a)
We use the formula :
m1v1i + m2v2i = m1v1f + m2v2f
Substituting the values in:
4.0kg*8.0m/s + 4.0kg*0m/s = 4.0kg*0m/s +4.0kg*v2f
Calculating this we get:
32.0kg*m/s + 0kg*m/s = 0kg*m/s + 4.0kg*v2f
Rearrange for v2f:
v2f = [tex]\frac{32.0kg*m/s}{4.0kg}[/tex]
This gives us 8.0 m/s as the final velocity of the second ball.
b)
Since the collision is assumed to be elastic it means that the kinetic energy must be equal before and after the collision.
This means we use the formula:
Ek = [tex]\frac{1}{2} *m*v^{2}[/tex]+ [tex]\frac{1}{2} *m*v^{2}[/tex] = [tex]\frac{1}{2} *m*v^{2}[/tex] + [tex]\frac{1}{2}*m*v^{2}[/tex]
Substituting in values:
Ek = 0.5*4.0kg*(8.0m/s)^2 + 0.5*4.0kg*(0m/s)^2 = 0.5*4.0kg*(0m/s)^2 + 0.5*4.0kg*(8.0m/s)^2
This simplifies to:
Ek= 128J + 0J = 0J + 128J
This shows us that the kinetic energy is equal on each side therefore the collision is Elastic and no energy has been lost.
The final velocity of the second ball after collision is 8 m/s.
The total kinetic energy of the balls before and after collision is the same.
The given parameters;
Apply the principle of conservation of linear momentum to determine the final velocity of the second ball.
[tex]m_1u_1 + m_2u_2 = m_1v_1 + m_2 v_2\\\\4(8) + 4(0) = 4(0) + 4(v_2)\\\\32 = 4v_2\\\\v_2 = 8 \ m/s[/tex]
The final velocity of the second ball after collision is 8 m/s.
The total kinetic energy of the balls before collision is calculated as follows;
[tex]K.E_i = \frac{1}{2} \times 4 \times 8^2 \ + \ \frac{1}{2} \times 4 \times 0^2\\\\K.E_i = 128 \ J[/tex]
The total kinetic energy of the balls after collision is calculated as follows;
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