Answer to Question #235192 in Mechanics | Relativity for ivy

Question

An inelastic collision occurs in one dimension, in which a 10 kg block traveling at

5 m/s collides with a 5 kg block traveling at 3 m/s in the same direction, and they

stick together. What are the velocities of the blocks immediately after the collision?

Answer 1

Solution

From the law of conservation of momentum

$m_1 V_{1i} + m_2. V_{2i} = m_1V_1 + m_2V_2$

where vii- velocities before collision are v1i=5 m/s Particles of $m_1=10 kg$

$V_{2i}-$ velocities before collision are $v_{2i}=3 m/s$ Particles of $m_2=5 kg$

$V_i -$ velocities after collision Particles of $m_1=10 kg$

$V_2 -$ velocities after collision Particles of $m_2=5 kg$

Energy conservation law

$m_1 V_{1i}^2/2 + m_2. V_{2i}^2/2 = m_1V_1^2/2 + m_2V_2^2/2$

from here:

$v_1=\frac{v_{1i}(m_1-m_2)}{m_1+m_2}+\frac{v_{2i}(2)(m_2)}{m_1+m_2}=\frac{5(10-5)}{10+5}+\frac{3\times2\times5}{10+5}=3.66m/s$

$v_2=\frac{v_{1i}(2)(m_1)}{m_1+m_2}+\frac{v_{2i}(m_2-m_1)}{m_1+m_2}=\frac{5\times2\times10}{10+5}+\frac{3(5-10)}{10+5}+=5.66m/s$

ANSWER:

$v_1=3.66m/s$

$v_2=5.66m/s$