Answer to Question #191530 in Mechanics | Relativity for steven

Question

A 50 kg mass moving to the right at 8 m s−1 collides with a 25 kg mass moving to the right at 3 m s−1. After the collision, the velocities of the masses are v1 and v2, where v2 > v1 and 15% of the original kinetic energy is lost during the collision. Assume that no external forces are acting.

(a) Show that v2 = 19 − 2v1.

(b) Show that 3v 2 1 − 38v1 + 100 ≈ 0 is a good approximation of the equation for v1.

(c) Calculate v1 and v2.

(d) What happened to the 15% lost energy? 

Answer 1

$\text{before collision}$

$m_1 = 50;\vec{u_1}=8$

$m_2=25;\vec{u_2}=3$

$m_1*\vec{u_1}+m_2*\vec{u_2}=475$

$E_{k1}=\frac{m_1u_1^2}{2}+\frac{m_2u_2^2}{2}=1712.5$

$\text{after collision}$

$m_1 = 50;\vec{v_1}$

$m_2=25;\vec{v_2}$

$m_1*\vec{v_1}+m_2*\vec{v_2}=m_1*\vec{u_1}+m_2*\vec{u_2}=475(1)$

$E_{k}=\frac{m_1v_1^2}{2}+\frac{m_2v_2^2}{2}$

$E_k= E_{k1}-0.15*E_{k1}=1455.625$

$E_{k}=\frac{m_1v_1^2}{2}+\frac{m_2v_2^2}{2}=1455.625(2)$

$(a)\text{ from the equation(1)}$

$m_1*\vec{v_1}+m_2*\vec{v_2}=475$

$50*\vec{v_1}+25*\vec{v_2}=475$

$2*\vec{v_1}+\vec{v_2}=19$

$\vec{v_2}=19-2*\vec{v_1}(3)$

$(b)\text{ from the equation(2)}$

$\frac{m_1v_1^2}{2}+\frac{m_2v_2^2}{2}=1455.625$

$m_1v_1^2+m_2v_2^2=2911.25$

$50v_1^2+25v_2^2=2911.25$

$2v_1^2+v_2^2=116.45$

$\text{ from the equation(3)}$

$2v_1^2+(19-2v_1)^2=116.45$

$6v_1^2-76v_1+244.55=0$

$3v_1^2-38v_1+122.75=0$

$3v_1^2-38v_1+100\approx0(4)$

$(c)\text{ from the equation(4)}$

$3v_1^2-38v_1+100=0$

$v_1 =3.8;v_2=19-2*v_1=11.4$

$\text{Decision } v_1'=8.9 \text{ is not taken by the condition of the task }v_2>v_1$

$\text{Answer: }v_1 =3.8;v_2=11.4$

$(d)\text{ Part of the translational kinetic energy is transferred into the internal energy}\newline \text{of the colliding bodies.}$

$\text{ As an example, deformation and heating of bodies.}$