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4 years ago

1. Sketch graphs of the momenta of asteroids A and B before the collision.

2 answers:

8 0

<span>First, there is no external force acting on the asteroids ( I don't think so, since they are travelling at constant velocity). Since there is no external force acting on the individual asteroids, their respective momenta are constant in time as well. So, in your momentum v/s Time graph, the momenta for each asteroid will correspond to a straight line parallel to time axis. This happens until they collide. On collision an impulse acts on each asteroid ( by Newton's third law, the same magnitude of impulse acts on the two bodies). Due to this impulse, their respective momenta will change. You just need a little mathematics to figure out the respective momenta after the collision. Keep in mind there is no external force on the system, so after collision, the asteroids would again move with constant velocities. (Conservation of linear momentum of the system would help! You just need another equation to find out the velocities)</span>

maria [59]4 years ago

3 0

a) The momentum of asteroid A is $\boxed{10,000\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/{\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}}$ and the momentum of asteroid B is $\boxed{2500\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/{\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}}$ .

b) At the time of collision, the magnitude of force of asteroid A on asteroid B is greater than the magnitude of force of asteroid B on asteroid A.

c) The total momentum of the two asteroids at the time of collision is $\boxed{12500\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/ {\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}}$ .

Further Explanation:

Given:

The mass of the asteroid A is $500\,{\text{kg}}$ .

The mass of the asteroid B is $250\,{\text{kg}}$ .

The speed of the asteroid A is $20\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}$ .

The speed of the asteroid B is $10\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}$ .

Concept:

The momentum of an asteroid is the product of its mass and its velocity in a particular direction. The momentum of the asteroids is expressed as:

$p = m \times v$

Here, $p$ is the momentum of the asteroid, $m$ is the mass of the asteroid and $v$ is the velocity of the asteroid.

Part (a):

For Asteroid A

Substitute ${p_A}$ for $p$ and the values of mass and velocity of asteroid A in above expression.

$\begin{aligned}{p_A} &= 500\times 20\\&= 10000\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/{\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\\\end{aligned}$

For Asteroid B

Substitute ${p_A}$ for $p$ and the values of mass and velocity of asteroid A in above expression.

$\begin{aligned}{p_B} &= 250 \times 10\\&= 2500\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/{\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\\\end{aligned}$

The plot of the momentum of asteroid A and asteroid B are shown in Figure 1.

Thus, the momentum of asteroid A is $\boxed{10,000\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/{\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}}$ and the momentum of asteroid B is $\boxed{2500\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/ {\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}}$ .

Part (b):

According to the Newton’s second law of motion, the rate of change of momentum is directly proportional to the force exerted by the body.

When the heavier asteroid collides with the lighter one, the change in momentum of the lighter asteroid will be more as it will gain speed in the forward direction. The change in momentum of the asteroid B is due to the higher force exerted by the asteroid A on asteroid B at the time of collision.

Thus, at the time of collision, the magnitude of force of asteroid A on asteroid B is greater than the magnitude of force of asteroid B on asteroid A.

Part (c):

As the two asteroids collide, at the time of collision, the momentum of the system remains conserved.

Thus,

$\begin{aligned}{p_{{\text{final}}}} = {p_{{\text{initial}}}} \hfill\\{p_{final}} = {p_A} + {p_B} \hfill\\\end{aligned}$

Substitute the values of initial momentum of the two asteroids

$\begin{aligned}{p_{final}} &= 10,000\, + 2500\\&= 12500\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/{\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\\\end{aligned}$

The plot of the momentum of the asteroids at the time of collision is shown in Figure 2.

Thus, the total momentum of the two asteroids at the time of collision is $\boxed{12500\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/{\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}}\right.\kern-\nulldelimiterspace} {\text{s}}}}$ .

Learn More:

1. Which of the following is not a component of a lever brainly.com/question/1073452

2. A 700-kg car, driving at 29 m/s, hits a brick wall and rebounds with a speed of 4.5 m/s brainly.com/question/9484203

3. It's been a great day of new, frictionless snow. Julie starts at the top of the 60 degree slope brainly.com/question/3943029

Answer Details:

Grade: High School

Subject: Physics

Chapter: Momentum

Keywords: Momentum, asteroid A, asteroid B, collision, before the collision, sketch a graph, total momentum, Newton’s law, magnitude of force, p=mv, 10000 kg.m/s, 2500 kg.m/s, 12500 kg.m/s.

