Answer to Question #317668 in Mechanics | Relativity for ann

Question #317668

Two balls A and B, weighing 49 N each, approach each other, with speeds of 20 m/s and 30 m/s respectively. Determine their speeds after the collision if (a) the coefficient of restitution is 0.80, and (b) they are completely inelastic. 


1
Expert's answer
2022-03-25T09:11:24-0400

Explanations & Calculations


  • Say the velocities of A and B just after the collision are vavb\small v_a \quad v_b .
  • By conservation of linear momentum we get,

(Right is taken positive)ma.(20)mb.(30)=ma.(va)+mb.(vb)va+vb=10\qquad\qquad \begin{aligned} &\small \to( \text{Right is taken positive)}\\ \small m_a.(20)-m_b.(30)&=\small m_a.(v_a)+m_b.(v_b)\\ \small v_a+v_b&=\small -10 \end{aligned}

  • By Newton's experimental law,

vavb=0.80[20(30)]vavb=40\qquad\qquad \begin{aligned} \small v_a-v_b&=\small -0.80\Big[20-(-30)\Big]\\ \small v_a-v_b&=\small -40 \end{aligned}

  • By solving the two simultaneous eq.s

va=25ms1vb=15ms1\qquad\qquad \begin{aligned} \small v_a&=\small -25\,ms^{-1}\\ \small v_b&=\small 15\,ms^{-1} \end{aligned}


  • They will move together attached as one after the collision in this case. Then by conservation of linear momentum,

ma(20)+mb(30)=(ma+mb)VV=5ms1\qquad\qquad \begin{aligned} \small m_a(20)+m_b(-30)&=\small (m_a+m_b)V\\ \small V&=\small -5\,ms^{-1} \end{aligned}


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