Answer:
<h2>C. <u>
0.55 m/s towards the right</u></h2>
Explanation:
Using the conservation of law of momentum which states that the sum of momentum of bodies before collision is equal to the sum of the bodies after collision.
Momentum = Mass (M) * Velocity(V)
BEFORE COLLISION
Momentum of 0.25kg body moving at 1.0m/s = 0.25*1 = 0.25kgm/s
Momentum of 0.15kg body moving at 0.0m/s(body at rest) = 0kgm/s
AFTER COLLISION
Momentum of 0.25kg body moving at x m/s = 0.25* x= 0.25x kgm/s
<u>x is the final velocity of the 0.25kg ball</u>
Momentum of 0.15kg body moving at 0.75m/s(body at rest) =
0.15 * 0.75kgm/s = 0.1125 kgm/s
Using the law of conservation of momentum;
0.25+0 = 0.25x + 0.1125
0.25x = 0.25-0.1125
0.25x = 0.1375
x = 0.1375/0.25
x = 0.55m/s
Since the 0.15 kg ball moves off to the right after collision, the 0.25 kg ball will move at <u>0.55 m/s towards the right</u>
<u></u>
Answer:
v = 88.89 [m/s]
Explanation:
To solve this problem we must use the principle of conservation of momentum which tells us that the initial momentum of a body plus the momentum added to that body will be equal to the final momentum of the body.
We must make up the following equation:

where:
F = force applied = 4000 [N]
t = time = 0.001 [s]
m = mass = 0.045 [kg]
v = velocity [m/s]
![4000*0.001=0.045*v\\v=88.89[m/s]](https://tex.z-dn.net/?f=4000%2A0.001%3D0.045%2Av%5C%5Cv%3D88.89%5Bm%2Fs%5D)
To solve this problem we will apply the concepts related to energy conservation, so the potential energy in the package must be equivalent to its kinetic energy. From there we will find the speed of the package in the vertical component. The horizontal component is given, as it is the same as the one the plane is traveling to. Vectorially we will end up finding its magnitude. So,


Here,
m = Mass
g = Gravity
h = Height
v = Velocity
Rearranging to find the velocity

Replacing,


Using the vector properties the magnitude of the velocity vector would be given by,



Therefore the package is moving to 66.2m/s