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Which of the following statements are true?

inessss [21]

Answer:

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FALSE

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$a = \frac{d\vec v}{dt}$

so we can not say anything about the acceleration when speed is given to as and no information is given about velocity

b. If an object's acceleration is zero, then its speed must be constant.

TRUE

As we know that acceleration is defined as the rate of change in velocity

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Since we know that if acceleration is 0 then velocity must be constant and hence speed is also constant

c. If an object's velocity is constant, then its speed must be constant.

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d. If an object's acceleration is zero, its velocity must be constant.

TRUE

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3 years ago

A 50-g cube of ice, initially at 0.0°C, is dropped into 200 g of water in an 80-g aluminum container, both initially at 30°C.

MakcuM [25]

Answer:

b. 9.5°C

Explanation:

$m_i$ = Mass of ice = 50 g

$T_i$ = Initial temperature of water and Aluminum = 30°C

$L_f$ = Latent heat of fusion = $3.33\times 10^5\ J/kg^{\circ}C$

$m_w$ = Mass of water = 200 g

$c_w$ = Specific heat of water = 4186 J/kg⋅°C

$m_{Al}$ = Mass of Aluminum = 80 g

$c_{Al}$ = Specific heat of Aluminum = 900 J/kg⋅°C

The equation of the system's heat exchange is given by

$m_i(L_f+c_wT)+m_wc_w(T-T_i)+m_{Al}c_{Al}=0\\\Rightarrow 0.05\times (3.33\times 10^5+4186\times T)+0.2\times 4186(T-30)+0.08\times 900(T-30)=0\\\Rightarrow 1118.5T-10626=0\\\Rightarrow T=\dfrac{10626}{1118.5}\\\Rightarrow T=9.50022\ ^{\circ}C$

The final equilibrium temperature is 9.50022°C

4 0

3 years ago

What is the velocity of the object?

dmitriy555 [2]

<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Answer:</h2><h2 /><h2> $\huge\boxed{\text{V = 9.5 m/s}}$ </h2><h2>_____________________________________</h2>

mass = m = 2kg

Distance = x = 6m

Force = 30N

TO FIND:

Work = W = ?

Velocity = V = ?

<h2>SOLUTION:</h2>

According to the object of mass 2 kg travels a distance when the force was exerted on it. The graph between the Force and position was plotted which shows that 30 N of force was used to push the object till the distance of 6.0m.

To find the work, I will use the method of determining the area of the plotted graph. As the graph is plotted in the straight line between the Force and work, THE PICTURE ATTCHED SHOWS THE AREA COVERED IN BLUE AS WORK DONE AND HEIGHT AS 30m AND DISTANCE COVERED AS 6m To solve for the area(work) of triangle is given as,

${\Longrightarrow}\qquad \qquad \qquad W\ =\ \frac{1}{2}\;(Base)\:(Height)$

Base is the x-axis of the graph which is Position i.e. 6m

Height is the y-axis of the graph which is Force i.e. 30N

So,

$W\ =\ \frac{1}{2}\:6\:x\:30$

W = 90 J

The work done is 90 J.

According to the principle of work and kinetic energy (also known as the work-energy theorem) states that the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle.

${\Longrightarrow}\qquad \qquad \qquad W\quad =\quad K.E\\\\{\Longrightarrow}\qquad \qquad \qquad W\quad =\quad \frac{1}{2}\ m\ V^2 \\\\{\Longrightarrow}\qquad \qquad \qquad W\quad =\quad \frac{1}{2}\ 2\ (V_f-V_i)^2\\\\{V_i\ is\ 0\ because\ the\ object\ was\ initially\ at\ rest}\\\\ {\Longrightarrow}\qquad \qquad \qquad W\quad\ =\ \frac{1}{2}\ x\ 2\ (V_f-0)^2 \\\\{\Longrightarrow}\qquad \qquad \qquad 90\quad = \frac{1}{2}\ x\ 2\ (V_f)^2$

$\\\\{\Longrightarrow}\qquad \qquad \qquad V_f\quad =\ \sqrt{\frac{2\ (90)\ }{2}}\\\\{\Longrightarrow}\qquad \qquad \qquad \boxed {V_f\quad =\ 9.48\ m/s}$