Question

NASA is launching a probe (m = 500 kg) at a small asteroid (m = 2000 kg). The asteroid is traveling at 100 m/s, and the probe is moving perpendicularly toward it with a speed of 400 m/s. Unfortunately, the landing gear does not work right, so the probe bounces off the asteroid. After the collision, the asteroid is knocked off its course by 71ᵒ and is now going 165 m/s. If the probe’s speed is now 300 m/s, in what direction is it going?

Answer 1

The direction of the probe's final velocity is approximately
`\(15.76^\circ\)` with respect to the x-axis.

To solve this problem, we can use the principles of conservation of linear momentum and conservation of kinetic energy. The linear momentum of the system (probe + asteroid) is conserved in the absence of external forces, and the kinetic energy is conserved if the collision is elastic.

Let's denote:

-
`\(m_p\)`: mass of the probe (500 kg)

-
`\(m_a\)`: mass of the asteroid (2000 kg)

-
`\(v_(p1)\)`: initial velocity of the probe (400 m/s)

-
`\(v_(a1)\)`: initial velocity of the asteroid (100 m/s)

-
`\(v_(p2)\)`: final velocity of the probe (300 m/s)

-
`\(v_(a2)\)`: final velocity of the asteroid (165 m/s)

The linear momentum before the collision is the sum of the linear momentum of the probe and the asteroid:

`\[m_p \cdot v_(p1) + m_a \cdot v_(a1)\]`

The linear momentum after the collision is the sum of the linear momentum of the probe and the asteroid in their final states:

`\[m_p \cdot v_(p2) + m_a \cdot v_(a2)\]`

Now, we can set up an equation using the conservation of linear momentum:

`\[m_p \cdot v_(p1) + m_a \cdot v_(a1) = m_p \cdot v_(p2) + m_a \cdot v_(a2)\]`

Plug in the known values:

`\[(500 \, \text{kg} \cdot 400 \, \text{m/s}) + (2000 \, \text{kg} \cdot 100 \, \text{m/s}) = (500 \, \text{kg} \cdot 300 \, \text{m/s}) + (2000 \, \text{kg} \cdot 165 \, \text{m/s})\]`

Now, solve for \(v_{a2}\) (final velocity of the asteroid):

`\[(200000 \, \text{kg} \cdot \text{m/s}) + (200000 \, \text{kg} \cdot \text{m/s}) = (150000 \, \text{kg} \cdot \text{m/s}) + (330000 \, \text{kg} \cdot \text{m/s})\]`

`\[400000 \, \text{kg} \cdot \text{m/s} = 480000 \, \text{kg} \cdot \text{m/s}\]`

This implies that the final velocity of the asteroid
`\(v_(a2)\) is \(400 \, \text{m/s}\)`.

Now, we can use trigonometry to find the direction of the final velocity of the probe. The angle \(\theta\) between the initial velocity of the asteroid and its final velocity is given as \(71^\circ\).

`\[\tan(\theta) = \frac{{v_(a2y)}}{{v_(a1)}}\]`

where
`\(v_(a2y)\)`is the component of the final velocity of the asteroid in the y-direction.

`\[\tan(71^\circ) = \frac{{v_(a2y)}}{{100 \, \text{m/s}}}\]`

Solve for
`\(v_(a2y)\)`:

`\[v_(a2y) = 100 \, \text{m/s} \cdot \tan(71^\circ)\]`

Now, we can find the component of the final velocity of the probe in the y-direction
`(\(v_(p2y)\))`:

`\[v_(p2y) = v_(p2) \cdot \sin(\theta)\]`

Substitute the values:

`\[v_(p2y) = 300 \, \text{m/s} \cdot \sin(71^\circ)\]`

Now, we can find the y-component of the final velocity of the probe
`(\(v_(p2y)\))` by equating it to the y-component of the final velocity of the asteroid
`(\(v_(a2y)\))`:

`\[v_(p2y) = v_(a2y)\]`

`\[300 \, \text{m/s} \cdot \sin(71^\circ) = 100 \, \text{m/s} \cdot \tan(71^\circ)\]`

Now, you can solve for the angle \(\phi\) between the x-axis and the final velocity of the probe
`(\(v_(p2x)\))`:

`\[\tan(\phi) = \frac{{v_(p2x)}}{{v_(p2y)}}\]`

`\[\phi = \tan^(-1)\left(\frac{{v_(p2x)}}{{v_(p2y)}}\right)\]`

Plug in the values:

`\[\phi = \tan^(-1)\left(\frac{{300 \, \text{m/s} \cdot \cos(71^\circ)}}{{300 \, \text{m/s} \cdot \sin(71^\circ)}}\right)\]`

Let's calculate the value of
`\(\phi\)`:

`\[\phi = \tan^(-1)\left(\frac{{300 \, \text{m/s} \cdot \cos(71^\circ)}}{{300 \, \text{m/s} \cdot \sin(71^\circ)}}\right)\]`

First, calculate the values of \
`(\cos(71^\circ)\)`and \
`(\sin(71^\circ)\)`:

`\[\cos(71^\circ) \approx 0.2756\]`

`\[\sin(71^\circ) \approx 0.9613\]`

Now, substitute these values into the expression for \(\phi\):

`\[\phi \approx \tan^(-1)\left(\frac{{300 \, \text{m/s} \cdot 0.2756}}{{300 \, \text{m/s} \cdot 0.9613}}\right)\]`

Now, calculate the value:

`\[\phi \approx \tan^(-1)(0.2864)\]`

Use a calculator to find the arctangent:

`\[\phi \approx 15.76^\circ\]`

So, the direction of the probe's final velocity is approximately
`\(15.76^\circ\)` with respect to the x-axis.

Answer 2

To find the direction of the probe after bouncing off the asteroid, we need to consider the conservation of momentum. By using vector addition and trigonometry, we can calculate that the probe is now moving at an angle of 71ᵒ with respect to the horizontal axis.

Step-by-step explanation:

To determine the direction in which the probe is moving after it bounces off the asteroid, we need to consider the conservation of momentum.

Before the collision, the total momentum of the system (probe + asteroid) is the sum of the individual momenta of the objects. After the collision, the total momentum must be the same, but now we need to account for the change in direction of the asteroid.

To find the final direction of the probe, we can use vector addition.

Given the magnitude of the probe's speed (300 m/s) and the angle at which the asteroid is knocked off its course (71ᵒ), we can find the x and y components of the probe's velocity.

Using trigonometry, we can calculate that the x-component of the probe's velocity is

300 m/s * cos(71ᵒ),

and the y-component is

300 m/s * sin(71ᵒ).

Therefore, the probe is now moving at an angle of 71ᵒ with respect to the horizontal axis.