We read your photo as:

Which of the following is required for the conservation of momentum in a system of colliding objects?

Physics

ScanSolve AI Answer

Step-by-step solution

9 steps

here are the solutions for questions 46-49.

46. The law of conservation of momentum applies to a system of colliding objects if there are no net external forces acting on the system. If there are no net external forces, then the net external impulse (which is the change in momentum due to external forces) is zero.

  • Answer: (D) the net external impulse is zero

47.

  1. Calculate the time for the first fall (t1t_1).

    The ball falls from rest (v0=0v_0 = 0) from a height h1=4.9 mh_1 = 4.9 \, \text{m}.
    Using the kinematic equation h=v0t+12gt2h = v_0 t + \frac{1}{2}gt^2:

    4.9=0⋅t1+12(9.8)t124.9 = 0 \cdot t_1 + \frac{1}{2}(9.8)t_1^2 4.9=4.9t124.9 = 4.9t_1^2 t12=1t_1^2 = 1 t1=1 st_1 = 1 \, \text{s}
  2. Calculate the velocity just before the first impact (v1v_1).

    Using v=v0+gtv = v_0 + gt:

    v1=0+(9.8)(1)=9.8 m/sv_1 = 0 + (9.8)(1) = 9.8 \, \text{m/s}
  3. Calculate the velocity just after the first impact (v1′v_1').

    The coefficient of restitution e=0.75e = 0.75 is given by e=v1′v1e = \frac{v_1'}{v_1}.

    v1′=e⋅v1=0.75⋅9.8 m/s=7.35 m/sv_1' = e \cdot v_1 = 0.75 \cdot 9.8 \, \text{m/s} = 7.35 \, \text{m/s}

    This is the initial upward velocity for the second bounce.

  4. Calculate the time for the second flight (up and down, t2t_2).

    The time to reach the peak height after bouncing is tup=v1′gt_{up} = \frac{v_1'}{g}.

    tup=7.35 m/s9.8 m/s2=0.75 st_{up} = \frac{7.35 \, \text{m/s}}{9.8 \, \text{m/s}^2} = 0.75 \, \text{s}

    The total time for the ball to go up and come back down to the surface for the second hit is t2=2⋅tupt_2 = 2 \cdot t_{up}.

    t2=2⋅0.75 s=1.5 st_2 = 2 \cdot 0.75 \, \text{s} = 1.5 \, \text{s}
  5. Calculate the total time until the second hit.

    Total time = t1+t2=1 s+1.5 s=2.5 st_1 + t_2 = 1 \, \text{s} + 1.5 \, \text{s} = 2.5 \, \text{s}.

    • Answer: (C) 2.5 sec\boxed{\text{2.5 sec}}

    48.

  6. Write the formulas for kinetic energy and momentum.

    Kinetic energy: KE=12mv2KE = \frac{1}{2}mv^2
    Momentum: p=mvp = mv

  7. Express kinetic energy in terms of momentum.

    From p=mvp = mv, we get v=pmv = \frac{p}{m}. Substitute this into the kinetic energy formula:

    KE=12m(pm)2=12mp2m2=p22mKE = \frac{1}{2}m \left(\frac{p}{m}\right)^2 = \frac{1}{2}m \frac{p^2}{m^2} = \frac{p^2}{2m}
  8. Use the condition that kinetic energies are equal (KE1=KE2KE_1 = KE_2).

    p122m1=p222m2\frac{p_1^2}{2m_1} = \frac{p_2^2}{2m_2}
  9. Solve for the ratio of momenta.

    p12m1=p22m2\frac{p_1^2}{m_1} = \frac{p_2^2}{m_2} p12p22=m1m2\frac{p_1^2}{p_2^2} = \frac{m_1}{m_2}

    Taking the square root of both sides:

    p1p2=m1m2\frac{p_1}{p_2} = \sqrt{\frac{m_1}{m_2}}

    So, the ratio of their momentum is m1:m2\sqrt{m_1} : \sqrt{m_2}.

    • Answer: (A) m1:m2\boxed{\sqrt{m_1} : \sqrt{m_2}}

    49. An inelastic collision is characterized by the conservation of momentum, but the kinetic energy of the system is not conserved. Some kinetic energy is converted into other forms of energy, such as heat, sound, or deformation.

    • Answer: (D) momentum is conserved but kinetic energy is not conserved
Was this helpful?

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question
Handwritten step-by-step solution preview

Handwritten Step-by-Step Solution

Get this as a handwritten image on WhatsApp — like a tutor wrote it for you.

Try on WhatsApp