A ball of mass 0.120 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.600 m. What impulse was given to the ball by the floor?
Answers
mass (m) = 0.120kginitial velocity (Vo) = 0distance traveled (s) = 1.25 m
We first calculate for the final velocity of the ball:
Vf^2 = 2gs + Vo^2
Vf^2 = 2(9.8m/s2)(1.25m)
Vf = 4.95 m/s
Impulse = m(Vf−Vo)
Impulse = 0.120(4.95)
Impulse = 0.59 Ns
To find the momentum of the ball before it hit the floor, we need to figure out its final velocity using kinematics.
Values we know:
acceleration(a) - 9.81m/s^2 [down]
initial velocity(vi) - 0m/s
distance(d) - 1.25m [down]
This equation can be used to find final velocity:
Vf^2 = Vi^2 + 2ad
Vf^2 = (0)^2 + (2)(-9.81)(-1.25)
Vf^2 = 24.525
Vf = 4.95m/s [down]
Now we need to find the velocity the ball leaves the floor at using the same kinematics concept.
What we know:
a = 9.81m/s^2 [down]
d = 0.600m [up]
vf = 0m/s
Vf^2 = Vi^2 + 2ad
0^2 = Vi^2 + 2(-9.81)(0.6)
0 = Vi^2 + -11.772
Vi^2 = 11.772
Vi = 3.43m/s [up]
Now to find impulse given to the ball by the floor we find the change in momentum.
Impulse = Momentum final - momentum initial
Impulse = (0.120)(3.43) - (0.120)(-4.95)
Impulse = 1.01kgm/s [up]
Related Questions
Planets in our solar system do not revolve around the sun in perfect circles. Their orbits are more like ovals that scientists describe as which of the following? A. revolutionary B. rotational C. periodical D. elliptical
Some vehicles, like _____, have different rules for registration and renewal. A. trailers B. coupes C. sedans D. pickup trucks
How much heat energy must be added to 52kg Of water at 68°F to raise the temperature to 212°F? The specific heat capacity for water is 4.186×10 to the third power J/kg times degrees Celsius
As a result of mitosis, the cels of a Molecular organism sure which of these properties select to correct answers. A. All cells have the same number of chromosomes. B. The number of chromosomes varies amongst the cells C. All cells have identical genetic information. D. Genetic information varies among the cells.
Answers
Weight is always (mass) times (gravity in the place where you are).
On Earth, acceleration due to gravity is 9.8 m/s² ,
so the person's weight is
(80 kg) x (9.8 m/s²) = 784 newtons .
(about 176.4 pounds)
Answers
Answer:
V = 2.79 m/s
Explanation:
Momentum before = momentum after
Erica's momentum = 0
Danny momentum = 49 kg * 4.9 m/s = 240.1 kg-m/s
240.1 + 0 = (37 + 49) kg * V
86 V = 240.1
V = 2.79 m/s
Answers
Answer:
3.75m/s²
Explanation:
g= GM/r²
For planet 1
= GM/r² (i)
= 15m/s²
for planet 2
radius= 2*r= 2r
g= GM/r
= GM/(2r)²
= GM/4r²
= GM/r² *1/4
from (i)
=
*1/4
= 15/4
= 3.75m/s²
Answers
Fx= 112 cos 42
Fx= 83.2 N
Diagram is attached
Answers
Answer:
da
Explanation:
b.T1 is ..... M1g.
c. T3 is ..... m1g + M2g
d.T1 is ..... T2
e.The magnitude of the acceleration of M2 is ..... the magnitude of the acceleration on m1.
f. T1 + T2 is ..... T3
Answers
Answer:
a. center of mass acceleration supposed to be acceleration due to gravity, 9.81 m/s^2,
b. T1 = 9.81m1 N; c. T3 =9.81(M1+M2) N; d. T3-T1, e. (T3-T1)/M2; f. (M1+M2)T3/M3
Explanation:
Final answer:
In this frictionless, massless pulley system, the center of mass accelerates downward with an acceleration equal to the acceleration due to gravity. The tension in the string connected to mass M1 is equal to M1g, and the tension in the string connected to mass M2 is equal to m1g + M2g. The magnitudes of the accelerations of M1 and M2 are equal, and the sum of the tensions T1 and T2 is equal to the tension T3.
Explanation:
a. The center of mass accelerates: When considering the system as a whole, the acceleration of the center of mass is determined by the net external force acting on the system. In this case, the only external force is the force due to gravity. Therefore, the center of mass accelerates downward with an acceleration equal to g, the acceleration due to gravity.
b. T1 is equal to M1g: The tension in the string connected to mass M1 is equal to the weight of M1, which is given by the formula T1 = M1g.
c. T3 is equal to m1g + M2g: The tension in the string connected to mass M2 is equal to the sum of the weights of M1 and M2, which is given by the formula T3 = m1g + M2g.
d. T1 is equal to T2: Since the pulley is assumed to be frictionless and massless, the tension in the string connected to mass M1 is the same as the tension in the string connected to mass M2.
e. The magnitude of the acceleration of M2 is equal to the magnitude of the acceleration on M1: This is due to the constraint imposed by the tension in the string. Since the tension in the string connecting M1 and M2 is the same, their accelerations must also be the same.
f. T1 + T2 is equal to T3: The sum of the tensions T1 and T2 is equal to the tension T3, as the total force acting on mass M2 is equal to the sum of the individual tensions.
Learn more about Dynamics here:
#SPJ12