Respuesta :
Initial conditions:
m1 = 1.0 ; v1 = 5
m2 = 4.0 ; v2 = 0
In the case where the second object (sometimes called the target) is at rest the velocities after the condition are
v1' = v1* (m1-m2)/(m1+m2)
v2' = 2v1*m1/(m1+m2)
For this we get
v1' = 5*(-3)/5 = -3m/s (moving in the opposite direction as before at 3m/s
v2' = 2*5*(1)/5 = 2m/s in the same direction as the original ball was moving
you can see these directions by looking at the signs. The momenta also add to the initial momentum as required.
m1 = 1.0 ; v1 = 5
m2 = 4.0 ; v2 = 0
In the case where the second object (sometimes called the target) is at rest the velocities after the condition are
v1' = v1* (m1-m2)/(m1+m2)
v2' = 2v1*m1/(m1+m2)
For this we get
v1' = 5*(-3)/5 = -3m/s (moving in the opposite direction as before at 3m/s
v2' = 2*5*(1)/5 = 2m/s in the same direction as the original ball was moving
you can see these directions by looking at the signs. The momenta also add to the initial momentum as required.
Answer:
final speed of blue ball = 2m/s
final speed of red ball = -3 m/s
Explanation:
As we know that for elastic type of collision the coefficient of elasticity is always 1 and it is given by
[tex]e = \frac{v_2 - v_1}{u_1 - u_2}[/tex]
here we know that
[tex]u_1 = 5 m/s[/tex]
[tex]u_2 = 0[/tex]
now we have
[tex]v_2 - v_1 = 5 - 0 = 5 m/s[/tex]
also we can use the momentum conservation for this type of collision
so we will have
[tex]1\times 5 + 4 \times 0 = 1\times v_1 + 4 \times v_2[/tex]
so we have
[tex]v_1 + 4v_2 = 5[/tex]
now from above two equations we will have
[tex]5 v_2 = 10 [/tex]
[tex]v_2 = 2 m/s[/tex]
also we have
[tex]v_1 = - 3m/s[/tex]
so final speed of blue ball = 2m/s
final speed of red ball = 3 m/s