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Work Energy and Power question

2010 · Q170
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Work Energy and Power question

2010 · Q170

NEETPhysicsWork Energy and PowerMCQ+4 / −1
A ball moving with velocity 2 m/s collides head on with another stationary ball of double the mass. If the coefficient of restitution is 0.5, then their velocities (in m/s) after collision will be
  1. A
    0, 1
  2. B
    1, 1
  3. C
    1, 0.5
  4. D
    0, 2
View written solutionFree

Correct answer: A

Here, m1 = m, m2 = 2m, u1 = 2 m/s, u2 = 0

Coefficient of restitution, e = 0.5
Let v1 and v2 be their respective velocities after collision.
Applying the law of conservation of linear momentum, we get

m1u1

  • m2u2 = m1 v1
  • m2 v2

    ∴\therefore∴ m × 2 + 2m × 0 = m × v1
  • 2m × v2

    or 2m = mv1 + 2mv2 or 2 = (v1 + 2v2) ...(i)

    By definition of coefficient of restitution,

    e=v2−v1u1−u2e = {{{v_2} - {v_1}} \over {{u_1} - {u_2}}}e=u1​−u2​v2​−v1​​

    or e(u1 – u2) = v2 – v1 ⇒\Rightarrow⇒ 0.5(2 –
  1. = v2 – v1 ...(ii)

    Solving equations (i) and (ii), we get
    v1 = 0 m/s, v2 = 1 m/s
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