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PhyMedAdv 2010 • Paper 1

Elastic Collision of Moving Mass with Stationary Mass

A point mass of 1 kg1\text{ kg} collides elastically with a stationary point mass of 5 kg5\text{ kg}. After their collision, the 1 kg1\text{ kg} mass reverses its direction and moves with a speed of 2 ms−12\text{ ms}^{-1}. Which of the following statement(s) is (are) correct for the system of these two masses ?

Options

A

Total momentum of the system is 3 kg ms−13\text{ kg ms}^{-1}

Correct
B

Momentum of 5 kg5\text{ kg} mass after collision is 4 kg ms−14\text{ kg ms}^{-1}

C

Kinetic energy of the centre of mass is 0.75 J0.75\text{ J}

Correct
D

Total kinetic energy of the system is 4 J4\text{ J}

Topics & Concepts

Work, Energy and PowerCollisionCenter of Mass

Step-by-Step Solution

To determine the correct statement(s), let us analyze the 1D elastic collision between the two masses.

1. Given Data:

  • Mass of the moving particle: m1=1 kgm_1 = 1\text{ kg}
  • Mass of the stationary particle: m2=5 kgm_2 = 5\text{ kg}
  • Initial velocity of mass m2m_2: u2=0 ms−1u_2 = 0\text{ ms}^{-1}
  • Final velocity of mass m1m_1: v1=−2 ms−1v_1 = -2\text{ ms}^{-1} (reverses direction with speed 2 ms−12\text{ ms}^{-1})

2. Finding the Initial Velocity (u1u_1) of Mass m1m_1:
For a one-dimensional elastic collision, the velocity of the first mass after the collision is given by:
v1=(m1−m2m1+m2)u1+(2m2m1+m2)u2v_1 = \left(\frac{m_1 - m_2}{m_1 + m_2}\right) u_1 + \left(\frac{2 m_2}{m_1 + m_2}\right) u_2

Substituting u2=0u_2 = 0 into the equation:
v1=(m1−m2m1+m2)u1v_1 = \left(\frac{m_1 - m_2}{m_1 + m_2}\right) u_1

Substitute m1=1 kgm_1 = 1\text{ kg}, m2=5 kgm_2 = 5\text{ kg}, and v1=−2 ms−1v_1 = -2\text{ ms}^{-1}:
−2=(1−51+5)u1-2 = \left(\frac{1 - 5}{1 + 5}\right) u_1
−2=−46u1=−23u1-2 = -\frac{4}{6} u_1 = -\frac{2}{3} u_1
u1=3 ms−1u_1 = 3\text{ ms}^{-1}

3. Evaluation of Options:

  • Option A: Total momentum of the system
    Since no external horizontal forces act on the system, total linear momentum is conserved:
    Ptotal=m1u1+m2u2=(1 kg)(3 ms−1)+(5 kg)(0 ms−1)=3 kg ms−1P_{\text{total}} = m_1 u_1 + m_2 u_2 = (1\text{ kg})(3\text{ ms}^{-1}) + (5\text{ kg})(0\text{ ms}^{-1}) = 3\text{ kg ms}^{-1}
    Therefore, Option A is correct.

  • Option B: Momentum of 5 kg5\text{ kg} mass after collision
    Using momentum conservation:
    Ptotal=m1v1+p2P_{\text{total}} = m_1 v_1 + p_2
    3=(1)(−2)+p2  ⟹  p2=5 kg ms−13 = (1)(-2) + p_2 \implies p_2 = 5\text{ kg ms}^{-1}
    Therefore, Option B is incorrect.

  • Option C: Kinetic energy of the centre of mass
    The velocity of the centre of mass (vcomv_{\text{com}}) is:
    vcom=Ptotalm1+m2=3 kg ms−11 kg+5 kg=0.5 ms−1v_{\text{com}} = \frac{P_{\text{total}}}{m_1 + m_2} = \frac{3\text{ kg ms}^{-1}}{1\text{ kg} + 5\text{ kg}} = 0.5\text{ ms}^{-1}

    The kinetic energy associated with the motion of the centre of mass is:
    Kcom=12(m1+m2)vcom2=12(6 kg)(0.5 ms−1)2=3×0.25=0.75 JK_{\text{com}} = \frac{1}{2} (m_1 + m_2) v_{\text{com}}^2 = \frac{1}{2} (6\text{ kg}) (0.5\text{ ms}^{-1})^2 = 3 \times 0.25 = 0.75\text{ J}
    Therefore, Option C is correct.

  • Option D: Total kinetic energy of the system
    Since the collision is perfectly elastic, the total kinetic energy is equal to the initial kinetic energy:
    Ktotal=12m1u12+12m2u22=12(1 kg)(3 ms−1)2+0=4.5 JK_{\text{total}} = \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} (1\text{ kg}) (3\text{ ms}^{-1})^2 + 0 = 4.5\text{ J}
    Therefore, Option D is incorrect.

Conclusion:
The correct statements are A and C.

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