A 5kg mass is sliding at a speed of 7 m/s on horizontal sheet of frictionless ice when it smacks in to a wall of the ice rink. There are bumpers made of elastic springs placed along the wall for protection. The mass hits the bumper, compressing it by 5cm and momentarily comes to rest before bouncing off
a) What is the force constant of the spring bumper?
b) If 10% of the mechanical energy is lost during the encounter with the bumper, what is the rebound speed of the mass
Initial KE = (1/2) m v^2 = 2.5*49 = 122.5 Joules
It lost 122.5 Joules coming to rest. Presumably that was stored as potential energy in the spring.
(1/2) k x^2 = 122.5
.5 k (.05)^2 = 122.5
k = 98,000 Newtons/meter
.9 * 122.5 = 110.25
so
.5 * 5 * v^2 = 110.25
v = 6.64 m/s
Thank you
To calculate the force constant of the spring bumper, we can use Hooke's Law. Hooke's Law states that the force applied by a spring is directly proportional to the displacement of the spring from its equilibrium position.
a) First, let's find the force applied to the bumper when it is compressed by 5cm (0.05m). The formula for the force applied by a spring is:
F = k * x
Where F is the force, k is the force constant (spring constant), and x is the displacement.
From the problem, we know that the mass of the object is 5kg and it momentarily comes to rest, which means the velocity before hitting the bumper is 0 m/s.
The kinetic energy of an object can be calculated using the formula:
KE = 1/2 * m * v^2
Where KE is the kinetic energy, m is the mass, and v is the velocity.
Since the velocity is 0, the initial kinetic energy is 0.
The potential energy stored in the spring when it's compressed by 5cm can be calculated using the formula:
PE = 1/2 * k * x^2
Where PE is the potential energy.
Since the object momentarily comes to rest, all of the kinetic energy is converted into potential energy. Therefore, we can equate the initial kinetic energy to the potential energy:
1/2 * m * v^2 = 1/2 * k * x^2
Substituting the given values: m = 5kg, v = 0, x = 0.05m
0 = 1/2 * k * (0.05)^2
Simplifying the equation, we can solve for k:
0 = 1/2 * k * 0.0025
0 = 0.00125k
k = 0 / 0.00125
k = 0 N/m
Therefore, the force constant of the spring bumper is 0 N/m.
b) To calculate the rebound speed of the mass, we need to consider the conservation of mechanical energy. Since 10% of the mechanical energy is lost during the encounter with the bumper, the rebound speed will be less than the initial speed of 7 m/s.
The mechanical energy before the encounter is given by the kinetic energy:
KE_before = 1/2 * m * v^2
Substituting the given values: m = 5kg, v = 7 m/s
KE_before = 1/2 * 5 * (7)^2
KE_before = 122.5 Joules
If 10% of the mechanical energy is lost, the remaining energy after the encounter is:
E_after = (1 - 10%) * KE_before
E_after = 0.9 * 122.5
E_after = 110.25 Joules
The mechanical energy after the encounter is equal to the kinetic energy:
KE_after = 1/2 * m * v_rebound^2
Where v_rebound is the rebound speed we are trying to find.
Substituting the given values: m = 5kg, KE_after = 110.25 Joules
110.25 = 1/2 * 5 * v_rebound^2
2.205 = v_rebound^2
v_rebound = √2.205
v_rebound ≈ 1.48 m/s
Therefore, the rebound speed of the mass is approximately 1.48 m/s.