Answer:
a) v3 = 1 m/s
c) K3 < K1
d) p3 = p1
Explanation:
a) To solve this problem you use the conservation of the linear momentum in elastic collision.
In the first case you have:
[tex]p_i=p_f\\\\m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}[/tex]
but the second block is at rest, then v2i = 0m/s:
[tex]m_1v_{1i}=m_1v_{1f}+m_2v_{2f}[/tex]
Furthermore, you can assume that the first object stops just after the collision with the second one. From this last expression you obtain the value of the second object:
[tex]v_{2f}=\frac{m_1v_{1i}}{m_2}\\\\m_2=2m_1\\\\v_{2f}=\frac{m_1(4m/s)}{2m_1}=2\ m/s[/tex]
Then, you use the conservation of momentum for the second case, in which the second objects impact the third one:
[tex]m_2v'_{2i}+m_3v_{3i}=m_2v'_{2f}+m_3v_{3f}\\\\v_{3i}=0\\\\m_2v'_{2i}=m_2v'_{2f}+m_3v_{3f}\\\\v_{2f}=0\\\\m_2v'_{2i}=m_3v_{3f}\\\\v_{3f}=\frac{m_2v'_{2i}}{m_3}[/tex]
where again it has assumed that the second object stops, just after the impact with the third object. v'_2i = v_2f (in order to distinguish). BY using the fact m3 = 2m2 you obtain:
[tex]v_{3f}=\frac{m_2(2m/s)}{2m_2}=1\ m/s[/tex]
Then, you obtain that v3 < v2 < v1
c) The kinetic energy is given by:
[tex]K=\frac{1}{2}mv^2[/tex]
you compute for all the three objects:
[tex]K_1=\frac{1}{2}m_1(4m/s)^2=8m_1\ m^2/s^2\\\\K_2=\frac{1}{2}m_2(2m/s)^2=\frac{1}{2}(2m_1)(4m^2/s^2)=4m_1\ m^2/s^2\\\\K_3=\frac{1}{2}m_3=(1m/s)^2=\frac{1}{2}(2m_2)(1\ m^2/s^2)=\frac{1}{2}(2(2m_1))(1 m^2/s^2)=2m_1\ m^2/s^2[/tex]
then, k3 < k2 < k1
d) For the momentum you have:
[tex]p_1=4m_1\ m/s\\\\p_2=m_2(2m/s)=(2m_1)(2m/s)=4m_1\ m/s\\\\p_3=m_3(1m/s)=(2m_2)(1m/s)=(2(2m_1))(1m/s)=4m_1\ m/s[/tex]
p1 = p2 = p3
