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Question Number 25058 by Tinkutara last updated on 02/Dec/17

Two objects slide over a frictionless horizontal surface. The first object, mass m_1 = 5 kg, is propelled with a speed u = 4.5 m/s towards the second object, mass m_2 = 5 kg, which is initially at rest. After the collision, both objects have velocities which are directed at θ = 60° on either side of the original line of motion of the first object. What can you say about the elasticity of collision?

$${Two}\:{objects}\:{slide}\:{over}\:{a}\:{frictionless} \\ $$$${horizontal}\:{surface}.\:{The}\:{first}\:{object}, \\ $$$${mass}\:{m}_{\mathrm{1}} \:=\:\mathrm{5}\:{kg},\:{is}\:{propelled}\:{with}\:{a} \\ $$$${speed}\:{u}\:=\:\mathrm{4}.\mathrm{5}\:{m}/{s}\:{towards}\:{the}\:{second} \\ $$$${object},\:{mass}\:{m}_{\mathrm{2}} \:=\:\mathrm{5}\:{kg},\:{which}\:{is} \\ $$$${initially}\:{at}\:{rest}.\:{After}\:{the}\:{collision}, \\ $$$${both}\:{objects}\:{have}\:{velocities}\:{which}\:{are} \\ $$$${directed}\:{at}\:\theta\:=\:\mathrm{60}°\:{on}\:{either}\:{side}\:{of} \\ $$$${the}\:{original}\:{line}\:{of}\:{motion}\:{of}\:{the} \\ $$$${first}\:{object}.\:{What}\:{can}\:{you}\:{say}\:{about} \\ $$$${the}\:{elasticity}\:{of}\:{collision}? \\ $$

Answered by jota+ last updated on 05/Dec/17

θ=60 m_(1 ) u=m_1 v_1 cosθ+m_2 v_2 cos θ (1) 0=m_1 v_1 sin θ−m_2 v_(2 ) sin θ (2) from(2) p_1 =m_1 v_1 =m_2 v_(2 ) =p_2 then m_1 u=2p_1 cos θ=p_1 T_1 =(1/2)m_1 u^2 =(p_1 ^2 /(2m_1 )) T_2 =(p_1 ^2 /(2m_1 ))+(p_2 ^2 /(2m_2 )) ΔT=(p_2 ^2 /(2m_2 )) ≠ 0 The collision is inelastic.

$$ \\ $$$$\theta=\mathrm{60} \\ $$$${m}_{\mathrm{1}\:} {u}={m}_{\mathrm{1}} {v}_{\mathrm{1}} \mathrm{cos}\theta+{m}_{\mathrm{2}} {v}_{\mathrm{2}} \mathrm{cos}\:\theta\:\:\:\left(\mathrm{1}\right) \\ $$$$\mathrm{0}={m}_{\mathrm{1}} {v}_{\mathrm{1}} \mathrm{sin}\:\theta−{m}_{\mathrm{2}} {v}_{\mathrm{2}\:} \mathrm{sin}\:\theta\:\:\:\:\:\left(\mathrm{2}\right) \\ $$$$\:\:{from}\left(\mathrm{2}\right)\:\:{p}_{\mathrm{1}} ={m}_{\mathrm{1}} {v}_{\mathrm{1}} ={m}_{\mathrm{2}} {v}_{\mathrm{2}\:} ={p}_{\mathrm{2}} \\ $$$${then} \\ $$$${m}_{\mathrm{1}} {u}=\mathrm{2}{p}_{\mathrm{1}} \mathrm{cos}\:\theta={p}_{\mathrm{1}} \\ $$$${T}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}{m}_{\mathrm{1}} {u}^{\mathrm{2}} =\frac{{p}_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{2}{m}_{\mathrm{1}} } \\ $$$${T}_{\mathrm{2}} =\frac{{p}_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{2}{m}_{\mathrm{1}} }+\frac{{p}_{\mathrm{2}} ^{\mathrm{2}} }{\mathrm{2}{m}_{\mathrm{2}} } \\ $$$$\Delta{T}=\frac{{p}_{\mathrm{2}} ^{\mathrm{2}} }{\mathrm{2}{m}_{\mathrm{2}} }\:\neq\:\mathrm{0} \\ $$$${The}\:{collision}\:{is}\:{inelastic}. \\ $$

Commented by jota+ last updated on 05/Dec/17

No.ΔT is not zero

$$\:{No}.\Delta{T}\:{is}\:{not}\:{zero} \\ $$$$ \\ $$

Commented by Tinkutara last updated on 05/Dec/17

Answer given is perfectly elastic.

$${Answer}\:{given}\:{is}\:{perfectly}\:{elastic}. \\ $$

Commented by Tinkutara last updated on 05/Dec/17

Thank you Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Sir}! \\ $$

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