Final answer:
By applying the principle of conservation of momentum, the skateboarder is propelled in the opposite direction at approximately 1.692 m/s after throwing the second weight.
Step-by-step explanation:
To find out how fast the skateboarder is propelled in the opposite direction after throwing the second weight, we need to apply the principle of conservation of momentum. The momentum before and after each throw must be equal because there are no external horizontal forces acting on the skateboarder and weights system.
The total initial momentum of the system is zero since the skateboarder is initially at rest. When the first 5 kg weight is thrown at 7 m/s, to conserve momentum, the skateboarder (with the board and remaining weight) must move in the opposite direction.
The combined mass after throwing the first weight is 40 kg (skateboarder) + 3 kg (board) + 5 kg (remaining weight) = 48 kg. Let v be the velocity of the skateboarder after throwing one weight:
- mweight × vweight = mskateboarder+board+weight × v1
- (5 kg) × (7 m/s) = (48 kg) × v1
- v1 = (5 kg × 7 m/s) / 48 kg = 0.729 m/s (opposite to the weight's)
After the second weight is thrown, the skateboarder and board's combined mass is 43 kg. Assuming the skateboarder throws the second weight with the same speed (7 m/s) as the first one, we can calculate the final velocity v2:
- (5 kg) × (7 m/s) = (43 kg) × (v2 - v1)
- (5 kg) × (7 m/s) = (43 kg) × (v2 - 0.729 m/s)
- v2 = [(5 kg × 7 m/s) / 43 kg] + 0.729 m/s = 1.692 m/s
The skateboarder is propelled in the opposite direction at approximately 1.692 m/s after throwing the second weight.