In the simulation, there are three balls on the floor. Drag each of them up off the floor, and then let go. See what happens to
1 answer:
Answer:
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A way the balls behave the same way is by bouncing about 1 time after throwing the balls up. A way the balls act differently is the blue ball is bouncier than all the balls, the red ball bounces about 2 times before stopping, and the green ball doesn’t really bounce except for one time.
Explanation:
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Answer:
1) Please find attached, created with Microsoft Visio
2) The acceleration of the masses connected by the light string is 0.00735 m/s²
3) The tension in the cord is 0.147 N
4) The time it would take the block to go 1.2 m to the edge of the table is approximately 18.07 s
5) The velocity of the cart as soon as it gets to the edge of the table is 0.042 m/s
Explanation:
1) Please find attached, the required free body diagram, showing the tension, weight and frictional (zero friction) forces acting on the cart and the mass created with Microsoft Visio
2) The acceleration of the masses connected by the light string is given as follows;
F = Mass, m × Acceleration, a
The mass of the truck, M = 20.0 kg
The mass attached to the string, hanging rom the pulley, m = 0.0150 kg
The force, F acting on the system = The pulling force on the cart = The tension on the cable = The weight of the hanging mass = 0.0150 × 9.8 = 0.147 N
The pulling force acting on the cart, F = M × a
∴ F = 0.147 N = 20.0 kg × a
a = 0.147 N/(20.0 kg) = 0.00735 m/s²
The acceleration of the truck = a = 0.00735 m/s²
3) The tension in the cord = F = 0.147 N
4) The time, t, it would take the block to go 1.2 m to the edge of the table is given by the kinematic equation, s = u·t + 1/2·a·t²
Where;
s = The distance to the edge of the table = 1.2 m
u = The initial velocity = 0 m/s (The cart is assumed to be initially at rest)
a = The acceleration of the cart = 0.00735 m/s²
t = The time taken
Substituting the known values, gives;
s = u·t + 1/2·a·t²
1.2 = 0 × t + 1/2 ×0.00735 × t²
1.2 = 1/2 ×0.00735 × t²
t² = 1.2/(1/2 ×0.00735) ≈ 326.5306
t = √(1.2/(1/2 ×0.00735)) ≈ 18.07
The time it would take the block to go 1.2 m to the edge of the table = t ≈ 18.07 s
5) The velocity, v, of the cart as soon as it gets to the edge of the table is given by the kinematic equation, v² = u² + 2·a·s as follows;
v² = u² + 2·a·s
u = 0 m/s
v² = 0² + 2 × 0.00735 × 1.2 = 0.001764
v = √(0.001764) = 0.042
The velocity of the cart as soon as it gets to the edge of the table = v = 0.042 m/s.
