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.
You might be interested in
The speed of the elevator at the beginning of the 8 m descent is nearly 4 m/s. Hence, option A is the correct answer.
We are given that-
the mass of the elevator (m) = 1000 kg ;
the distance the elevator decelerated to be y = 8m ;
the tension is T = 11000 N;
let us determine the acceleration 'a' by using Newton's second law of motion.
∑Fy = ma
W - T = ma
(1000kg x 9.8 m/s² ) - 11000N = 1000 kg x a
9800 - 11000 = 1000
a = - 1.2 m/s²
Using the equation of kinematics to determine the initial velocity.
² =
² + 2ay
= √ ( 2 x 1.2m/s² x 8 m )
= √19.2 m²/s²
= 4.38 m/s ≈ 4 m/s
Hence, the initial velocity of the elevator is 4m/s.
Read more about the Equation of kinematics:
#SPJ4
Answer:
.
Explanation:
When the ball is placed in this pool of water, part of the ball would be beneath the surface of the pool. The volume of the water that this ball displaced is equal to the volume of the ball that is beneath the water surface.
The buoyancy force on this ball would be equal in magnitude to the weight of water that this ball has displaced.
Let denote the mass of this ball. Let
denote the mass of water that this ball has displaced.
Let denote the gravitational field strength. The weight of this ball would be
. Likewise, the weight of water displaced would be
.
For this ball to stay afloat, the buoyancy force on this ball should be greater than or equal to the weight of this ball. In other words:
.
At the same time, buoyancy is equal in magnitude the the weight of water displaced. Thus:
.
Therefore:
.
.
In other words, the mass of water that this ball displaced should be greater than or equal to the mass of of the ball. Let denote the density of water. The volume of water that this ball should displace would be:
.
Given that while
:
.
In other words, for this ball to stay afloat, at least of the volume of this ball should be under water. Therefore, the volume of this ball should be at least
.