Answer to Question #317668 in Mechanics | Relativity for ann

Question

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. 

Answer 1

Explanations & Calculations

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

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

• By Newton's experimental law,

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

• By solving the two simultaneous eq.s

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

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

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