An object of mass 2.0 kg starts from rest at the top of a frictionless curved ramp. At the bottom,
it collides with and sticks to another 2.0-kg mass object. Together the objects then cross a small
rough section where there is friction before leaving the edge of the table together.
3.6 m
m = 2.0 kg
at rest
at rest
m = 2.0 kg
1.5 m
= 0.19
4.2 m
y
(a) Find the height of the table, y.
(b) Find the velocity (magnitude and direction) at which the objects reach the ground.
Presentation instructions:
Write a solution that looks like one of the examples of the book: fully explain your reasoning in
words, show all your computations clearly and correctly written (format). Explain any and all
assumptions that you make as you solve the problem.
To solve this problem, we need to apply the principles of conservation of energy and momentum. Let's break it down step by step:
(a) Find the height of the table, y:
Initially, the object is at rest at the top of the ramp, which means it has gravitational potential energy. At the bottom, it collides and sticks to another object before leaving the table. Therefore, the total mechanical energy is conserved (assuming no external forces).
The total mechanical energy can be expressed as the sum of the gravitational potential energy and kinetic energy:
Initial energy = Final energy
m1gh = (m1 + m2)v^2 / 2
Where:
m1 and m2 are the masses of the objects (both 2.0 kg)
g is the acceleration due to gravity (approximately 9.8 m/s^2)
h is the height of the table (what we need to find)
v is the final velocity of the objects
Rearranging the equation, we can solve for h:
h = (m1 + m2)v^2 / (2m1g)
Substituting the given values:
h = (2.0 kg + 2.0 kg)(v^2) / (2(2.0 kg)(9.8 m/s^2))
h = v^2 / 9.8
We need to find the velocity v first in order to calculate the height. Let's proceed to part (b).
(b) Find the velocity (magnitude and direction) at which the objects reach the ground:
To find the velocity, we will use the principle of conservation of momentum.
Initial momentum = Final momentum
0 = m1u1 + m2u2
Where:
m1 and m2 are the masses of the objects (both 2.0 kg)
u1 and u2 are the initial velocities of the objects (u1 = 0, as the first object is at rest)
Simplifying the equation:
m2u2 = -m1u1
Since m1 = m2, we have:
m2u2 = -m2u1
u2 = -u1
This tells us that the initial velocities are equal in magnitude and opposite in direction. Thus, we can say:
u1 = -u
Now, let's consider the friction on the rough surface of the table. The friction force (Ff) can be calculated using the equation:
Ff = μN
Where:
μ is the coefficient of friction (given as 0.19)
N is the normal force, which is equal to the weight of the objects (m1g + m2g)
The work done by friction can be calculated as the product of the friction force and the distance traveled on the rough surface (1.5 m):
Work by friction (Wf) = Ff * D
Since work done by friction converts mechanical energy into thermal energy, the change in mechanical energy can be expressed as the negative of the work done by friction:
ΔE = -Wf
From the law of conservation of energy:
Initial energy - Wf = Final energy
m1gh - μ(m1g + m2g)(1.5 m) = (m1 + m2)v^2 / 2
Substituting the given values, we have:
v^2 / 9.8 - 0.19(2.0 kg + 2.0 kg)(9.8 m/s^2)(1.5 m) = (2.0 kg + 2.0 kg)(v^2) / (2(2.0 kg)(9.8 m/s^2))
Simplifying and solving for v:
v^2 / 9.8 - 0.19(4.0 kg)(9.8 m/s^2)(1.5 m) = 4v^2 / (2 * 9.8 m/s^2)
v^2 - 0.19(4.0 kg)(9.8 m/s^2)(1.5 m) = 4v^2 / 4.9
v^2 - 2.49 (m/s^2) = v^2 / 1.225
1.225 v^2 - 1v^2 = 2.49 (m/s^2)
0.225v^2 = 2.49
v^2 = 2.49 / 0.225 (m/s^2)
v^2 = 11.0667
v ≈ √11.0667 (m/s)
Thus, the magnitude of the velocity at which the objects reach the ground is approximately 3.326 m/s.
Now that we have the value of v, we can substitute it back to calculate the height, h:
h = v^2 / 9.8
h = (3.326 m/s)^2 / 9.8 m/s^2
h = 1.128 m
Therefore, the height of the table, y, is approximately 1.128 meters.