Question

Finally: Create your own problem, show your math work solving for it, and compare Force acting upon a

mass(kg) given two different speeds(time in seconds) in which a collision takes place:

Pick a mass in
kilograms
Pick a Velocity in
meters per second to
decelerate from
Calculate initial
momentum of the
object
Pick a fast
deceleration
Pick a slower
deceleration
What Force acts on
the mass in the faster
deceleration? (Show
your work and include
correct final units)
What Force acts on
the mass in the slower
daralaration? (Show
E.g. I have a mass of about 82.0kg, a
new popular cell phone has a mass of
0.204kg.
E.g. I'm going 55m/h which equals 24.6
m/s
P=mv
P=82kg x 24.6 m/s
P=2020 kg*m/s
E.g. I'm NOT wearing my seatbelt and I
crash into a wall coming to 0 m/s in just
0.20 seconds
E.g. I am wearing my seatbelt and my
velocity changes over 0.91 seconds.
Fet=P-P₁/t
F=(0 kg*m/s-2020 kg*m/s)/0.2s
F=-2020 kg*m/s +0.20s
Force =-10,000 Newtons (or
kg*m/s)
Fret=Pr-P₁/t
F=(0 kg*m/s-2020 kg*m/s)/0.91s
C-30306/001

Answer 1

Problem:

A mass of 10 kg is moving at a velocity of 5 m/s. The mass undergoes a deceleration of 10 m/s^2, resulting in a collision. Calculate the force acting upon the mass in the collision.

Solution:

First, we need to calculate the initial momentum of the object before the collision. Momentum is defined as the product of an object's mass and its velocity, so the initial momentum of the mass is 10 kg * 5 m/s = 50 kg*m/s.

Next, we can use the formula for force, which is defined as the product of an object's mass and its acceleration. In this case, the force acting upon the mass in the collision is 10 kg * 10 m/s^2 = 100 N.

To compare the force acting upon the mass in a faster deceleration, we can assume that the mass undergoes a deceleration of 20 m/s^2 instead of 10 m/s^2. In this case, the force acting upon the mass would be 10 kg * 20 m/s^2 = 200 N, which is twice as large as the force in the original scenario.

Similarly, in a slower deceleration of 5 m/s^2, the force acting upon the mass would be 10 kg * 5 m/s^2 = 50 N, which is half as large as the force in the original scenario.