Is it possible for the momentum of a system consisting of two carts on a low-friction track to be zero even if both carts are mo
1 answer:
Answer:
Yes, if the two carts are moving into opposite directions
Explanation:
The total momentum of the system of two carts is given by:
where
m1, m2 are the masses of the two carts
v1, v2 are the velocities of the two carts
Let's remind that v (the velocity) is a vector, so its sign depends on the direction in which the cart is moving.
We want to know if it is possible that the total momentum of the system can be zero, so it must be:
From this equation, we see that this condition can only occur if v1 and v2 have opposite signs. Opposite signs mean opposite directions: therefore, the total momentum can be zero if the two carts are moving into opposite directions.
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This question involves the concepts of the equations of motion, kinetic energy, and potential energy.
a. The kinetic energy of the rocket at launch is "3.6 J".
b. maximum gravitational potential energy of the rocket is "3.6 J".
<h3>a. KINETIC ENERGY AT LAUNCH</h3>The kinetic energy of the rocket at launch is given by the following formula:
where,
- K.E = initial kinetic energy = ?
- m = mass of rocket = 0.05 kg
= initial speed = 12 m/s
Therefore,
K.E = 3.6 J
<h3>b. MAXIMUM GRAVITATIONAL POTENTIAL ENERGY</h3>
First, we will use the third equation of motion to find the maximum height reached by rocket:
where,
- g = -9.81 m/s²
- h = maximum height = ?
- vf = final speed = 0 m/s
Therefore,
2(-9.81 m/s²)h = (0 m/s)² - (12 m/s)²
h = 7.34 m
Hence, the maximum gravitational potential energy will be:
P.E = mgh
P.E = (0.05 kg)(9.81 m/s²)(7.34 m)
P.E = 3.6 J
Learn more about the equations of motion here: