Physics of Momentum: Exploring Mass, Force, and Collisions Concepts

This assignment solution delves into the physics of momentum, exploring the relationship between mass, force, and collisions. It begins by examining the concept of mass in relation to volume and density, explaining how different materials with the same volume can have different masses. The solu...Read More

RUNNING HEAD: PHYSICS OF MOMENTUM
Momentum
Name
Institutional affiliation

Physics of Momentum 2
1a. Mass of different solids with an equal volume
Different solids having the same volume also have the same mass and this can be explained by
the fact that mass is the measure of an object's resistance to change in its motion state. Necessary
force is used to measure mass so that the object can have acceleration and the space that matter
occupies is the volume. For example, the force exerted by gravity can be used to measure the
object's mass on earth and can be referred to as the weight of the object (Teates, 2010).
Figure 1: Relationship between mass, density, and volume
The figure above demonstrated that volume and mass are related both through the density. The
density which is the substance's mass per unit volume is the property matter that is usually used
by the chemists as the tag for identifying a substance. The diagram also demonstrated that the
mass, density and the object compact are directly proportional in that when the mass is increased,

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both density and object compact becomes greater. Volume and density are inversely proportional
in that the density reduces when the volume increased.
Solids have increased densities and only compressible to a very small extent. They are rigid since
the solid is required to maintain the shape irrespective of its container. These characteristics
show that the solid components are close and exert strong forces on each other. Hence, some of
the solid materials are compacted together when compared with others. Therefore, dissimilar
materials will have dissimilar mass even if their solids materials are of the same volume (Larkin,
2011).
1b. with the use of equation F=a*m where by 'm' is equal to 3.50kg and 'a' is equal to 2 m
per second square. When 3.50 kg is multiplied by 2m per the second square the product
becomes 7N.
The relationship between the mass of the object and force required to accelerate the object is
described by the motion's second law of newton. This law is usually stated as F=ma, whereby the
F (Force) being applied to the object is the same as the objects' m (mass) when it's an
(accelerating). This can be explained that the more object's mass, the higher the required force
for acceleration and the higher the force, the faster the acceleration of the object. The force
required is directly proportional to objects' acceleration and inversely proportional mass. If the
doubling of the net force occurred, the object's acceleration will be twice as large. Also when the
object’s mass is doubled, the acceleration becomes half as large. The SI units of mass are the Kg
and m per s2 for acceleration and as a result, 1Kkg /s2 is the same as 1 Newton (Holzschuh,
2016).

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Based on the above equation, the acceleration and mass are not dependent variable and only
force is a dependent variable. Independent variables are the ones that are controlled or changed
in science-based experiments. The equation illustrated that 2m/s2*3.50kg is equal to 7N. It can be
explained from the equation that, the mass has three significant figures and the acceleration has
one significant figure. The principle of the law that the answer or the product must have at least
one significant figure is applied in this equation where the result is 7N.
2. The ball has a mass of 2kg and is set to motion in a straight line. It strikes a 2nd unmoving
ball with a mass of 5kg. The speed of the 1st ball when striking the 2nd ball is 20m/s. the 1st
ball becomes a complete halt because of collision. This experiment was done for
0.01seconds. Calculate the applied force by the 1st ball to the 2nd ball.
When the force is applied to an object or a body, the proportional change must result in its
motion. The motion's second law of newton illustrates that the force applied (F) generates the
proportional change in the body's momentum and can be represented mathematically as;
(∆ₚ):∆ ∞ F). Because the momentum is relative, the significant quantity is not relative and isₚ
dependent on the observers’ motion. This is according to the motions’ second law of newton that
the applied force to the object is the same as the change that results in its momentum per unit
time. The newton law is usually [variable⃗ F]⃗ Fav = ∆⃗ p
∆ t over the finite time at (∆t) when the force
can change (Wroldsen, 2010).
The above equation illustrated that the average force and momentum are directly proportional to
each other and the same force and the changing time are inversely proportional to each other.
This means that, when the average force is increased or raised, the change in momentum will
also be increased in the same period. Also if the force is increased or is high, the change in time

