Question

A steel ball of radius 2 cm is initally at rest It is struck head on by another steel ball of radius 4 cm travelling with a velocity of 81 cm/s the common velocity if it is perfectly inelastic collisions
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Answer 1

Answer:

Common velocity = 72 cm/s

Explanation:

Given:

• Radius of the first steel ball = 2 cm
• Radius of the second steel ball = 4 cm
• Velocity of the first ball = 0 cm/s
• Velocity of the second ball = 81 cm/s

To Find:

• Their common velocity if their collision is perfectly inelastic

Solution:

First we have to find the ratio of masses of the two balls (m₁ and m₂).

Let us assume their densities as ρ

We know that,

Mass = Density × Volume

Volume of a sphere is given by,

Volume of a sphere = 4/3 π r³

Hence,

$\tt \dfrac{Mass\:of\:first\:ball(m_1)}{Mass\:of\:second\:ball(m_2)} =\dfrac{\rho \times 4/3\times \pi \times 2^{3} }{\rho \times 4/3 \times \pi \times 4^{3} }$

$\tt \dfrac{m_1}{m_2} =\dfrac{ 2^{3} }{ 4^{3} }$

$\tt \dfrac{m_1}{m_2} =\dfrac{ 8 }{ 64 }=\dfrac{1}{8}$

Now the common velocity when two bodies undergo a perfectly inelastic collision is given by,

$\tt V=\dfrac{m_1v_1+m_2v_2}{m_1+m_2}$

where V is the common velocity, m₁ is the mass of the first object, v₁ is the velocity of the first object, m₂ is the mass of the second object, v₂ is the velocity of the second object.

Here v₁ = 0, hence,

$\tt V=\dfrac{m_2v_2}{m_1+m_2}$

Rearranging the equation,

$\tt V =\dfrac{m_2}{m_1} \times \dfrac{v_2}{(1+\dfrac{m_2}{m_1} )}$

We know that m₁/m₂ = 1/8

Hence, m₂/m₁ = 8

Substitute the data,

$\tt V= 8\times \dfrac{81}{1+8}$

V = 648/9

V = 72 cm/s

Hence the common velocity of the object is 72 cm/s.