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Center of Mass question

2023 · 10 Apr · Shift 1 · Q65
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Center of Mass question

2023 · 10 Apr · Shift 1 · Q65

JEE MainPhysicsCenter of MassMCQ+4 / −1
A particle of mass m moving with velocity v collides with a stationary particle of mass 2m. After collision, they stick together and continue to move together with velocity
  1. A
    vvv
  2. B
    v3\frac{v}{3}3v​
  3. C
    v4\frac{v}{4}4v​
  4. D
    v2\frac{v}{2}2v​
View written solutionFree

Correct answer: B

  1. Identify the type of collision

    Since the two particles stick together after collision, this is a perfectly inelastic collision.

  2. Apply conservation of linear momentum

    Initial momentum:

    • First particle: mass mmm, velocity vvv p1=mvp_1 = mvp1​=mv
    • Second particle: mass 2m2m2m, initially at rest p2=2m⋅0=0p_2 = 2m \cdot 0 = 0p2​=2m⋅0=0

    So total initial momentum is pinitial=mvp_{\text{initial}} = mvpinitial​=mv

  3. Find the total mass after collision

    Since they stick together, combined mass is m+2m=3mm + 2m = 3mm+2m=3m

    Let the common velocity after collision be VVV.

    Then final momentum is pfinal=3mVp_{\text{final}} = 3mVpfinal​=3mV

  4. Equate initial and final momentum

    mv=3mVmv = 3mVmv=3mV

    Cancel mmm from both sides: v=3Vv = 3Vv=3V

    Therefore, V=v3V = \frac{v}{3}V=3v​

  5. Match with the options

    The correct option is: v3\boxed{\frac{v}{3}}3v​​

    So, Option B is correct.

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