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17 May, 04:39

A block of mass 1.3 kg is resting at the base of a frictionless ramp. A bullet of mass 50 g is traveling parallel to the ramp surface at 250 m/s. It collides with the block, enters it, and exits the other side at 100 m/s. How far up the ramp will the block travel?

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  1. 17 May, 04:55
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    s = 2.65 m

    Explanation:

    given,

    mass of block, M = 1.3 Kg

    mass of bullet, m = 50 g = 0.05 Kg

    speed of bullet, u = 250 m/s

    speed of bullet after collision, v = 100 m/s

    distance traveled by the block = ?

    Assuming the angle of inclination of ramp equal to 40°

    calculating the speed of the block

    using conservation of momentum

    M u' + m u = m v + M v'

    initial speed of the block is equal to zero

    0 + 0.05 Kg x 250 = 0.05 x 100 + 1.3 x v'

    1.3 v' = 7.5

    v' = 5.77 m/s

    now, calculation of acceleration

    equation the horizontal component

    -mg sin θ = ma

    a = - g sin θ

    a = - 9.8 x sin 40°

    a = - 6.29 m/s²

    using equation of motion for the calculation of distance moved by the block

    v² = u² + 2 a s

    0² = 5.77² + 2 x (-6.29) x s

    12.58 s = 33.29

    s = 2.65 m

    hence, the distance moved by the block is equal to 2.65 m
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