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will reduce at the same rate it is important to combine the time and forced into the one notion
that is equal to the momentum change. One concept that is formed by force and time becomes (⃗ F
av∆ t ¿ and it is referred to as the force impulse and the alternative of the newton’s second law becomes
the impulse equals to change in momentum or ∆⃗ p=¿⃗ F ¿av∆ t CITATION Fab17 \l 1033 (Fabrocini, Fantoni, &
Polls, 2017)
Based on the above equation, it is seen clearly that the impulse and the change of impulse are directly
proportional to each other. This can be described that when the momentum change is increased then the
impulse will be raised or increased at the same time rate. The equation has demonstrated that both
momentum change and impulses are the vectors and which means that both of them have directions. For
the scalar forms for the motion along the straight line will meet the needs; [motion along the straight line]
∆ p=∆ t Fav
The impulse will generate the specific momentum change irrespective of the speed and mass of the
recipient body or the object. The object that is initially at rest will move in the direction of the applied net
force getting the momentum of ∆ p=∆ t Fav which is what usually occurs when the dart is thrown or the
golf ball is stroked. When the club is in contrast with the ball and force applied, there is some ongoing
momentum in the force ball direction (Ninno, Cantele, & Trani, 2018).
SOLUTION
1. Translation
The small mass that is known to move in the motion of straight line at a specific velocity collides with
mass at rest. The 1st ball come to the complete halt because of collision. Determine the applied force to
the second ball by the first ball.
2. Take the initial motion’s direction to be positive.

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Given that: Mass of the 1st ball is 2kg, a mass of the 2nd ball is 5kg, and the initial speed of the 1st ball is
20m /s, final speed of the 1st ball is 0m/s and the initial speed of the 2nd ball is 0m/s.
Calculate the applied force by the 1st ball to the 2nd ball.
3. Problem type
Linear momentum and the inelastic collision of the momentum conservation.
4. Procedure:
This collision is inelastic completely since the two balls slide away together. It is good to know that the
momentum is always conserved but E or KE which is kinetic is not conserved.
5. Calculations
The equation is J n=∆ p=¿ ∆ t × Fn and it can be used to calculate many variables;
The net impulse from very objects’ forces is Jn and like in this ball, it should be equal to the momentum
change of the object ∆ p. The net force is Fn and time period when the net force is acting is ∆ t .
(1) The initial and final momentum is equal; pi= pf ,where the momentum is dependent on the variables
of velocity and mass. From the equation, the objects' momentum is equal to the objects' mass multiplied
by the object's velocity. m vf - m vi from this formula, the values from the question can be inserted for the
calculations to be done. ¿)(0m/s) – (2kg)(-20m/s). 20m/s is negative since the momentum is the vector
with directions. The two velocities much have dissimilar sides because the balls can always change the
directions and the 1st velocity must be negative (-20m/s) with the assumption that moving right-side is
positive. When substituted and calculated the answer will is= 2kgm/s – (-40kgm/s) and the final answer is⃗
J= 40kgm/s.

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After realizing that the product of the balls’ net impulse, it means that the product of the 1st ball’s net
force can be calculated by the formula of¿ ¿); Jn=¿ ∆ t × Fn; 40 kg m/s = ( 0.01 s × Fn ). The equation can
be rearranged by using the time on the equation’s other side; Fn= 40 kgm/s
0.01 s = 4000 N
REFERENCES
Fabrocini, A., Fantoni, S., & Polls, A. (2017). FHNC calculations of momentum distribution. Lettere al
Nuovo Cimento, 283-288.
Holzschuh, E. (2016). The convergence of momentum space, pseudopotential calculations for Si.
Physical Review B, 7346-7348.
Larkin, E. (2011). ANCIENT PLANETARY RINGS, VOLUME, MASS, AND DENSITY. Science, 350-351.
Ninno, D., Cantele, G., & Trani, F. (2018). Real-space grid representation of momentum and kinetic
energy operators for electronic structure calculations. Journal of Computational Chemistry,
1406-1412.
Teates, T. (2010). Measurement Corner Mass, Volume, Density, and Matter. School Science and
Mathematics, 351-352.
Wroldsen, J. (2010). An Easy Way of Doing Angular Momentum Calculations. Physica Scripta, 111-115.

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