Question

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

Answer 1

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)

Answer 2

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 $\boxed{12500\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/{\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}}\right.\kern-\nulldelimiterspace} {\text{s}}}}$.

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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.