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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Physics
Chapter
Center of Mass, Momentum and Collision

NTA Abhyas 2020 · Physics · Center of Mass, Momentum and Collision

Three identical blocks A , B and C are placed on a horizontal...

Three identical blocks AA, BB and CC are placed on a horizontal frictionless surface. The blocks BB and CC are at rest but AA is approaching towards BB with a speed 10ms−110ms^{- 1}. The coefficient of restitution for all collisions is 0.5\text{0.5}. The speed of the block CC just after the collision is

Three identical blocks A , B and C are placed on a horizontal frictionless surface. The blocks B and C are at rest but

Options

  1. A5.6ms−1\text{5.6}ms^{- 1}
  2. B6ms−16ms^{- 1}
  3. C8ms−18ms^{- 1}
  4. D10ms−110ms^{- 1}

Correct answer

A. 5.6ms−1\text{5.6}ms^{- 1}

Step-by-step solution

For collision between blocks AA and BB,

e=vB−vAuA−uB=vB−vA10−0=vB−vA10e = \dfrac{{v}_{B} - {v}_{A}}{{u}_{A} - {u}_{B}} = \dfrac{{v}_{B} - {v}_{A}}{10 - 0} = \dfrac{{v}_{B} - {v}_{A}}{10}

∴vB−vA=10e=10×0.5=5….(i)∴ {v}_{B} - {v}_{A} = 10e = 10 \times 0.5 = 5 \ldots. ( i )

From principle of momentum conservation,

mAuA+mBuB=mAvA+mBvB{m}_{A}{u}_{A} + {m}_{B}{u}_{B} = {m}_{A}{v}_{A} + {m}_{B}{v}_{B}

m×10+0=mvA+mvBm \times 10 + 0 = {mv}_{A} + {mv}_{B}

∴vA+vB=10….(ii)∴ {v}_{A} + {v}_{B} = 10 \ldots. ( ii )

Adding eqs. (i) and (ii), we get

vB=7.5ms−1…(iii){v}_{B} = \text{7.5}ms^{- 1} \ldots ( iii )

Similarly for collision between BB and CC,

vC−vB=7.5e=7.5×0.5=3.75{v}_{C} - {v}_{B} = 7.5 e = 7.5 \times 0.5 = 3.75

∴vC−vB=3.75ms−1∴ {v}_{C} - {v}_{B} = 3. 75ms^{- 1}...(iv)

Adding Eqs. (iii) and (iv) we get

2vC=11.252{v}_{C} = \text{11.25}

∴vC=11.252=5.6ms−1∴ {v}_{C} = \dfrac{11.25}{2} = \text{5.6}ms^{- 1}
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