g A 10.0 g bullet is fired into and embeds itself in a 1.65 kg block attached to a spring with a spring constant of 17.2 N/m and whose mass

Question

g A 10.0 g bullet is fired into and embeds itself in a 1.65 kg block attached to a spring with a spring constant of 17.2 N/m and whose mass is negligible. How far is the spring compressed if the bullet has a speed of 300 m/s just before it strikes the block, and the block slides on a frictionless surface? (Note: You must use conservation of momentum in this problem.

in progress 0
Xavia 3 years 2021-08-24T02:24:34+00:00 1 Answers 0 views 0

Answers ( )

  1. Vodka
    0
    2021-08-24T02:26:09+00:00
    Reply

    Answer:

    the compression of the spring is 0.5613 m

    Explanation:

    Given;

    mass of bullet, m₁ = 0.01 kg

    mass of block, m₂ = 1.65 kg

    initial velocity of the block, u₂ = 0

    initial velocity of the bullet before hitting the block, u₁ = 300 m/s

    Final speed of the bullet-block system after collision, v = ?

    spring constant, K = 17.2 N/m

    Apply the principle of conservation of linear momentum;

    m₁u₁ + m₂u₂ = v( m₁ + m₂)

    0.01 x 300 + 1.65 x 0 = v(0.01 + 1.65)

    3 = 1.66v

    v = 3 / 1.66

    v = 1.807 m/s

    Apply the principle of conservation of mechanical energy to determine the compression of the spring;

    KE₁ + PE₁ = KE₂ + PE₂

    ¹/₂mu² + ¹/₂Kx₁² = ¹/₂mv² + ¹/₂Kx₂²

    where;

    m is mass of bullet and block embedded together

    u is the initial velocity of the  bullet-block system = 1.807 m/s

    v is the final velocity of the  bullet-block system = 0

    x₁ is the initial compression of the spring = 0

    x₂ is the final compression of the spring = ?

    ¹/₂(1.65 + 0.01) (1.807)² + ¹/₂(17.2)(0)² = ¹/₂(1.65 + ).01)(0)² + ¹/₂(17.2)(x₂)²

    2.71 + 0 = 0 + 8.6(x₂)²

    (x₂)² = 2.71 / 8.6

    (x₂)² = 0.3151

    x₂ = √0.3151

    x₂ = 0.5613 m

    Therefore, the compression of the spring is 0.5613 m

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )