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NEET PHYSICSWORK, ENERGY AND POWERMedium

Question

A ball moving with velocity 2 m s12\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 s1\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 s1u_1 = 2\text{ m s}^{-1}.
  • Ball 2: Mass m2=2mm_2 = 2m, Initial velocity u2=0 m s1u_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=v2v1u1u2e = \frac{v_2 - v_1}{u_1 - u_2} 0.5=v2v1200.5 = \frac{v_2 - v_1}{2 - 0} 1=v2v1    v1=v21— (ii)1 = v_2 - v_1 \implies v_1 = v_2 - 1 \quad \text{--- (ii)}
  3. Solve for Final Velocities: Substitute (ii) into (i): (v21)+2v2=2(v_2 - 1) + 2v_2 = 2 3v2=3    v2=1 m s13v_2 = 3 \implies v_2 = 1\text{ m s}^{-1} Substitute v2v_2 back into (ii): v1=11=0 m s1v_1 = 1 - 1 = 0\text{ m s}^{-1} Thus, the velocities after collision are 0 m s10\text{ m s}^{-1} and 1 m s11\text{ m s}^{-1}.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from WORK, ENERGY AND POWER. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSWORK, ENERGY AND POWERmovingvelocitycollidesanotherstationary

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