6. A frustrated physics student blew up her Physics textbook using a small amount of explosive. It broke into three pieces, which miraculous

Question

6. A frustrated physics student blew up her Physics textbook using a small amount of explosive. It broke into three pieces, which miraculously flew off in directions that were all in the same plane. A 0.200 kg piece flew off at 20.0 m/s and a 0.100 kg piece went off at 90° to the first piece, at 30.0 m/s. a) What was the momentum of the third piece? b) If the mass of the third piece was 0.150 kg, what was its velocity right after the explosion?

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Dulcie 5 months 2021-08-03T00:54:07+00:00 1 Answers 15 views 0

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    2021-08-03T00:55:50+00:00

    Answer:

    a) p₃ = 5 kg*m/s

    b) v₃ = 33.33 m/s

    Explanation:

    Given

    m₁ = 0.200 kg

    v₁ = 20.0 m/s i

    m₂ = 0.100 kg

    v₂ = 30.0 m/s j

    m₃ = 0.150 kg

    a) Using the equation  p = m*v

    p₃ = ?

    If p₁ = m₁*v₁

    and

    p₂ = m₂*v₂

    v = 0 m/s  (speed of the the Physics textbook before the explosion)

    m = mass of the Physics textbook before the explosion

    We apply the following Principle

    m*v = p₁ + p₂ + p₃

    then

    m*0 = m₁*v₁ + m₂*v₂ + p₃

    ⇒ 0 = (0.200 kg)*(20.0 m/s i) + (0.100 kg)*(30.0 m/s j) + p₃  

    ⇒ 0 = 4 kg*m/s i + 3 kg*m/s j + p₃

    ⇒ p₃ =   4 kg*m/s i – 3 kg*m/s j   (i)

    and its module is

    p_{3} =\sqrt{(-4)^{2} +(-3)^{2} }  kg*\frac{m}{s} = 5 kg*\frac{m}{s}

    b) Using equation (i):

    p₃ =   4 kg*m/s i – 3 kg*m/s j

    m₃*v₃ = – 4 kg*m/s i – 3 kg*m/s j

    (0.150 kg)*v₃ = – 4 kg*m/s i – 3 kg*m/s j

    ⇒ v₃ = (- 4 kg*m/s/(0.150 kg)) i – (3 kg*m/s/(0.150 kg)) j

    ⇒ v₃ = – 26.67 m/s i – 20 m/s j

    and its module is

    v_{3} =\sqrt{(- 26.67)^{2} +(- 20)^{2} }  \frac{m}{s} = 33.33 \frac{m}{s}

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