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:
2) center mass
3) when work is done by a system, energy is released
Explanation:
Answer:
8) 709.8875 J
9) The object is at 7.24375 m from the ground
10) Kinetic energy increases as the object falls.
Explanation:
We use the expression for the displacement h(t) as a function of time of an object experiencing free fall:
h(t) = hi - (g/2) t^2
hi being the initial position of the object (10m) above ground, g the acceleration of gravity (9.8 m/s^2), and t the time (in our case 0.75 seconds):
h(0.75) = 10 - 4/9 (0.75)^2 = 7.24375 m
This is the position of the 10 kg object after 0.75 seconds (answer for part 9)
Knowing this position we can calculate the potential energy of the object when it is at this height, using the formula:
U = m g h = 10kg * 9.8 (m/s^2) * 7.24375 m = 709.8875 J (answer for part 8)
Part 10)
the kinetic energy of the object increases as it gets closer to ground, since its velocity is increasing in magnitude because is being accelerated in its motion downwards.
Answer:
A. 0N
Explanation:
assume you're pulling in the positive direction, then we have 5 + (-5) = netforce = 0N
Sitting. Because its no move at all
