Answer to Question #245098 in Molecular Physics | Thermodynamics for Kelani

Question

A 2.0-g particle moving at 6.2 m/s makes a perfectly elastic head-on collision with a resting 1.0-g object.

(a) Find the speed of each particle after the collision.

2.0 g particle ___    m/s

1.0 g particle ___   m/s

(b) Find the speed of each particle after the collision if the stationary particle has a mass of 10 g.

2.0 g particle  ___   m/s

10.0 g particle ___   m/s

(c) Find the final kinetic energy of the incident 2.0-g particle in the situations described in parts (a) and (b).

KE in part (a) ___  J

KE in part (b) ___    J

Answer 1

a)

$V_1 = (\frac{m_1-m_2}{m_1+m_2})v_1 + (\frac{2m_2}{m_1+m_2})v_2 \\ V_2 = (\frac{m_1-m_2}{m_1+m_2})v_2 + (\frac{2m_1}{m_1+m_2})v_1 \\ m_1=2g \\ m_2 = 1g \\ v_1 = 6.2 \;m/s \\ v_2 =0 \\ V_1 = (\frac{2-1}{2+1})6.2 +0 = 2.07 \;m/s \\ V_2 = 0 + \frac{2 \times 2}{2+1}6.2 = 8.27 \;m/s$

b)

$m_1=2g \\ m_2= 10g \\ v_1 = 6.2 \;m/s \\ v_2 = 0 \\ V_1 = (\frac{2-10}{2+10})6.2+0 = -4.13 \;m/s \\ V_2 = 0 +(\frac{2 \times 2}{2+10})6.2 = 2.07 \;m/s$

c)

$KE_a = \frac{1}{2}m_1v_1^2 \\ = \frac{1}{2} \times 2 \times 10^{-3} \times (2.07)^2 = 4.37 \times 10^{-3} \;J \\ KE_b = \frac{1}{2}m_1v_1^2 \\ = \frac{1}{2} \times 2 \times 10^{-3} \times (-4.13)^2 = 17.1 \times 10^{-3} \;J$