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Korvikt [17]
3 years ago
6

An air gun fires 2 gram pellets at 60 m/s at a 45 gram golfball initially at rest. The pellets rebound at a speed 40 m/s and are

fired at a rate of 4 per second, How much momentum is delivered to the golf ball each second in kg*m/s
Physics
1 answer:
snow_lady [41]3 years ago
8 0

Answer:

0.16 kgm/s

Explanation:

m_1 = Mass of pellet = 2 g

m_2 = Mass of golf ball = 45 g

u_1 = Initial Velocity of pellet  = 60 m/s

u_2 = Initial Velocity of golf ball = 0 m/s

v_1 = Final Velocity of pellet = 40 m/s

v_2 = Final Velocity of golf ball

In this system the linear momentum is conserved

m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\Rightarrow m_2v_2=m_1u_1+m_2u_2-m_1v_1\\\Rightarrow m_2v_2=0.002\times 60+0-0.002\times 40\\\Rightarrow m_2v_2=0.04

As the golf ball is being shot 4 times every second the momentum delivered to the golf ball each second is 4\times 0.04=0.16\ kgm/s

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Answer:

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Explanation:

In order to get the value of the x component of the resultant force, we need to get the value of the x component of each force.

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We can repeat the process for F₂, as follows:

For F₂, as it is directed at an angle of 53.3º below the negative x axis, we can find F₂ₓ just applying the definition of cosine of an angle, as follows:

cos θ = \frac{x}{r}

In this case, x = F₂ₓ, and r = F₂

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Fₓ = F₁ₓ + F₂ₓ = -5.28 N + -4.18 N = -9.46 N

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