A man of mass m 1 5 70.0 kg is skating at v1 5 8.00 m/s behind his wife of mass m 2 5 50.0 kg, who is skating at v2 5 4.00 m/s.
Instead of passing her, he inadvertently collides with her. He grabs her around the waist, and they maintain their balance. (a) Sketch the problem with before - and - after diagrams, representing the skaters as blocks. (b) Is the collision best described as elastic, inelastic, or perfectly inelastic? Why? (c) Write the general equation for conservation of momentum in terms of m 1, v 1, m 2, v 2, and final velocity vf . (d) Solve the momentum equation for vf . (e) Substitute values, obtaining the numerical value for vf , their speed after the collision.
1 answer:
Answer:
A. Kindly find attached free body diagram for your reference (smiles I guess I will make a terrible artist)
B. The collision is inelastic because both the husband and the wife moved together with same velocity as he grabs her on the waist
C. The general equation for conservation of momentum in terms of m 1, v 1, m 2, v 2, and final velocity vf
Say mass of husband is m1
Mass of the wife is m2
Velocity of the husband is v1
Velocity of the wife is v2
According to the conservation of momentum principle momentum before impact m1v1+m2v2 =momentum after impact Common velocity after impact (m1+m2)vf
The momentum equation is
m1v1+m2v2= (m1+m2)vf
D. To solve for vf we need to make it subject of formula
vf= {(m1v1) +(m2v2)}/(m1+m2)
E. Substituting our given data
vf=
{(1570*58)+(2550*54)}/(1570+2558)
vf=91060+137700/4120
vf=228760/4120
vf=55.52m/s
Their speed after collision is 55.52m/s
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Answer:
A = 1.4 m/s²
B = -0.10493 m/s³
a = 1.29507 m/s²
T = 28095.8271 N
T = 1.13198 W
Explanation:
t = Time taken
g = Acceleration due to gravity = 9.81 m/s²
The equation
Differentiating with respect to time
At t = 0
Hence, A = 1.4 m/s²
B = -0.10493 m/s³
At t = 5 seconds
a = 1.29507 m/s²
T = 28095.8271 N
Weight of rocket
T = 1.13198 W
Answer:
Explanation:
Let's assume that an object is launched straight upward in a gravitational field. Its initial kinetic energy is given by
(1)
where m is the mass and v is the initial speed.
As the object goes higher, its kinetic energy decreases and it is converted into gravitational potential energy, since the total mechanical energy (sum of kinetic and potential energy) must remain constant:
At the highest point of the trajectory, the speed of the object is zero (v=0), so the kinetic energy is also zero (K=0), which means that all the kinetic energy has been converted into potential energy:
(2)
where g is the gravitational acceleration and h is the maximum height of the object.
Due to conservation of energy, we can write that (1) and (2) are equal, so:
from which we can derive an expression for the maximum height reached by the object