m
3.
A person falls flat on their back on some slippery ice and is unconscious for a second. The
horizontal position of their legs torso and head from their feet is given by 0.41m, 1.11m
and 1.8m respectively and their weights are 30kg 50kg and 10kg. They try to sit up on the
ice which is so slippery they are not able to move their horizontal centre of mass. The new
positions of their legs, torso and head from their feet are 0.23m, 0.56m and 0.8m. What is
the change in position of the person’s feet?
all you have to do is find the horizontal center of mass(X) two different times
A) 0.39 B) 0.48 C) 0.62 D) 0.72
use this equation (center of mass equation)
X=(mass1*position1)+(mass2*position2)+
(mass3*position3) / (mass 1+2+3)
now find X using the numbers above ( masses are the 30,50,10) and their respective
positions (.41,1.11,1.18)
after u find X
find X again, this time substitute the positions values with the new set (.23,.56,.8)
after u get these two X values, basically u have found two different center of masses for
the person, so by subtracting these two values u should get his change in position
A person falls flat on their back on some slippery ice and is unconscious for a second.
The horizontal position of their legs torso and head from their feet is given by 0.31m,
1.39m and 1.65m respectively and their weights are 30kg 50kg and 10kg. They try to
sit up on the ice which is so slippery they are not able to move their horizontal centre of
mass. The new positions of their legs, torso and head from their feet are 0.2m, 0.57m
and 0.78m. What is the change in position of the person’s feet?
A) 0.76
B) 0.43
C) 0.59
D) 0.83
all you have to do is find the horizontal center of mass(X) two different times
use this equation (center of mass equation)
X=(mass1*position1)+(mass2*position2)+
(mass3*position3) / (mass 1+2+3)
now find X using the numbers above ( masses are the 30,50,10) and their respective
positions (.31,1.39,1.65)
after u find X
find X again, this time substitute the positions values with the new set (.2,.57,.78)
after u get these two X values, basically u have found two different center of masses for
the person, so by subtracting these two values u should get his change in position
u should get .59 as ur answer