Answer: a) 6.67cm/s b) 1/2
Explanation:
According to law of conservation of momentum, the momentum of the bodies before collision is equal to the momentum of the bodies after collision. Since the second body was initially at rest this means the initial velocity of the body is "zero".
Let m1 and m2 be the masses of the bodies
u1 and u2 be their velocities respectively
m1 = 5.0g m2 = 10.0g u1 = 20.0cm/s u2 = 0cm/s
Since momentum = mass × velocity
The conservation of momentum of the body will be
m1u1 + m2u2 = (m1+m2)v
Note that the body will move with a common velocity (v) after collision which will serve as the velocity of each object after collision.
5(20) + 10(0) = (5+10)v
100 + 0 = 15v
v = 100/15
v = 6.67cm/s
Therefore the velocity of each object after the collision is 6.67cm/s
b) kinectic energy of the 10.0g object will be 1/2MV²
= 1/2×10×6.67²
= 222.44Joules
kinectic energy of the 5.0g object will be 1/2MV²
= 1/2×5×6.67²
= 222.44Joules
= 111.22Joules
Fraction of the initial kinetic transferred to the 10g object will be
111.22/222.44
= 1/2
Answer: hello question b is incomplete attached below is the missing question
a) attached below
b) V = 0.336 ft/s
Explanation:
Elongation ( Xo) = 16/ 7 feet
mass attached to 4-foot spring = 16 pounds
medium has 9/2 times instanteous velocity
<u>a) Find the equation of motion if the mass is initially released from the equilibrium position with a downward velocity of 2 ft/s</u>
The motion is an underdamped motion because the value of β < Wo
Wo = 3.741 s^-1
attached below is a detailed solution of the question
The added weight of the sand puts more downward pressure on the wheels contacting the rails, which would cause the trains speed to decrease.
It means that they were set earlier therefor they are older.
