Two blocks can collide in a one-dimensional collision. The block on the left hass a mass of 0.30 kg and is initially moving to the right at

Question

Two blocks can collide in a one-dimensional collision. The block on the left hass a mass of 0.30 kg and is initially moving to the right at 2.4 m/s toward a second block of mass 0.80 kg that is initially at rest. When the blocks collide, a cocked spring releases 1.2 J of energy into the system. (For velocities, use to mean to the right, – to mean to the left).A) What is the velocity of the first block after the collision?
B) What is the velocity of the second block after the collision?

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Kiệt Gia 3 years 2021-09-02T19:38:50+00:00 1 Answers 0 views 0

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  1. trungdung22
    0
    2021-09-02T19:39:58+00:00
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    Answer:

    a) 3.632 m/s

    b) 0.462 m/s

    Explanation:

    Using the law of conservation of momentum:

    m_{1} u_{1} + m_{2} u_{2}= m_{1} V_{1} + m_{2} V_{2}……….(1)

    m_{1} = 0.30 kg\\u_{1} = 2.4 m/s\\m_{2} = 0.80 kg\\u_{2} = 0 m/s

    Substituting the above values into equation (1) and make V2 the subject of the formula:

    0.3(2.4) + 0.80(0)= 0.3 V_{1} + 0.8 V_{2}\\

    V_{2} = \frac{0.72 - 0.3 V_{1}}{0.8}………………(2)

    Using the law of conservation of kinetic energy:

    0.5m_{1} u_{1} ^{2} + 1.2 = 0.5m_{1} V_{1} ^{2} + 0.5m_{2} V_{2} ^{2}\\0.5(0.3) (2.4) ^{2} + 1.2 = 0.5(0.3) V_{1} ^{2} + 0.5(0.8)V_{2} ^{2}\\

    2.064 = 0.15 V_{1} ^{2} + 0.4V_{2} ^{2}…….(3)

    Substitute equation (2) into equation (3)

    2.064 = 0.15 V_{1} ^{2} + 0.4(\frac{0.72 - 0.3V_{1} }{0.8}) ^{2}\\2.064 = 0.15 V_{1} ^{2} + 0.4(\frac{0.5184 - 0.432V_{1} + 0.09V_{1} ^{2} }{0.64}) \\1.32096 = 0.096 V_{1} ^{2} + 0.20736 - 0.1728V_{1} + 0.036V_{1} ^{2} \\0.132 V_{1} ^{2} - 0.1728V_{1} - 1.1136 = 0\\V_{1} = 3.632 m/s

    Substituting V_{1} into equation(2)

    V_{2} = \frac{0.72 - 0.3 *3.632}{0.8}\\V_{2} = \frac{0.72 - 0.3 *(3.632)}{0.8}\\V_{2} = 0.462 m/s

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