Physics, 28.03.2020 21:55, mya032

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

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answered: unknown9263

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:

$\Delta p = F\Delta t$

where

$\Delta p$ is the change in momentum

F is the force

$\Delta t$ is the time during which the force is applied

In this problem:

$\Delta t = 2.5 s$

For the boy, the change in momentum is:

$\Delta p = m_2 (v_2 - u_2)$

And since

$m_2 = 50 kg\\u_2 = 0 m/s\\v_2 = 3 m/s$

We have

$\Delta p = (50)(3-0)=150 kg m/s$

So, the force exerted between the boy and the girl is:

$F=\frac{\Delta p}{\Delta t}=\frac{150}{2.5}=60 N$

answered: Guest

ammonium chloride & tallow

answered: Guest

answer: at 24 february, 2014 the population gets twice of its initial population.

explanation:

since we have given that

the population for alpha city, t years after january 1, 2004 is given as

$p(t)=0.3t^2+6t+80$

first we find out the initial population, i.e. t=0,

so, our quadratic equation becomes,

$p(0)=0+0+80=80$

according to question, we have said that if the population has twice its initial population, then it becomes

$p(t)=2\times 80=160$

so, time taken to reach this above mentioned population is given by

$160=0.3t^2+6t+80\\\\160-80=0.3t^2+6t\\\\80=0.3t^2+6t\\\\0.3t^2+6t-80=0\\\\\text{using quadratic formula ,we get }\\\\t_{1,\: 2}=\frac{-60\pm \sqrt{60^2-4\cdot \: 3\cdot \: 800}}{2\cdot \: 3}\\\\t=-29.14\ and\ t=9.149$

now, we know that time can't be negative so, we take t=9.149 years.

hence, at 24 february, 2014 the population gets twice of its initial population.

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