Respuesta :
Here we can use momentum conservation as in this type of collision there is no external force on it
[tex]m_1v_{1i} + m_2v_{2i} = m_1 v_{1f} + m_2v_{2f}[/tex]
now here we can say
[tex]m_1 = 10 g[/tex]
[tex]v_{1i} = 0[/tex]
[tex]m_2 = 550 g[/tex]
[tex]v_{2i} = 3.5 m/s[/tex]
now here we can say
[tex]10*0 + 550 * 3.5 = 10 v_{1f} + 550 v_{2f}[/tex]
[tex]192.5 = v_{1f} + 55 v_{2f}[/tex]
now by coefficient of restitution
for elastic collision we know that e = 1
[tex]v_{2f} - v_{1f} = e(v_{1i} - v_{2i})[/tex]
[tex]v_{2f} - v_{1f} = 0 - 3.5[/tex]
now by solving the two equation
[tex]56v_{2f} = 189[/tex]
[tex]v_{2f} = 3.375 m/s[/tex]
also we know that
[tex]v_{1f} = v_{2f} + 3.5 = 3.375 + 3.5 = 6.875 m/s[/tex]
so final speed of the nail is 6.875 m/s
Answer:
0.236J
Explanation:
Given: collision between a hammer and a nail, it is approximately elastic.a 550 g hammer moving with an initial speed of 3.5 m/s struck a 10g nail.
To Find: Kinetic Energy acquired by nail.
Solution: Let mass and initial speed of hammer be=[tex]\text{m}_{1} , \text{v}_{1i}[/tex]
final speed of hammer=[tex]\text{v}_{1f}[/tex]
mass and initial speed of nail be=[tex]\text{m}_{2} , \text{v}_{2i}[/tex]
final speed of nail=[tex]\text{v}_{2f}[/tex]
momentum before collision
[tex]\text{m}_{1} \text{v}_{1i}[/tex] + [tex]\text{m}_{2} \text{v}_{2i}[/tex]
momentum after collision
[tex]\text{m}_{1} \text{v}_{1f}[/tex] + [tex]\text{m}_{2} \text{v}_{2f}[/tex]
as collision is elastic momentum is conserved
momentum before collision = momentum after collision
[tex]\text{m}_{1} \text{v}_{1i}[/tex] + [tex]\text{m}_{2} \text{v}_{2i}[/tex] = [tex]\text{m}_{1} \text{v}_{1f}[/tex] + [tex]\text{m}_{2} \text{v}_{2f}[/tex]
as nail was at rest initially , [tex]\text{v}_{2i}[/tex] = [tex]0[/tex]
[tex]\text{m}_{1} v_{1i}[/tex]=[tex]\text{m}_{1} v_{1f} +\text{m}_{2} v_{2f}[/tex]
[tex]\text{m}_{1}(\text{v}_{1i} -\text{v}_{1f})[/tex] = [tex]m_{2} v_{2f}[/tex]
[tex]\frac{\text{m}_{1}}{\text{m}_{2}} = \frac{\text{v}_{2f}^2}{\text{v}_{1i}-\text{v}_{1f}}[/tex]
kinetic energy before collision
[tex]\frac{1}{2}\text{m}_{1}\text{v}_{1i}^{2} +\frac{1}{2}\text{m}_{2}\text{v}_{2i}^{2}[/tex]
kinetic energy after collision
[tex]\frac{1}{2}\text{m}_{1}\text{v}_{1f}^{2} +\frac{1}{2}\text{m}_{2}\text{v}_{2f}^{2}[/tex]
As in elastic collision Kinetic energy remains conserved
kinetic energy before collision= kinetic energy after collision
[tex]\frac{1}{2}\text{m}_{1}\text{v}_{1i}^{2} +\frac{1}{2}\text{m}_{2}\text{v}_{2i}^{2}[/tex] = [tex]\frac{1}{2}\text{m}_{1}\text{v}_{1f}^{2} +\frac{1}{2}\text{m}_{2}\text{v}_{2f}^{2}[/tex]
given, [tex]v_{2i}[/tex] = [tex]0[/tex]
[tex]\text{m}_{1}(\text{v}_{1i}^2-\text{v}_{1f}^2) = \text{m}_{2}\text{v}_{2f}^2[/tex]
[tex]\frac{\text{m}_{1}}{\text{m}_{2}}[/tex]= [tex]\frac{\text{v}_{2f}^{2}}{(\text{v}_{1i}-\text{v}_{1f})^{2}}[/tex]
putting value of [tex]\frac{\text{m}_{1}}{\text{m}_{2}}[/tex] from previous equation
[tex]\frac{\text{v}_{2f}}{\text{v}_{1i}-\text{v}_{1f}}[/tex]=[tex]\frac{\text{v}_{2f}^{2}}{(\text{v}_{1i}-\text{v}_{1f})^{2}}[/tex]
[tex]\text{v}_{2f} = \text{v}_{1i} + \text{v}_{1f}[/tex]
putting it in equation of momentum, we get
[tex]\frac{\text{v}_{1i}}{\text{v}_{1f}}=\frac{\text{m}_{1}+\text{m}_{2}}{\text{m}_{1}-\text{m}_{2}}[/tex]
putting values [tex]\text{v}_{1f}= 3.375\text{m}\setminus\text{s}[/tex]
[tex]\text{v}_{2f} = \text{v}_{1i} + \text{v}_{1f}[/tex]
[tex]\text{v}_{2f} = \text{3.5} + \text{3.375}[/tex]
[tex]\text{v}_{2f} = 6.875[/tex]
Kinetic energy acquired by nail =[tex]\frac{1}{2}\text{m}\text{v}_{2f}^2[/tex]
[tex]\frac{1}{2}\times 0.01\times 6.875^2[/tex]
0.236 J
Hence Kinetic Energy acquired by nail is 0.236 J
