A 75-g projectile traveling at 600 m /s strikes and becomes embedded in the 40-kg block, which is ini-tially stationary. Compute

Question

A 75-g projectile traveling at 600 m /s strikes and becomes embedded in the 40-kg block, which is ini-tially stationary. Compute the energy lost during the impact. Express your answer as an absolute value ΔE| and as a percentage n of the original system energy E.

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Huyền Thanh 5 months 2021-08-28T08:55:56+00:00 2 Answers 1 views 0

Answers ( )

    0
    2021-08-28T08:57:29+00:00

    Explanation:

    The given data is as follows.

              Mass, m = 75 g

             Velocity, v = 600 m/s

    As no external force is acting on the system in the horizontal line of motion. So, the equation will be as follows.

              m_{1}v_{1_{i}} = (m_{1} + m_{2})vi

    where,  m_{1} = mass of the projectile

                m_{2} = mass of block

                  v = velocity after the impact

    Now, putting the given values into the above formula as follows.

                  m_{1}v_{1_{i}} = (m_{1} + m_{2})vi

             75(10^{-3}) \times 600 = [(75 \times 10^{-3}) + 50] \times v

                                      = \frac{45}{50.075}

                                  v = 0.898 m/s

    Now, equation for energy is as follows.

                   E = \frac{1}{2}mv^{2}

                      = \frac{1}{2} \times (75 \times 10^{-3} + 50) \times (600)^{2}

                      = 13500 J

    Now, energy after the impact will be as follows.

                 E’ = \frac{1}{2}[75 \times 10^{-3} + 50](0.9)^{2}

                     = 20.19 J

    Therefore, energy lost will be calculated as follows.

               \Delta E = E  E’

                           = (13500 – 20) J

                           = 13480 J

    And,   n = \frac{\Delta E}{E}

                 = \frac{13480}{13500} \times 100

                 = 99.85

                 = 99.9%

    Thus, we can conclude that percentage n of the original system energy E is 99.9%.

    0
    2021-08-28T08:57:30+00:00

    Answer:

    Explanation:

    Answer:

    Explanation:

    m1 = 75 g = 0.075 kg

    m2 = 40 kg

    u1 = 600 m/s

    u2 = 0

    Let the velocity of combined mass is v.

    Use conservation of momentum

    m1 x u1 + m2 x u2 = ( m1 + m2) x v

    0.075 x 600 + 0 = ( 40 + 0.075) x v

    v = 1.12 m/s

    Initial energy, E = 0.5 x 0.075 x 600 x 600 = 13500 J

    Final energy, E’ = 0.5 x 40.075 x 1.12 x 1.12 = 25.14 J

    Loss in energy, ΔE = E – E’ = 13500 – 25.14 = 13474.86 J

    % loss of energy = ( 13474.86 x 100) / 13500 = 99.8 %  

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