Question

SPH4U - Assignment 2 Momentum and Energy
Full Solution Assignments (40%) require thorough solutions and calculations presented in the GRASS format, which stands for Given, Required, Analysis, Solution, and Statement. Please refer to my notes for a detailed explanation of this format. Additionally, please note that all Dynamics problems must include a Free Body Diagram (FBD).



Problem:
A 3000 kg space capsule is travelling in outer space with a velocity of 200 m/s. In an effort to alter its course, it fires a 25.0 kg projectile perpendicular to its original direction of motion at a speed of 2000 m/s.
(a) What is the new speed of the space capsule?
(b) By what angle has the space capsule direction changed?

Answer 1

Sure! Let's solve the problem step by step using the GRASS format.

### Given
- Mass of the space capsule, \( m_{c} = 3000 \, \text{kg} \)
- Initial velocity of the capsule, \( \vec{v_{c}} = 200 \, \text{m/s} \) (taking the positive x-direction)
- Mass of the projectile, \( m_{p} = 25.0 \, \text{kg} \)
- Velocity of the projectile relative to the capsule, \( \vec{v_{p}} = 2000 \, \text{m/s} \) (perpendicular to the direction of the capsule)

### Required
(a) New speed of the space capsule, \( v_{c}' \).

(b) Angle of direction change, \( \theta \).

### Analysis
Since the projectile is fired perpendicular to the original direction of motion of the capsule, we can use the principle of conservation of momentum in two dimensions (horizontal and vertical).

**Initial momentum of the system (before firing the projectile):**
- The capsule has momentum in the x-direction.
\[
\text{Initial momentum of capsule} = m_{c} \cdot v_{c} = 3000 \, \text{kg} \cdot 200 \, \text{m/s} = 600000 \, \text{kg m/s} \, \text{(in x-direction)}
\]

**Final momentum of the system (after firing the projectile):**
- The final momentum of the capsule in the x-direction remains unaltered (initial x-direction).
- The projectile has momentum in the y-direction (since it is fired perpendicularly).
\[
\text{Momentum of projectile} = m_{p} \cdot v_{p} = 25.0 \, \text{kg} \cdot 2000 \, \text{m/s} = 50000 \, \text{kg m/s} \, \text{(in y-direction)}
\]

The final momentum of the capsule in the x-direction is:
\[
\text{Final momentum of capsule in x-direction} = m_{c} \cdot v_{c}' + 0 \Rightarrow m_{c} \cdot v_{c}'
\]

### Solution
Using conservation of momentum in the x-direction:

\[
600000 \, \text{kg m/s} = 3000 \, \text{kg} \cdot v_{c}' \Rightarrow v_{c}' = \frac{600000}{3000} = 200 \, \text{m/s} \, \text{(x-component of velocity)}
\]

In the y-direction, the momentum after firing the projectile will equal the initial momentum transferred to the projectile:

\[
0 = 0 + 50000 \, \text{(in y-direction)}
\]

So, we can use Pythagoras' theorem to find the new speed of the capsule:

\[
v_{c}' = \sqrt{(200 \, \text{m/s})^2 + \left(\frac{50000}{3000}\right)^2}
\]

Calculating \( \frac{50000}{3000} \):

\[
\frac{50000}{3000} \approx 16.67 \, \text{m/s}
\]

\[
v_{c}^{\text{new}} = \sqrt{(200)^2 + (16.67)^2} = \sqrt{40000 + 278.89} = \sqrt{40278.89} \approx 200.69 \, \text{m/s}
\]

#### Angle of Direction Change
Using the tangent of the angle:

\[
\tan \theta = \frac{\text{Opposite (y-direction)}}{\text{Adjacent (x-direction)}} = \frac{v_{p}}{v_{c}}
\]

Where \( v_{p} = \frac{50000}{3000} \) and \( v_{c} = 200 \).

So:

\[
\tan \theta = \frac{16.67}{200}
\]

Calculating \( \theta \):

\[
\theta = \tan^{-1}(0.08335) \approx 4.76^\circ
\]

### Statement
(a) The new speed of the space capsule is approximately **200.69 m/s**.

(b) The angle by which the space capsule's direction has changed is approximately **4.76°**.

### Free Body Diagram (FBD)
In the FBD, depict the space capsule initially traveling horizontally, with an arrow pointing to the right (200 m/s). After firing the projectile, draw an arrow in the vertical direction (representing the projectile firing). Finally, draw the resultant vector representing the new direction of the capsule slightly upwards towards the right, indicating the change in direction from the original path.

### Conclusion
The final calculations and diagram demonstrate the dynamic system behavior post-firing the projectile in space, confirming the principles of momentum conservation in two dimensions.