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April sits at rest on a skateboard. She has a mass of 55 kg. Her friend throws her a watermelon (m = 2 kg) at a speed of 5 m/s. If she catches it, how fast will she, the skateboard, and the watermelon move?

A. 5 m/s
B. 11 m/s
C. 0.18 m/s
D. 0.09 m/s


Sagot :

To solve this problem, we need to apply the principle of conservation of momentum. Momentum is the product of mass and velocity, and for a closed system, the total momentum before an event must equal the total momentum after the event.

1. Step 1: Identify the masses and velocities.
- Mass of April: [tex]\( m_{April} = 55 \, \text{kg} \)[/tex]
- Mass of the watermelon: [tex]\( m_{watermelon} = 2 \, \text{kg} \)[/tex]
- Velocity of the watermelon before being caught: [tex]\( v_{watermelon} = 5 \, \text{m/s} \)[/tex]
- Initial velocity of April (and the skateboard) before catching the watermelon: [tex]\( v_{April,initial} = 0 \, \text{m/s} \)[/tex]

2. Step 2: Calculate the total initial momentum of the system.
Since April is initially at rest, her initial momentum is zero.
- Initial momentum of April: [tex]\( p_{April,initial} = m_{April} \times v_{April,initial} = 55 \times 0 = 0 \, \text{kg} \cdot \text{m/s} \)[/tex]
- Initial momentum of the watermelon: [tex]\( p_{watermelon} = m_{watermelon} \times v_{watermelon} = 2 \times 5 = 10 \, \text{kg} \cdot \text{m/s} \)[/tex]

Therefore, the total initial momentum of the system is:
[tex]\[ p_{total,initial} = p_{April,initial} + p_{watermelon} = 0 + 10 = 10 \, \text{kg} \cdot \text{m/s} \][/tex]

3. Step 3: Set up the conservation of momentum equation.
After catching the watermelon, April and the watermelon (plus the skateboard) will move together with a common final velocity [tex]\( v_{final} \)[/tex].
- The total mass after catching the watermelon: [tex]\( m_{total} = m_{April} + m_{watermelon} = 55 + 2 = 57 \, \text{kg} \)[/tex]

The total initial momentum must equal the total final momentum:
[tex]\[ p_{total,initial} = m_{total} \times v_{final} \][/tex]

4. Step 4: Solve for the final velocity [tex]\( v_{final} \)[/tex].
[tex]\[ 10 \, \text{kg} \cdot \text{m/s} = 57 \, \text{kg} \times v_{final} \][/tex]

Solving for [tex]\( v_{final} \)[/tex]:
[tex]\[ v_{final} = \frac{10}{57} \approx 0.175 \, \text{m/s} \][/tex]

5. Step 5: Compare the calculated final velocity with the provided choices.
- A. [tex]\( 5 \, \text{m/s} \)[/tex]
- B. [tex]\( 11 \, \text{m/s} \)[/tex]
- C. [tex]\( 0.18 \, \text{m/s} \)[/tex]
- D. [tex]\( 0.09 \, \text{m/s} \)[/tex]

The closest choice to [tex]\( 0.175 \, \text{m/s} \)[/tex] is C. [tex]\( 0.18 \, \text{m/s} \)[/tex].

Therefore, the correct answer is:
C. [tex]\(0.18 \, \text{m/s}\)[/tex]
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