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Wewaii [24]
3 years ago
7

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

ving?
Physics
1 answer:
tamaranim1 [39]3 years ago
8 0

Answer:

Yes, if the carts are travelling into opposite directions

Explanation:

The total momentum of a two carts system is the sum of the momenta of the individual carts:

p=p_1 +p_2 = m_1 v_1 + m_2 v_2

where

m1, m2 are the masses of the two carts

v1, v2 are the velocities of the two carts

In order to have a total momentum of zero, we must have

m_1 v_1 + m_2 v_2 = 0\\m_1 v_1 = - m_2 v_2 (1)

Let's remind that velocity (and so, momentum as well) is a vector quantity: this means that it has a direction, so when summing together the momenta, we must also consider the sign of the velocity, depending on its direction.

Therefore, if the two carts are moving in opposite directions, the signs of the two velocities will be opposite. For example, we can have

v_1 > 0 \\v_2 < 0

This means that the condition in eq.(1) can be satisfied, provided that the two carts are travelling into opposite directions.

You might be interested in
A car is driving at 85 km/h and the driver spots a stop sign ahead. What coefficient of friction is needed to stop the car at th
almond37 [142]

Answer:

μ = 0.0315

Explanation:

Since the car moves on a horizontal surface, if we sum forces equal to zero on the Y-axis, we can determine the value of the normal force exerted by the ground on the vehicle. This force is equal to the weight of the cart (product of its mass by gravity)

N = m*g (1)

The friction force is equal to the product of the normal force by the coefficient of friction.

F = μ*N (2)

This way replacing 1 in 2, we have:

F = μ*m*g (2)

Using the theorem of work and energy, which tells us that the sum of the potential and kinetic energies and the work done on a body is equal to the final kinetic energy of the body. We can determine an equation that relates the frictional force to the initial speed of the carriage, so we will determine the coefficient of friction.

\frac{1}{2} *m*v_{i}^{2}-(F*d)=  \frac{1}{2} *m*v_{f}^{2}

where:

vf = final velocity = 0

vi = initial velocity = 85 [km/h] = 23.61 [m/s]

d = displacement = 900 [m]

F = friction force [N]

The final velocity is zero since when the vehicle has traveled 900 meters its velocity is zero.

Now replacing:

(1/2)*m*(23.61)^2 = μ*m*g*d

0.5*(23.61)^2 = μ*9,81*900

μ = 0.0315

5 0
3 years ago
Anne applies a force on a toy car and makes it move forward. What can be said about the forces acting on the toy car at the mome
tester [92]

The correct answer to the question is unbalanced .

EXPLANATION:

As per Newton's first laws of motion, we know that a body at rest will be at rest or a body moving with uniform velocity along a straight line will keep on moving with the same velocity along that line unless and until it is compelled by some external unbalanced forces acting on.

Hence, it is the unbalanced force which is responsible for creating the motion on the body.

As per the question, Anne applied some force on the toy car. It is called applied force. There is also frictional force between car and the surface which opposes the motion of the car. The toy car will move due to the net force acting on it. The net force is known as unbalanced force.

5 0
4 years ago
Read 2 more answers
jack be nimble jack be quick jack jumped over the candlestick with a velocity of 5.0 m/s at an angle of 30.0 degrees to horizont
Mrrafil [7]

Answer:

No

Explanation:

The vertical component of Jack's initial velocity is:

5.0

⋅

sin

30

∘

=

5.0

⋅

1

2

=

2.5

m/s

With gravitational acceleration

9.8

m/s

2

, he will reach the highest point of his trajectory after:

2.5

9.8

≈

0.255

s

The average vertical component of his velocity in that

0.255

s

will be:

1

2

⋅

2.5

=

1.25

m/s

So the highest point of his trajectory will be:

0.255

⋅

1.25

≈

0.32

m

So he will pass approximately

7

cm

above the top of the candle.

The horizontal component of his velocity will be a constant:

5.0

⋅

cos

30

∘

=

5.0

⋅

√

3

2

≈

4.33

m/s

So Jack's trajectory will be substantially longer than it is high and he will spend little time anywhere near above the candle.

4 0
3 years ago
If you travel from Yakima to Ellensburg (Yakima to Ellensburg is 50 miles) with a speed of 60 miles/hour for half of the
koban [17]

Answer:

\displaystyle \frac{480}{7}\approx 68.6\; \rm mph.

Explanation:

The average speed of an object is equal to total distance over total time.

  • Distance traveled: \rm 50 \; mi.

How much time is taken? This trip is divided into two halves, each of distance \displaystyle \frac{50}{2} = 25\;\rm mi.

Time spent on the first half of the trip:

\displaystyle t_1 = \frac{s_1}{v_1} = \frac{25}{60} = \frac{5}{12}\; \rm hours.

Similarly, time spent on the second half of the trip:

\displaystyle t_2 = \frac{s_2}{v_2} = \frac{25}{80} = \frac{5}{16}\; \rm hours.

In total:

\displaystyle \frac{5}{12} + \frac{5}{16} = \frac{35}{48} \; \rm hours.

Average speed:

\begin{aligned} \text{Average speed} &= \frac{\text{Total Distance}}{\text{Total Time}}\\ &= 50 \left/\frac{35}{48}\right.\\ &= 50 \cdot \frac{48}{35} \\&= \frac{480}{7}\approx 68.6\; \rm mph \end{aligned}.

This value turned out to be slightly different from the average of the speed during the two halves of the journey. The reason is that the object traveled at each speed for a different amount of time. It spent more time at the slower speed, which gives that speed a greater weight in the average. That explains why the average speed is closer to \rm 60\; mph rather than \rm 80\; mph.

3 0
3 years ago
5. The volume of physical activity attained by an individual who exercises at a level of 7 METs for 35 min · day-1, 4 days · wk-
Airida [17]
D. 980, this is the best answer because 35 x 7 is 980 :)

3 0
3 years ago
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