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3 years ago

A kayaker moves 26 meters southward, then 18 meters

Physics

1 answer:

Svetllana [295]3 years ago

8 0

Total distance: 56 meters. Magnitude and direction of displacement: 20 meters South.

Explanation:

The term distance refers to space between one point and other, or the total space a body or object covered while moving. In the case presented, this can be calculated by adding the partial distances given. This means the total distance is 56 meters as 26 meters + 18 meters + 12 meters = 56 meters.

On the other hand, displacement considers the distance from the initial position to the final position, and the direction of movement. This means partial distances should not be added but each movement should be considered according to the direction. The process is shown below:

-The first movement was 26 meters southward; this means by the end of this movement the distance between the initial position was 26 meters south.

- The second movement was 18 northward; this means the kayaker moved 18 meters towards the position. This changes the displacement to 8 meters South as 26 meters south - 18 meters north = 8 meters to the South.

-The last movement was 12 meters sound; this means the kayaker increased the distance from the original position 8 meters to the South + 12 meters to the South = 20 meters South (total displacement.)

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Two people stand facing each other at roller skating rink then push off each other

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a) 0 kg m/s

b) 0 kg m/s

c) +3 m/s

d) 60 N

Explanation:

The momentum of an object is a vector quantity given by:

$p=mv$

where

m is the mass of the object

v is the velocity of the object

In this problem, we have a system of two people, so the total momentum will be the sum of the individual momenta of the two people:

$p=p_1 + p_2$

Which can be rewritten as

$p=m_1 u_1 + m_2 u_2$

where $m_1,m_2$ are the masses of the two people and $u_1,u_2$ their initial velocities.

However, the two people are initially at rest, so

$u_1 = 0\\u_2 = 0$

Therefore the total momentum is

$p=0+0=0$

The principle of conservation of momentum states that when there are no external forces acting on a system, the total momentum of the system is conserved, so we can write:

$p_i = p_f$

where

$p_i$ is the total momentum of the system before

$p_f$ is the total momentum of the system after

In this problem,

$p_i = 0$

As we calculated in part a: this is because the total momentum of the two people before they push off each other is zero.

Therefore, according to the law of conservation of momentum,

$p_f = p_i = 0$

So the total momentum is zero also after they push off each other.

The total momentum of the girl and the boy after they push off each other can be written as:

$p_f = m_1 v_1 + m_2 v_2$ (1)

where:

$m_1 = 30 kg$ is the mass of the girl

$v_1 = -5 m/s$ is her velocity (she moves backward, so the negative sign)

$m_2 = 50 kg$ is the mass of the boy

$v_2$ is the velocity of the boy

As calculated in part b), we also know that the total momentum of the girl and the boy is

$p_f = 0$ (2)

By combining eq(1) and eq(2) we get

$0=m_1 v_1 + m_2 v_2$

And solving for v2 we find the velocity of the boy:

$v_2=-\frac{m_1 v_1}{m_2}=-\frac{(30)(-5)}{50}=+3 m/s$

and the positive sign means he is moving forward.

We can solve this part by applying the impulse theorem, which states that the change in momentum of an object is equal to the product between the force applied on it and the duration of the collision:

$\Delta p = F\Delta t$