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A) In your own words, define linear momentum and impulse. b) Explain the relationship between Impulse and change in momentum, giving one real-life example for each.

Physics

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Step-by-step solution

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The problem statement does not specify the mass of the second stationary ball. For this solution, it is assumed that the second ball has the same mass as the first ball, i.e., m2=0.2 kgm_2 = 0.2 \text{ kg}.

  1. Identify the given values and the assumption.

    • Mass of the first ball: m1=0.2 kgm_1 = 0.2 \text{ kg}
    • Initial velocity of the first ball: u1=5 m/su_1 = 5 \text{ m/s}
    • Initial velocity of the second ball (stationary): u2=0 m/su_2 = 0 \text{ m/s}
    • Assumption: Mass of the second ball: m2=0.2 kgm_2 = 0.2 \text{ kg}
  2. Apply the principle of conservation of linear momentum.

    For an elastic collision, the total momentum before the collision equals the total momentum after the collision:

    m1u1+m2u2=m1v1+m2v2m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2

    Substitute the known values:

    (0.2 kg)(5 m/s)+(0.2 kg)(0 m/s)=(0.2 kg)v1+(0.2 kg)v2(0.2 \text{ kg})(5 \text{ m/s}) + (0.2 \text{ kg})(0 \text{ m/s}) = (0.2 \text{ kg}) v_1 + (0.2 \text{ kg}) v_2 1.0 kg⋅m/s=0.2v1+0.2v21.0 \text{ kg} \cdot \text{m/s} = 0.2 v_1 + 0.2 v_2

    Divide the entire equation by 0.2:

    5=v1+v2(∗)5 = v_1 + v_2 \quad (*)
  3. Apply the relative velocity equation for an elastic collision.

    For an elastic collision, the relative speed of approach equals the relative speed of separation:

    u1−u2=v2−v1u_1 - u_2 = v_2 - v_1

    Substitute the known values:

    5 m/s−0 m/s=v2−v15 \text{ m/s} - 0 \text{ m/s} = v_2 - v_1 5=v2−v1(∗∗)5 = v_2 - v_1 \quad (**)
  4. Solve the system of equations for v1v_1 and v2v_2.

    We have two equations:

    1. v1+v2=5v_1 + v_2 = 5
    2. −v1+v2=5-v_1 + v_2 = 5

    Add equation (1) and equation (2):

    (v1+v2)+(−v1+v2)=5+5(v_1 + v_2) + (-v_1 + v_2) = 5 + 5 2v2=102v_2 = 10 v2=5 m/sv_2 = 5 \text{ m/s}

    Substitute the value of v2v_2 into equation (1):

    v1+5=5v_1 + 5 = 5 v1=0 m/sv_1 = 0 \text{ m/s}

    The velocities of both balls after the collision are:

    • Velocity of the first ball: v1=0 m/s\boxed{\mathbf{v_1 = 0 \text{ m/s}}}
    • Velocity of the second ball: v2=5 m/s\boxed{\mathbf{v_2 = 5 \text{ m/s}}}
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