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NEET PHYSICSMedium

A ball moving with velocity 2 m s−12\text{ m s}^{-1} collides head on with another stationary ball of double the mass. If the coefficient of restitution is 0.50.5, then their velocities (in m s−1\text{m s}^{-1}) after collision will be:

A

0, 1

B

1, 1

C

1, 0.5

D

0, 2

Step-by-Step Solution

  1. Identify Given Values:
  • Ball 1: Mass m1=mm_1 = m, Initial velocity u1=2 m s−1u_1 = 2\text{ m s}^{-1}.
  • Ball 2: Mass m2=2mm_2 = 2m, Initial velocity u2=0 m s−1u_2 = 0\text{ m s}^{-1} (stationary).
  • Coefficient of restitution: e=0.5e = 0.5.
  1. Apply Conservation of Linear Momentum: Total momentum before collision equals total momentum after collision [Class 11 Physics, Ch 6, Sec 5.11.1]. m1u1+m2u2=m1v1+m2v2m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 m(2)+2m(0)=mv1+2mv2m(2) + 2m(0) = m v_1 + 2m v_2 2m=mv1+2mv22m = m v_1 + 2m v_2 v1+2v2=2— (i)v_1 + 2v_2 = 2 \quad \text{--- (i)}
  2. Apply Coefficient of Restitution Formula: The coefficient of restitution (ee) relates the relative velocity of separation to the relative velocity of approach. e=v2−v1u1−u2e = \frac{v_2 - v_1}{u_1 - u_2} 0.5=v2−v12−00.5 = \frac{v_2 - v_1}{2 - 0} 1=v2−v1  ⟹  v1=v2−1— (ii)1 = v_2 - v_1 \implies v_1 = v_2 - 1 \quad \text{--- (ii)}
  3. Solve for Final Velocities: Substitute (ii) into (i): (v2−1)+2v2=2(v_2 - 1) + 2v_2 = 2 3v2=3  ⟹  v2=1 m s−13v_2 = 3 \implies v_2 = 1\text{ m s}^{-1} Substitute v2v_2 back into (ii): v1=1−1=0 m s−1v_1 = 1 - 1 = 0\text{ m s}^{-1} Thus, the velocities after collision are 0 m s−10\text{ m s}^{-1} and 1 m s−11\text{ m s}^{-1}.
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