In a demonstration of skill, an archer shoots an apple which their assistant has thrown intothe air. The apple is thrown straight upwards and is still rising at 2.5 m/s when the arrowhits it. The arrow is travelling at 18 m/s at a 45 degree angle to horizontal the instant before itskewers the apple. The arrow has a mass of 20 g and the apple a mass of 180 g.(a) Determine the initial momentum of the combined apple-arrow system immediately before the arrow hits.(b) What is the momentum of this system immediately after the collision?
Question
In a demonstration of skill, an archer shoots an apple which their assistant has thrown intothe air. The apple is thrown straight upwards and is still rising at 2.5 m/s when the arrowhits it. The arrow is travelling at 18 m/s at a 45 degree angle to horizontal the instant before itskewers the apple. The arrow has a mass of 20 g and the apple a mass of 180 g.(a) Determine the initial momentum of the combined apple-arrow system immediately before the arrow hits.(b) What is the momentum of this system immediately after the collision?
Solution
(a) To determine the initial momentum of the combined apple-arrow system immediately before the arrow hits, we first need to calculate the momentum of each object separately.
Momentum is given by the formula p = mv, where m is mass and v is velocity.
For the apple: Mass (m1) = 180g = 0.18 kg (converted from grams to kilograms) Velocity (v1) = 2.5 m/s (upwards) So, momentum (p1) = m1v1 = 0.18 kg * 2.5 m/s = 0.45 kgm/s (upwards)
For the arrow: Mass (m2) = 20g = 0.02 kg Velocity (v2) = 18 m/s at a 45 degree angle to the horizontal. We need to find the vertical component of the velocity, which is v2sin(45) = 18 m/s * sin(45) = 12.73 m/s (upwards) So, momentum (p2) = m2v2 = 0.02 kg * 12.73 m/s = 0.2546 kg*m/s (upwards)
The total initial momentum (p_initial) is the vector sum of the individual momenta, which in this case is just the sum since they are both moving upwards. p_initial = p1 + p2 = 0.45 kgm/s + 0.2546 kgm/s = 0.7046 kg*m/s (upwards)
(b) The momentum of the system immediately after the collision is the same as the momentum immediately before the collision, assuming no external forces act on the system (principle of conservation of momentum). So, p_final = p_initial = 0.7046 kg*m/s (upwards).
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