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Question: Answered & Verified by Expert
A ball moving with velocity $2 \mathrm{~m} / \mathrm{s}$ collides head on with another stationary ball of double the mass. If coefficient of restitution is 0.5 , then their velocities (in $\mathrm{m} / \mathrm{s}$ ) after collision will be :
PhysicsCenter of Mass Momentum and CollisionAP EAMCETAP EAMCET 2023 (16 May Shift 1)
Options:
  • A $0, 1$
  • B $1, 1$
  • C $1, 0.5$
  • D $0, 2$
Solution:
1263 Upvotes Verified Answer
The correct answer is: $0, 1$
Let mass of first ball be $m$ i.e. $m_1=m$
Mass of second ball $\mathrm{m}_2=2 \mathrm{~m}$
Velocity of first ball before collision $\mathrm{u}_1=2 \mathrm{~m} / \mathrm{s}$
Velocity of second ball before collision $\mathrm{u}_2=\mathrm{om} / \mathrm{s}$
Given : $\mathrm{e}=0.5$
So, velocity of first ball after collision $\mathrm{v}_1=\frac{\left(\mathrm{m}_1-\mathrm{em}_2\right) \mathrm{u}_1+(1+\mathrm{e}) \mathrm{m}_2 \mathrm{u}_2}{\mathrm{~m}_1+\mathrm{m}_2}$
$\therefore \mathrm{v}_1=\frac{[\mathrm{m}-0.5(2 \mathrm{~m})](2)+(1+0.5)(2 \mathrm{~m})(0)}{\mathrm{m}+2 \mathrm{~m}}=\mathrm{om} / \mathrm{s}$
Velocity of second ball after collision $\mathrm{v}_2=\frac{\left(\mathrm{m}_2-\mathrm{em}_1\right) \mathrm{u}_2+(1+\mathrm{e}) \mathrm{m}_1 \mathrm{u}_1}{\mathrm{~m}_1+\mathrm{m}_2}$
$\therefore \mathrm{v}_2=\frac{[2 \mathrm{~m}-0.5 \mathrm{~m}](0)+(1+0.5)(\mathrm{m})(2)}{\mathrm{m}+2 \mathrm{~m}}=1 \mathrm{~m} / \mathrm{s}$
Correct answer is option A.

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