Respuesta :
Answer:
a) [tex]v'=-0.227\ m.s^{-1}[/tex]
b) [tex]v=1.36\ m.s^{-1}[/tex]
Explanation:
Given:
mass of the lighter block, [tex]m'=0.02\kg[/tex]
velocity of the lighter block, [tex]u'=0.68\ m.s^{-1}[/tex]
mass of the heavier block, [tex]m=0.04\ kg[/tex]
velocity of the heavier block, [tex]u=0\ m.s^{-1}[/tex]
a)
Using conservation of linear momentum:
[tex]m'.u'+m.u=m'.v'+m.v[/tex]
where:
[tex]v'=[/tex] final velocity of the lighter block
[tex]v=[/tex] final velocity of the heavier block
[tex]m'.u'=m'.v'+m.v[/tex]
[tex]m'(u'-v')=m.v[/tex] ........................(1)
Since kinetic energy is conserved in elastic collision:
[tex]\frac{1}{2}m'.u'^2=\frac{1}{2}m'.v'^2+\frac{1}{2}m.v^2[/tex]
[tex]m'(u'^2-v'^2)=m.v^2[/tex]
[tex]m'(u'-v')(u'+v')= m.v^2[/tex]
divide the above equation by eq. (1)
[tex]v=u'+v'[/tex] .............................(2)
now we substitute the value of v from eq. (2) in eq. (1)
[tex]m'(u'-v')=m(u'+v')[/tex]
[tex]\frac{m'+m}{m'-m} =\frac{u'}{v'}[/tex]
[tex]\frac{0.02+0.04}{0.02-0.04} =\frac{0.68}{v'}[/tex]
[tex]v'=-0.227\ m.s^{-1}[/tex] (negative sign denotes that the direction is towards left)
b)
now we substitute the value of v' from eq. (2) in eq. (1)
[tex]m'(u'-v+u')=m.v[/tex]
[tex]2m'.u'=(m-m')v[/tex]
[tex]2\times 0.02\times 0.68=(0.04-0.02)\times v[/tex]
[tex]v=1.36\ m.s^{-1}[/tex]
(a) the speed of the 20g block is -0.227 m/s
(b) the speed of the 40g block is 0.453 m/s
Elastic collision:
In elastic collision two objects collide, there is no deformation and the objects move separately. The momentum and kinetic energy of the system remains conserved.
Let the masses be m₁ = 20g = 0.02kg and m₂ = 40g = 0.04kg,
the corresponding initial velocities be u₁ = u = 0.68m/s and u₂ = 0
and the final velocities be v₁ and v₂
(a) the speed of the 20g block after the collision is given by:
[tex]v_1=\frac{m_1-m_2}{m_1+m_2}u\\\\v_1=\frac{0.02-0.04}{0.02+0.04}\times0.68\\\\v_1=-0.227\;m/s[/tex]
the 20g block bounces back along the same line with a speed of 0.227 m/s
(b) the speed of the 40g block after the collision is given by:
[tex]v_2=\frac{2m_1}{m_1+m_2}u\\\\v_2=\frac{2\times0.02}{0.02+0.04}\times0.68\\\\v_2=0.453\;m/s[/tex]
Learn more about elastic collision:
https://brainly.com/question/2356330?referrer=searchResults
