Question

Two balls collide in a closed system. The first ball has a mass of 5kg and is moving at 10m/s when it crashes into a 2 kg ball at rest. After the collision, the 5 kg ball is moving with a velocity of 3m/s. What is the velocity of the 2kg ball?

Answers:
17.5m/s
3.1m/s
2.9m/s

Answer 1

Answer:

17.5 m/s

Explanation:

Formula:

m1 (v1 - v'1) = m2 (v'2 - v2 )

Answer 2

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: Given :- }}}$

$\implies\textsf{ Two balls collide in a closed system.}\\\implies\textsf{ Mass of first ball is 5kg and second is of 2kg.}\\\implies\textsf{ Velocity of first is 10m/s and second is 0m/s.}\\\implies\textsf{ After collision Velocity of 5kg ball is 3m/s.}$

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: To \ Find :- }}}$

$\implies\textsf{ The velocity of second ball .}$

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: Answer :- }}}$

According to the Law of Conservation of Momentum , for an isolated system Initial Momentum of the system is equal to the final Momentum of the system . Here since the second ball was at rest , so its initial Velocity will be 0m/s.

$\underline{\boldsymbol{\purple{ According \ to \ Law\ of \ Conservation\ of\ Momentum:-}}}$

$\sf:\implies\pink{ m_1u_1 + m_2u_2 = m_1 v_1+m_2v_2}\\\\\sf:\implies m_{1st \ ball} u_{1st\ ball} + m_{2nd\ ball} u_{2nd\ ball} = m_{1st \ ball} v_{1st\ ball} + m_{2nd\ ball} v_{2nd\ ball}\\\\\sf:\implies 5kg \times 10m/s + 2kg\times 0m/s = 5kg\times 3m/s + 2kg\times v_{2nd \ ball} \\\\\sf:\implies 50 kg-m/s + 0 = 15kg-m/s + v_{2nd \ ball} ( 2kg) \\\\\sf:\implies v_{2nd \ ball} ( 2) = ( 50 - 15) m/s \\\\\sf:\implies v_{2nd \ ball} ( 2kg)= 35 \\\\\sf:\implies v_{2nd \ ball} =\dfrac{35}{2} m/s \\\\\sf:\implies\boxed{\pink{\mathfrak{ v_{2nd \ ball} = 17.5 \ m/s }}}$

$\underline{\blue{\sf \therefore Hence \ the \ velocity \ of \ second\ ball \ is \ \textsf{\textbf{17.75 \ m/s }}. }}$